zotero-db/storage/WZN7N59R/.zotero-ft-cache

Ramanujan’s Notebooks
Part II

Bust of Ratnanujan by Paul Granlund

Bruce C. Berndt
RLamanujan’sNotebooks
Part II
Springer-Verlag New York Berlin Heidelberg
London Paris Tokyo

Bruce C. Berndt

Department

of Mathematics

University of Illinois

Urbana, IL 61801

USA

The following journals have published earlier versions of chapters in this book: LEnseignement Mathématique 26 (1980), l-65. Journal of the Indian Mathematical Society 46 (1982), 31-16. Bulletin London Mathematical Society 15 (1983), 273-320. Expositiones Marhematicae 2 (1984) 289-347. Journal fur die reine und angewandte Mathematik 361(1985), Rocky Mountain Journal of Mathematics 15 (1985), 235-310. Acta Arithmetica 47 (1986) 123-142.

118-134.

Mathematics

Subject Classification (1980): 1 l-03, 1 lP99

Library of Congress Cataloging-in-Publication

Data

(Revised for volume 2)

Ramanujan Aiyangar, Srinivasa, 1887-1920.

Ramanujan’s notebooks.

Includes bibliographies and index.

1. Mathematics.

1. Berndt. Bruce C., 1939-

11. Title.

QA3.R33 1985

510

84-20201

Printed on acid-free paper.

0 1989 by Springer-Verlag New York Inc.

Al1 rights reserved. This work may not be translated or copied in whole or in part without the

written permission of the publisher (Springer-Verlag,

175 Fifth Avenue, New York, NY 10010,

USA), except for briefexcerpts in connection with reviews or scholarly analysis. Use in connection

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by similar or dissimilar methodology now known or hereafter developed is forbidden.

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by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone.

Typeset by Asco Trade Typesetting Ltd., Hong Kong. Printed and bound by R. R. Donnelley and Sons, Harrisonburg, Printed in the United States of America.

Virginia.

987654321

ISBN O-387-96794-X ISBN 3-540-96794-X

Springer-Verlag Springer-Verlag

New York Berlin Heidelberg Berlin Heidelberg New York

Dedicated to my mother Helen
and the memory of my father Harvey

The relation between Hardy and Ramanujan is unparalleled in scientific history. Each had enormous respect for the abilities of the other. Mrs. Ramanujan told the author in 1984 of her husband’s deep admiration for Hardy. Although Ramanujan returned from England with a terminal illness, he never regretted accepting Hardy’s invitation to visit Cambridge.

Photograph reprinted with permission from Collected Papers of G. H. Hardy, Vol. 1, Oxford University Press, Oxford, 1969.

Preface

During the years 1903-1914, Ramanujan recorded many of his mathematical disco,veries in notebooks without providing proofs. Although many of his results were already in the literature, more were not. Almost a decade after Ramanujan’s death in 1920, G. N. Watson and B. M. Wilson began to edit his notebooks, but never completed the task. A photostat edition, with no editing, was published by the Tata Institute of F’undamental Research in Bombay in 1957.
This book is the second of four volumes devoted to the editing of Ramanujan’s notebooks. Part 1, published in 1985, contains an account of Chapters l-9 in the second notebook as well as a description of Ramanujan’s quarterly reports. In this volume, we examine Chapters 10-15 in Ramanujan’s second notebook. If a result is known, we provide references in the literature where proofis may be found; if a result is not known, we attempt to prove it. Except in a few instances when Ramanujan’s intent is not clear, we have been able to establish each result in these six chapters.
Chapters 10-15 are among the most interesting chapters in the notebooks. Not only are the results fascinating, but for the most part, Ramanujan’s methods remain a mystery. Much work still needs to be done. We hope readers Will strive to discover Ramanujan’s thoughts and further develop his beautiful ideas.

Urbana, Illinois Novernber 1987

Bruce C. Berndt

Contents

Preface

ix

Introduction

CHAPTER 10

Hypergeometric Series, 1

7

CHAI’TER 11

Hypergeometric Series, II

48

CHAI’TER 12

Conlinued Fractions

103

CHAI’TER 13

Integrals and Asymptotic Expansions

185

CHAI’TER 14

Inhite Series

240

CHAI’TER 15

Asymptotic Expansions and Modular Forms

300

References

339

Index

355

Introduction
We ta ke up something--we know it is fmite; but as soon as we begin to analyze it, it leads us beyond our reason, and we never find an end to all its qualities, its possibilities, its powers, its relations. It has become intïnite.
Vivekananda In a certain sense, mathematics has been advanced most by those who are distinguished more for intuition than for rigorous methods of proof.
Felix Klein For now we see through a glass, darkly; but then face to face: now 1 know in part; but then a,hall 1 know even as also 1 am known.
First Corinthians 13 : 12
The quoted passagesof Vivekananda, Klein, and St. Paul each point to a certain facet of Ramanujan’s work. First, on June 1-5, 1987,the centenary of Ramanujan’sbirth wascelebrated at the University of Illinois with a seriesof 28expository lecturesand severalcontributed papersthat traced Ramanujan’s influence to many areas of current research; seethe conference Proceedings edited by Andrews et al. [l]. Thus, Ramanujan’s mathematics continues to generatea vast amount of researchin a variety of areas.Second,in the sequel, we shiallseemany instanceswhere Ramanujan made profound contributions but for which he probably did not have rigorous proofs; for example, seeEntry 10of Chapter 13.Third, although St. Paul’spassageis eschatological in nature, it points to the great need to learn how Ramanujan reasonedand made his discoveries.Perhaps wecari prove Ramanujan’s claims,but we may not know the well from which they sprung. These three aspects of Ramanujan’s work Will frequently be made manifest in the pagesthat follow.

2

Introduction

In this book, we examine Chapters 10-15 in Ramanujan’s second notebook. In many respects, these chapters contain some of Ramanujan’s most fascinating and enigmatic discoveries. Our goal has been to prove each claim made by Ramanujan. With a few possible exceptions where the meaning is obscure, we either give a proof or indicate where in the literature proofs cari
be found. We emphasize that many (perhaps most) of our proofs are undoubtedly different from those found by Ramanujan. In particular, we have often employed the theory of functions of a complex variable, a subject with which Ramanujan had no familiarity. In no way should our proofs, or this book, be regarded as delïnitive. In many instances, more transparent proofs, especially those that might give insight into Ramanujan’s reasoning, should be sought.
Each of Chapters 10-13 and 15 contains 12 pages, while Chapter 14 encompasses 14 pages in Ramanujan’s second notebook. The number of theorems, corollaries, and examples in each chapter is listed in the following table.

Chapter
10
11
12 13 14 15
Total

Number of Results
116
103
113 92 87 94
605

In the sequel, we have employed Ramanujan’s designations of corollary, example, and SOon, although the appellations may not be optimal. Generally, we have adhered to Ramanujan’s notation SO that the reader following our account with a copy of Ramanujan’s notebooks at hand Will have an easier task. At times, for clarity, we have changed notation, especially in Chapter 14 where we make heavy use of complex function theory. Except for some minor alterations, especially in Chapter 15, we have also preserved Ramanujan’s order of presentation.
Many of the theorems communicated by Ramanujan in his famous letters to G. H. Hardy on January 16, 1913 and February 27,1913 may be found in Chapters 10-15. In the table below, we list these results.

Introduction

3

Location in Collected Papers
p. xxvi, V, (2) p. xxvi, V, (3) p. xxvi, V, (4) p. xxvi, V, (5) p. xxvi, V, (6) p. xxvi, VI, (3) p. xxvi, VII, (2)
p. xxvi, VII, (3) p. xxvii, VII, (7) p. xxvii, IX, (1) p. xxviii, (3) p. xxviii, (10) p. xxix, (14) p. 349, v, (7) P. 349, V, (8) p. 350,VI, (4) p. 350,VI, (5) p. 350,IX, (2) p. 35 1, last formula in lïrst letter p. 352, penultimate paragraph
of 3 p. 352, last paragraph of 3 p. 353, (16)

Location in Notebooks

Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter

10, Section 7, Example 15 10, Section 7, Example 14 14, Section 13, Corollary (iii) 14, Entry 25(ii) 14, Entry 25(vii) 11, Section 20, Example 2 12, Entry 48, Corollary of Entry 48 13, Entry 6 13, Corollary (ii) of Entry 10 15, Section 2, Example (iv) 12, Section 25, Corollary 1 10, Equation (31.1) 11, Entry 29(i) 12, Entry 27 14, Entry 25(xi) 14, Entry 25(xii) 13, Corollary of Entry 21 13, Example for Corollary of Entry 21 12, Entry 34 10, Entry 29(b) 15, Section 2, Example (ii)

Chapter 15, Section 2, Example (iv) Chapter 12, Corollary to Entry 34

Several of Ramanujan’s publishedpapersand problemsposedin the Journal of the Indian Mathematical Society have their origins in the notebooks. In most cases,only a small portion of the published paper is actually found in the notebooks. We list below those papers with their genesesin Chapters 10--l& together with the respective locations in the notebooks.

Pape1 On question 330 of Prof. Sanjana Modular equations and approximations
to ïl On the product fl
::[1+(&i)l] Some delïnite integrals
Some delïnite integrals connected with Gauss’s sums

Location in Notebooks Chapter 10, Section 13 Chapter 14, Section 8, Example
Chapter 13, Section 27
Chapter 13, Entries 14, 15, lO(iii), Corollary of Entry 19, Entry 21, Corollary of Entry 21, Entry 22
Chapter 14, Section 6 Chapter 14, Entry 22(ii)

4
Paper On certain arithmetical functions
On certain trigonometrical sums and their applications in the theory of numbers
Asymptotic formulae in combinatory analysis (with G. H. Hardy)
A class of defïnite integrals Question 289
Question 294
Question 296 Question 358 Question 387 Question 769

Introduction
Location in Notebooks
Chapter 15, Sections 9, 10, 12, 13, and 14
Chapter 14, Entry 13
Chapter 15, Section 2, Example (iv)
Chapter 13, Sections 23-25 Chapter 12, Section 4, Examples
(9, (ii) Chapter 12, Section 48 Chapter 13, Entry 6 Chapter 13, Section 21, Example Chapter 14, Corollary of Entry 14 Chapter 14, Section 8, Example Chapter 13, Entry 11(iii)

We now provide brief summaries for each of Chapters 10-15. More detailed descriptions may be found at the beginning of each chapter.
Of a11the topics examined by Ramanujan in his notebooks, only modular equations received more attention than hypergeometric series. Chapter 10 is the lïrst of two chapters devoted almost entirely to the latter subject. In 1923, Hardy [l], [7, pp. 505-5161 published a brief overview of the corresponding chapter in the lïrst notebook. Ramanujan rediscovered most of the classical formulas in the subject, including those attached to the names of Gauss, Kummer, Dougall, Dixon, and Saalschütz. Ramanujan possessed the uncanny ability for finding the most important examples of theorems, and Chapter 10 contains many elegant examples of infinite series summed in closed form. Ramanujan was the lïrst to discover identities for certain partial sums of hypergeometric series, and these may be found in the latter parts of Chapter 10. Ramanujan continues his study of hypergeometric series in Chapter 11. Two topics dominate the chapter. The lïrst concerns products of hypergeometric series, and most of these results are original with Ramanujan. Second, Ramanujan offers several beautiful asymptotic formulas for hypergeometric functions. By far, the most interesting is Corollary 2 in Section 24. Quadratic transformations of hypergeometric series are also featured in Chapter 11.
Chapter 12 is almost entirely devoted to continued fractions and is one of the most fascinating chapters in the notebooks. Ramanujan’s published papers contain only one continued fraction! However, Ramanujan submitted some continued fractions as problems to the Journal of the Indian Mathematical Society, and his letters to Hardy contain some of his most beautiful theorems on continued fractions. Nonetheless, the great majority of the results in Chapter 12 are new. Perhaps the most exquisite theorems are the many

Introduction

5

continued fraction expansions for products and quotients of gamma functions. We have no idea how Ramanujan discovered these formulas. Especially awe inspiring is Entry 40 involving several parameters.
Equally astonishing is Chapter 13. In the first 11 sections, one finds several beautiful, deep asymptotic expansions for integrals and series. Entries 7 and 10 are perhaps highlights. Ramanujan left us no clues of how he discovered these fascinating theorems. Are these results prototypes for further yet undiscovered theorems? Although we have given proofs, we do not have a firm understanding of how these wonderful theorems fit with the rest of mathematics.
Those readers who are fascinated by elegant series evaluations and identities will take great pleasure in reading Chapter 14. Here, one cari find several series identities that have a symmetry that one often associates with certain applications of the Poisson summation formula, which, however, does not seem to be applicable in most cases here. Several closed form evaluations of series involving hyperbolic functions are given. Some of the results in this chapter cari be established by employing partial fraction decompositions. We have utilized two additional primary tools: contour integration and
some: theorems of the author on transformations of Eisenstein series. Since neither of these techniques was in Ramanujan’s arsenal, we do not know how Ramanujan discovered most of the results in Chapter 14.
C%apter 15 is the most unorganized of a11 the chapters in the second notebook. The first seven sections are primarily devoted to interesting asymptotic expansions of several series. Entry 8 offers an elegant transformation formula for a modified theta-function.
In. the sequel, equation numbers refer to equations in the same chapter, unless another chapter is indicated. Unless otherwise stated, page numbers refer to pages in Ramanujan’s second notebook [ 151 in the pagination of the
Tata Institute. Part 1 refers to the author’s account [9] of Chapters 1-9, and Part III refers to his account [l l] of Chapters 16-21.
In what follows, the principal value of the logarithm is always denoted by Log. The set of a11(fïnite) complex numbers is denoted by %ZT.he residue of a function f at an isolated singularity a Will be denoted by R(a). (The identity off ,will always be clear.)
A small portion of this book has been aided by notes left by G. N. Watson
and B. M. Wilson in their efforts to edit Ramanujan’s notebooks. We are grateful to the Master and Fellows of Trinity College, Cambridge, for providing a copy of these notes and for permission to use this material in this book.
We sincerely appreciate the collaboration of Robert L. Lamphere on Chapter 12 and Ronald J. Evans on Chapters 13 and 15. Because of their
efforts, our accounts of these chapters are decidedly better than what we would bave: accomplished without their help. Most of the material in this book appeared in previously published versions of these chapters. We are grateful

6

Introduction

for the cooperation shown by each of the journals publishing our earlier accounts. A table below indicates the bibliographie data for the original publications. (Portions of Chapter 15 were published in two parts.)

Chapter
10 11

Coauthors

12 R. L. Lamphere, B. M. Wilson 13 R. J. Evans
14
15 R. J. Evans

15 R. J. Evans

Publication
J. Indian Math. Soc. 46 (1982),31-76
Bull. London Math. Soc. 15 (1983), 273-320
Rocky Mt. J. Math. 15 (1985), 235-310 Expos. Math. 2 (1984), 289-347 L’Enseign. Math. 26 (1980), l-65 J. Reine Angew. Math. 361 (1985),
118-134 Acta Arith. 47 (1986), 123-142

Although only one author is listed on the caver of this book, several mathematicians have made valuable contributions. We are very grateful to George Andrews,Richard Askey, Henri Cohen, Ronald Evans, Jerry Fields, P. Flajolet, M. L. Glasser, Mourad Ismail, Lisa Jacobsen, Robert Lamphere, David Masser, F. W. J. Olver, R. Sitaramachandrarao, and Don Zagier for the many proofs and suggestions that they have contributed. In particular, Askey, Evans, and Jacobsen have each supplied several proofs and offered many
helpful comments, and we are especially indebted to them. Others, not named, have made helpful comments, and we publicly offer them our thanks as well.
The author bears the responsibility for a11 errors and would like to be notified of such, whether they be minor or serious.
The manuscript was typed by the three best technical typists in ChampaignUrbana-Melody Armstrong, Hilda Britt, and Dee Wrather. We thank them for the superb quality of their typing.
Lastly, we express our deep gratitude to James Vaughn and the Vaughn Foundation for the generous funding that they have given the author during summers. This book could not have been completed without the support of
the Vaughn Foundation.

CHAPTER 10
Hypergeometric Series, 1
In 1923, Hardy published a paper [l], [7, pp. 505-5163 providing an overview of the contents of Chapter 12 of the lïrst notebook. This chapter, which corresponds to Chapter 10 of the second notebook, is concerned primarily with hypergeometric series. It should be emphasized that Hardy gave only a brief survey of Chapter 12; this chapter contains many interesting results not mentioned by Hardy, and Chapter 10 of the Se#cond notebook possesses material not found in the first. Quite remarkably, Kamanujan independently discovered a great number of the primary classical theorems in the theory of hypergeometric series. In particular, he rediscovered well-known theorems of Gams, Kummer, Dougall, Dixon, Saalschütz, and ‘Thomae, as well as special cases of Whipple’s transformation. Unfortunately, Ramanujan left us little knowledge as to how he made his beautiful discoveries about hypergeometric series. The lïrst notebook contains a few brief sketches of proofs, but the only sketch in the second notebook is found after E,ntry 8, which is Gauss’s theorem. We shall present this argument of Ramanujan in the sequel.
As the reader Will see, this chapter contains a wealth of beautiful evaluations of hypergeometric functions, usually at the argument + 1 or - 1. In this connection, we mention the recent work of R. Wm. Gosper, 1. Gessel, and D. Stanton. By employing “splitting functions” a.nd the computer algebra system MACSYMA, Gosper discovered many new hypergeometric function evaluations. Most of these, in the terminating cases, were ingeniously proved by Gesse1 and Stanton [l]. Two conjectures of Gosper were established by P. W. Karlsson [l].
Many elegant and useful binomial coefficient. sums cari be evaluated, usually quite simply, by employing the theorems of Gauss, Dixon, Saalschütz, Kummer, and others. See the paper by R. Roy [2] for many illustrations.

8

10. Hypergeometric Series, 1

We now offer several remarks about notation. As usual, we put r(a + k)
(4 = r(a)-7

where k is any complex number. The generalized hypergeometric defined by
1 2, ...,
pFq ;:: ;,, . ..) zix

series pFq is
(0.1)

where p and q are nonnegative integers and c(r, CI~, .. , c(~and &, BZ, . . . , B, are complex numbers. If the number of parameters is “small,” we may sometimes use the notation pF&~l, c(*, . . . , clp; Pr, &, . . , &; x) in place of the notation on the left side of (0.1). In this chapter, we are concerned only with the cases when p = q + 1. In these instances, the series defining pFq converges when (xl < 1 for a11choices of the parameters cli, ~j, 1 I i 5 q + 1, 1 I j 5 q. However, q+l F4 cari be continued analytically into the complex plane tut at [l, 00). If x = 1, the series converges for Re(a, + ... + a,+i) < ReUA + . . . + &); if x = - 1, there is convergence for Re(a, + . .. + ~,+i) < Re(/A + ... + p,) + 1. In a11 the theorems and examples that follow, when x = f 1, we state the conditions for convergence, but without further com-
ment. Moreover, as is customary, if x = 1, we omit the argument in the notation (0.1). It should be remarked that Ramanujan has no notation for hypergeometric series. Al1 formulas are stated by writing out the first few terms in each series. This practice has one distinct advantage in that the elegance of formulas involving series is often more easily discerned. Frequently, a compact notation obscures the aesthetic beauty of a series relation. For brevity, we usually use a compact notation, but, at times, in particularly elegant instances, we follow Ramanujan’s practice. TO aid readers examining this chapter in conjunction with the second notebook, we have usually adhered to Ramanujan’s notations for the parameters.
For the most part, we refer only to primary sources. For example, we give a reference to Dougall’s paper wherein his famous theorem is initially proved,
but we do not usually offer further references to other proofs, applications, and SOon. The classical texts of Appel1 and Kampé de Fériet [l], Klein [l], Bailey [4], and Slater [l] contain excellent bibliographies on which it would be diflïcult to elaborate. In the sequel, Bailey’s well-known tract [4] Will be our basic reference. We also indicate which formulas have been discussed by Hardy [l] in his overview. For those readers wishing to learn more about the history of hypergeometric functions, we recommend the papers of Askey [ 11, Dutka [3], and Bühler [l].
In the sequel, always, il/(z) = P(~)/I(Z). Frequent use is made of the classical representation (e.g., see Luke’s text [l, p. 121)

ll/(z+l)= -Y+kzOl o(’g-kq+,z

WJ

10. Hypergeometric Series, 1

9

where y denotes Euler’s constant. We also often employ the simple differentiation formulas

= +(k - l)!, k r 1.

(0.3)

u=o

Entry 1. Suppose that ut least one of the quantities x., y, z, u, or -x - y - z u - 211- 1 is a positive integer. Then

n,$n + 1, -.x, -y, -z, -u,x+y+z+u+2n+l F 7 6 [ *n,x+n+l,y+n+l,z+n+l,u+n+l,-x-y-z-u-n

1

T(x+n+l)T(y+n+l)F(z+n+l)F(u+n-tl)I(x+y+z+n+l) = IY(n+l)I(x+y+n+l)F(y+z+n+l)F(x+u+n+l)F(z+u+n+l)

r(y + z + u + n + ~)I(X + u + z + n + l)F(x + y + u + n + 1) ' r-(x +Z + n + i)r(y + U+ n + i)ryx +y+~+ u + n + 1) ’
(1.1)

Ramanujan did not indicate that (1.1) holds when - x - y - z - u - 2n 1 is a positive integer.
Entry 1 is originally due to Dougall [l] in 1907, which is probably less than three years before Ramanujan discovered the: theorem. Hardy [l, Eq. (2.1)] has thoroughly discussed Entry 1 and gives Dougall’s proof, as does Bailey [4, p. 341.

Entry 2. Zf either x, y, or z is a positive integer, then

-x, -y, --z F 3 2 [ n+l,-x--y-z-n

1

r(n+ i)r(x +y+ n + i)r(y+z + n + i)r(z + x + n + 1)

=r(x+ n+ l)r(y+ n+ iv++ n+ i)r(x+y+z+ n+l)’

Entry 2 is known as Saalschütz’s theorem Cl], [2], although according to Jacobi [l], [2] and Askey [l], the result was lïrst established by Pfaff [l] in 1797. In Hardy? paper [l], Entry 2 corresponds to Eq. (5.1) there. It should be mentioned that Hardy’s formulation is incorrect. For a proof of Entry 2, see Bailey’s tract [4, p. 91.

Entry 3. If x, y, z, or ---x .- y - z - 2n is a positive integer, then

*n + 1, 1, -x, -y, -z, x + y t z + 2n 6F5 [ +n,x+n+l,y+n+l,z+n+l,-x-y-z-n+1

1

(x + n)(y + n)(z + n)(x + y + z + n) = n(x + y t n)(,y + z + n)(x t z + n) ’

10

10. Hypergeometric Series, 1

PROOF. Set u = - 1 in Entry 1.

0

Entry 4. If either x, y, z, or -x - y - z - 2n - 1 is a positive integer, then

(n + 2k)(-x),(-y),(-~)&

+ y + z + 2n + l)k

kf=k-l- k(n + k)(x + n + l),(y + n + l),(z + n + l),( -x - y - z - n)k

= +(x + n + 1) + @(y + n + 1) + @(z + n + 1) + @(x + y + z + n + 1)

-$(n+l)-$(x+y+n+l)-+(y+z+n+l)-+(z+x+n+l).

PROOF. Logarithmically differentiate both sides of (1.1) with respect to u and then set u = 0. Using (0.3), we complete the proof after a little simplification.
q

Example (i). Zf x is a positioe integer, then

l--4(.x + 1)1-4(3x - 1) r6(2x)l-(4x - 2) ’

PROOF. In Entry 1, put n = 1, replace x by x - 1, and set y = z = u = x - 1.

After some simplification, the desired equality follows.

cl

Example (ii). If x is an odd, positive integer, then (x - 1)3(3x - 1) 1 (x - 1)(x - 3) 3(3x - 1)(3x + 1) (X + 1)3(3x - 3) + Z ( cx + I)(X + 3j > (3x - 3)(3x - 5) + ‘..
=;{ti(*I=1)+3~(~)-31(x)-i(l)].

PROOF. In Entry 4, put n = 0, replace x by i(x - 1) and set y = z = 4(x - 1).

The proposed equality now readily follows.

cl

Example (iii). Zf x is a positive integer, then

x3(3x - 2) (2x - 1)3 .

PROOF. In Entry 3, set n = 1, replace x by x - 1, and let y = z = x - 1. The

displayed equality now easily follows.

0

10. Hypergeometric Series, 1

11

Example (iv). Zj- x is a nonnegatiue integer, then

1 + (;)‘c+

(~~)‘,$---~;)

+ . . . =: r3(xr;‘f;;3;J+

l).

PROOF.In Entry 2, setn = 0 and x = y = z to achieve the desiredresult. 0

In the notebooks (p. 118),Ramanujan hasmistakenly put I-(3x + 1)in the numerator instead of the denominator in Example (iv).

Example (v). Zf x is a positive integer, then

xx-l

x

l+jÏx+p-----

x(x - 1) (x - 1)(x - 2) 2! (x+ l)(x+2)i:4x-

8r3(3x + 1)1-(x + 1) = 9r3(2x + l)I(4x+’

x(x - 1) 1)(4x-2)+“’

PROOF.In Entry 2, put n = z = x and y = x - 1. The proposed equality

readily follows.

cl

Entry 5. If Re(x + y + z + n + 1) > 0, then
1 +n + 1, n, --x, -y, -z
St4f +n, x + n + 1, y + n + 1, z + n + 1
r(x + n + i)r(y + n + i)r(z + n + i)r(x + y + z + n + 1) = r(n + i)r(x + y + n + i)r(y + z + n + i)r(x + z + n + 1)’ (5.1)

Entry 5 is again due to Dougall [l]. Hardy [l] discussesEntry 5 ((3.1) in hispaper) and gives a proof basedon a theorem of Carlson. For another proof, seeBailey’s monograph [4, p. 271. It is interesting that a q-analogue of Entry 5 was establishedby L. J. Rogers [l] in 1895,twelve years before Dougall’s discovery.
Wilson [l] hasshown that Dougall’s theorem is intimately connected with the orthogonality of certain orthogonal polynomials. Moreover [l, p. 6941,

SIm r(a + ix)r(b + ix)r(c + ix)r(d + ix) 2 dx

0

r(2ix)

is a continuous analogue of the sum in Entry 5. The specialcasec = 0, d = 3 was, in fact, evaluated by Ramanujan [S], [16, p. 571. For further related comments, seeSection 22 of Chapter 13.
For brevity, let

denote the sum of the tint m + 1 terms of ,+,F,(a,, . . . , aptl; PI,. . . , BP; 1).

12

10. HypergeometricSeries1,

Entry 6. Zf c1+ fi + y + 1 = n, then

Un + 2)Ua + l)r(p + l)I(y + 1) r(n - CI+ l)I(n - p + l)I(n - y + 1)

I 1 x5&

+<n+ [ z1(n + l),

3), n + 1, ~1+ n - tL + 1, n -

1,j3 + 1, y + 1 j3 + 1, n - y +

1

m

N 2 Log m - $(cr + 1) - $(fi + 1) - $(y + 1) - C,

asm tendsto CO,where C denotesEuler’s constant.

In our originally published account of khapter 10(seethe reference in the Introduction), we gave a proof of Entry 6 supplied to us by J. L. Fields based on his paper [l]. R. J. Evans [l] has sincefound a much simpler proof of a slightly stronger result. We reformulate this stronger version of Entry 6 and give Evans’s proof.

Entry 6 (Second Version). Zf a, b, c, and a + b + c are not nonpositive integers,
then as m tendsto CO,

II T(a + b + c)r(a)r(b)r(c)

+(a + b + c + l), a + b + c - 1, a, b, c

T(b + c)T(a + c)T(a + b) 5F4 +(a + b + c - l), b + c, a + c, a + b m

,

where y denotesEuler’s constant.

PROOF. Recall Whipple’s transformation [l] (Bailey [4, p. 25]),

1 a, 1 +$a,b,c,d,e, -m
F 7 6 $a, 1 + a - b, 1 + a - c, 1 + a - d, 1 + a - e, 1 + a + m

= (1 + a),(1 + a - d - e),

l+a-b-c,d,e,

(1 + a - d),( 1 + a - e), 4F3 l+a-b,l+a-c,d+e-a-m

-m

1 ’

(6.1)

where m is a nonnegative integer. Replacing a, d, and e by a + b + c - 1, a, and a + b + c + m + E,respectively, in (6.1), where E> 0, we fïnd that

1 a + b + c - 1,+(a + b + c + l), b, c, a, a + b + c + m + E, -m
i(a+b+c-l),a+c,a+b,b+c,-m-E,a+b+c+m

= (a + b + c),( -a - m - E), a,a,a+b+c+m+E, (b + c),( -m - e), 4F3 a+c,a+b,a+

1-m
l+.s *

Letting Etend to 0, we deduce that

10. H,ypergeometric Series, 1

13

Il + b + c + l), a + b + c - 1, a, b, c
5F4 $(u + b + c - l), a + c, a + b. b + c m

(a + b + ~),,,(a + l),

a,a,a+b+c+m,-m

= TF&(l),

4F3 a+c,a+b,a+l

1 ’

Thus, the left side of Entry 6 is equal to

I-(a + b + C)l-(a)I-(b)I-(c) (a + b + ~),,,(a + l),
T(b+c)T(a + C)r(u+ b) (b + 4,(l),

a,a,a+b+c+m,

x 4F3

a+c,a+b,a+

1-m
1 ’

(6.2)

We now apply a transformation for 1-balanced terminating 4F3 series found. in Bailey’s tract [4, p. 561. If u + u + w = x -k y + z - m + 1, then

4F~~~~~;~m]=~~~~~~~~z~m~~~[l~u~~~~~,~~~~~~m,~~>

Lettingx=a+b+c+m,y=a,z=a,u=a+b,u=a+c,andw=a+1, we lïnd that

a,a,a+b+c+m,-m

4 F3

a+c,a+b,a+l

1

_ Mn(l)nl

a,b, -c-m

(a + c),(a + l), 3Fz a+b,l-c-m

Ilm .

Using this equality in (6.2), we lïnd that the left side in Entry 6 equals

-T(a + b + c + m)r(c + m)R T(b + c + m)r(a + c + m) m’

(6.4)

where

Il R _ ww a, b, -c - m

-[ m r(a + b) 3Fz a+b,l-c-m

m

By Slirling’s formula, the coefficient of R, in (6.4) equals 1 + O(l/m). Examining Entry 6, we see that it remains to show that

R, = 2 Log m - y - +(a) - $(b) - e(c) + 0

(6.5)

Let R, = U,,, + v,,

14

10. Hypergeometric Series, 1

where r(a + k)lT(b + k)
u,= f k=lJ I-(a + b + k)r(l + k)
and
~(C+I k)r(b + k) v*= kF=lr(a + b + k)T(k)(m+ c- k) *

From Luke’s book [l, p. 110, Eq. (391,

u, = Log m - y - $(a) - $(b) + O(l/m),

(6.6)

as m tends to CO. (A slightly weaker version is given in Entry 15 below. See also (24.5) of Chapter 11 for (6.6)) By Stirling’s formula and (0.2),

v,= k$=lm+ ; _ k(l + O(W))

=$C+L-1 +“(&,,$(m+:-k+;))

= $(m + c) - $(c) + O(l/m)

= Log m - $(c) + O(l/m),
by Stirling’s formula for $(z) (Luke [l, p. 33-J). Combining deduce (6.5) to complete the proof.

(6.7) (6.6) and (6.7), we
ci

Corollary.

Let 0 < x < 1. Then us x tends to 0,

$4

5’1’1
[ 4;24;2;,2;2; 3 3

1 1 -x

- -Logx+3Log2.

PROOF. Let n = CI= fi = y = -3 in Entry 6 to obtain the formula

as m tends to CO,where on the right side above y now denotes Euler’s constant. Since $(*) = -2 Log 2 - y (see Luke’s book [l, p. 13]), we lïnd that

as m tends to 00. It follows that
T(k + ;)T4(k + $) T(k + a)(k!)”

1
- = 3 Log2. k

Hence,
lim
x+1-

T(k + $)l+(k + ;-i)-- 1 Xk = 3 Log2.

T(k + $)(k!)4

k 1)

10. HypergeometricSeries1,

15

Therefore, as x tends to 1-,

1 $Fd

45,”2‘1729232 [ ;, 1, 1, 1 ; x

-

-Log(l-x)+3Log2.

The corollary now follows.

0

For further expansions of hypergeometric functions in the neighborhoods of logarithmic singularities, seeSection 15 of this chapter and Sections24-26 in Chapter 11.B. C. Carlson [l] hasestablishedexpansionsabout logarithmic branch points for several classesof related functions.

Entry 7. Zf Re(x + y + trt + 1) > 0, then

n, -.x, --y 3F2[ x+n+ t,y+n+

1 1

I-(x + n + l)r(y + n + l)r(*n + i)r(x + y + +n + 1)

= T(n + l)T(x + y + n + i)r(x + +n + i)r(y + +n + 1) ’

PROOF. Set z = -3n in Entry 5.

Cl

Entry 7 is a famous theorem of Dixon [l]. In HaLrdy’spaper [l], see(3.2). A terrninating version of Dixon’s theoreni cari be used to evaluate Selberg’s integr,al in two dimensions (Andrews [3]). The caLsen = 3 of the DysonGunson-Wilson identity cari also be establishedfrom a terminating caseof Dixon’s theorem (Andrews Cl]). Gesse1and Stanton [2] have found new short proofs of both Saalschütz’s theorem (Entry 2) and Dixon’s theorem by computing the constant terms in certain Laurent serieslintwo variables.

Corollary 1. If Re(x + y + n + 1) > 0, then

( - x)k( - y)k
(x + n + l),(y + n + l)k
= $(x + n + 1) + $(y + n + 1) - $(n + 1) - $(x + y + n + 1). (7.1)

PROOF. Logarithmically differentiate both sidesof (5.1) with respect to z and

then set z = 0. With the aid of (0.3), we obtain the identity above after a little

simplification.

cl

Corolllary 2. Zf Re(x + y + 1) > 0, then
1 +n + 1, n, n, -x, -y
5F4[ in, x + n + 1, y + n + 1,l
=-- ryx + n + i)qy + n + l)rtx + y + 1) r(n + I)~(X + y + n + i)r(x + i)r(y + 1)’

16

10. Hypergeometric Series, 1

PROOF. Set z = -n in Entry 5.

Cl

Corollary 3. If Re(x + y + n) > 0, then

(x+ +n) $l + 1, -x, -y, 1

d(Y

1 4F3 i fn, x + n + 1, y + n + 1 = n(x + y + n) .

PROOF. Put z = - 1 in Entry 5.

cl

Corollary 4. If Re(x + y + *(n + 1)) > 0, trien
1 $l f 1, n, -x, -y
4F3 [ jn, x + n + 1, y + n + 1 I-(x + n + l)r(y + n + l)r($(n + l))T(x + y + +(n + 1))
= r(n + i)r(x + y + n + i)r(x -t +(n + i))r(y + +(n + 1))’

PROOF. Set z = -+(n + 1) in Entry 5.

Ci

Corollary 5. For Re(2x + 2y + n + 2) > 0,

1 +n + 1,n, -x, -y . -1 = y:(~ + n + i)r(y + n + 1)

4F3 [ +n, x + n + 1,y + n + 1’

Ir(n + i)rtx + y + n + 1)’ (7.2)

PROOF.Corollary 5 follows from Entry 5 by letting z tend to 00. The details

are easily justified by using Stirling’s formula.

0

Bailey [4, p. 281 gives a proof of Corollary 5 based on Whipple’s transformation (6.1).

Corollary 6. Zf Re(x + n + 1) > 0, then

( -x)k(k - l)!

F

1

k + n + k> (x + n + l),(n + l)k = k” (k + x + n)’ (7.3)

PROOF.Differentiate both sidesof (7.1) with respect to y and then set y = 0.

With the use of (0.2) and (0.3), we complete the proof.

0

On the left sideof(7.3), Ramanujan (p. 119)haswritten k! instead of(k - l)!.

Corollary 7. Zf Re(x - n + 1) > 0, then
_W7M+y+;r-;n2+;2.(7.4 +n+ l,n,n,n, -x 1 5F4[ )n, x + n + 1, 1, 1

PROOF.Set y = z = -n in Entry 5.

0

10. Hypergeometric Series, 1

17

Corolllary 8. If Re(x -- $I + 1) > 0, then
3Ffx+;+cl]=l-(x + n + l)r(+n + l)l-(x - *n + 1) r(n + i)r(x + i)r(i -+n)r(x + +n + 1)’

PROW. Put y = -n in Entry 7.

0

Corolllary 9. Zf Re(x -- tn + 3) > 0, then

$n + 1, n, n, -x-

r(x + n + l)r(+n -F +)r(x - in + 3)

SF’ [ in, x + n + 1, l- 1 = r(n + i)r(x + i)r(+ -+I)~(X + +n + 3)’

PFUXW.In Entry 5, sety = -n and .z= -+(n + 1).

0

Corolllary 10. Zj Re(2x - n + 2) > 0, then

1 in -t 1, n, n, -x .-1
4F3[ in,x+n+l,l’

=- r(x -t n + 1) r(n + i)r(x + 1)'

Pnoot:. Let y = -n in Corollary 5.

0

Corolllary 11. Zf Re x > -- 1, then

1 in, n, -x

r(x + n + l)P(+n + i)r(x + 1)

3F2 fn + 1,x + n + 1 = r(n + i)P(x + +n + 1) ’

PROW. Put y = -in in Entry 7.

0

Corolllary 12. Zf Re x > -3, then

,F,(n, -x;

x +

n +

1) =

r(x+n+ibr(2x+i) -r(22x + n + i)r(x

+ 1)'

Ralmanujan probably deduced Corollary 12 from Entry 7 by setting y = --i(n + 1)and then using Legendre’sduplication formula to simplify the resulting evaluation. However, in fact, Corollary 12is a specialcaseof Gauss’s theorem, which is given by Ramanujan in complete generality in Entry 8 below. SeeBailey’s monograph [4, pp. 2, 31for a proof.

Corolllary 13. Zf Re x > -- 1, then

2Fl(n,

-x;

x $- n + 1; -

r(x 1) = r(x

+ n +, l)r(+n + +n t l)r(n

+ 1) + 1)’

Coroilary 13 is known as Kummer’s theorem Cl], [2, pp. 75-1661 and is most commonly proved by using a quadratic transformation also due to Kummer. SeeBailey’s tract [4, pp. 9, 101for details. Ramanujan evidently derived Corollary 13 by letting y tend to cc in Entry 7.

18

10. Hypergeometric Series, 1

Corollary 14. Zf Re x > -f, then

1 F )n + 1, n, -x ;-1
3 * [ $l,x+n+l

=- r(x + n + l)I-($I + 4) qn + l)l-(x + +n + 3) *

PROOF.Put y = -$(n + 1)in Corollary 5.

0

Corollary 15. Zf Re n < 5, then

1 +n + 1, n, n, n, n r*(n) sin(7cn)tan(7rn)

F
5 4

*n, 1, 1, 1

= _ dr(2n + I)

'

PROOF.Let x = y = z = -n in Entry 5 and use the reflection formula to

simplify the resulting evaluation.

Cl

Corollary 15 is Eq. (3.33) in Hardy’s paper [l].

Corollary 16. Zf Re n < 3, then

1 +n+ l,n,n,n

4 F3 [

+n, 1, 1

=--s- in(r()n

+ +)r(* - $I)

d?(f -in)

.

PROOF.Put x = y = -n in Corollary 4.

q

Corollary 16is (3.31)in Hardy’s paper [l] and cari also befound in Bailey’s text [4, p. 961.

Corollary 17. Zf Re n < 5, then

1 in+ l,n,n,n

4 F3 [

+FI, 1, 1

;-1

=- sin(7rn) nn ’

PROOF.Set x = y = -n in Corollary 5.

0

Corollary 17 is Eq. (3.32) in Hardy’s paper [l] and is recorded by Bailey
C4,P. 961.

Corollary 18. Zf Re n < 1, then

[ l- in, n, n
3F2 fn + 1,1

2 tan($cn)r4(+r + 1) 7tnlr*(n + 1) ’

PROOF.Put x = -II in Corollary 11 and usethe reflection principle to sim-

plify the resulting equality.

0

Corollary 19. Zf Re n < 2, then 3F2

1 n:nr*(+n + 1) = sin(+nn)I(n + 1)’

10. Hypergeometric Series, 1

19

PROOF. Put x = -4~ in Corollary 11.

0

For the evaluation of certain other classesof $2 and 4F3 seriesat the argument 1, seethe papersby Lavoie Cl], [2].
Corollary 20. If Re(2x + n + 2) > 0, then

PROOF. Take the logarithmic derivative of both sidesof (7.2) with respectto

y and then set y = 0. Simplifying with the aid of (0.3), we achieve the desired

equahty.

Cl

Corollary 21. Zf Re x > - 1, then

1

1

k+x+n

3k .

PR~OF. In Entry 5, setz = -n, logarithmically differentiate both sidesof (5.1)

with respect to y, and then set y = 0. Using (0.2) and (0.3), we deduce the

desired result.

0

Ramanujan (p. 120)neglected to record the summands - l/k, 1 < k < a~, in Corollary 21.

Corollary 22. If Re n > 0, then

----(=1% (n + 1):

zn2 y

1

ffi (k + n)3’

(7.5)

PROOF. In (7.3), replace II by n - 1,differentiate both sideswith respect to x,

and then set x = 0. Use (0.3) in completing the proof.

0

Corollary 23. If Re n > - 2, then

-z k=i (k + l $)”

PROOF. In Corollary 6, set x = -$n. After a little simplification, the desired

result follows.

cl

Corollary 24. Zf Re n < 1, then

20

10. Hypergeometric Series, 1

PROOF. Differentiate both sides of (7.4) in Corollary 7 with respect to x

and then set x = 0. Using (0.2) and (0.3) and simplifying, we complete the

proof.

0

Example 1. Zf Re x > 4, then

PROOF. InEntry5,letn=

l,replacexbyx-

l,andsety=z=x-

1. 0

Example 1 has been given by both Harldy [l, Eq. (3.45)] and Bailey [4, p. 961. The following example is also recorded by Hardy [l, Eq. (3.43)].

Example 2. Zf Re x > i, then

kz’ (2k + l)e

= L xk 2x - 1.

PROOF. In Corollary 2, let n = 1, replace x by x - 1, and set y = x - 1. 0

Example 3. Zf Re x > 4, then

PROOF.In Entry 7, put n = 1, replace x by x - 1, and let y = x - 1. After

using Legendre’s duplication formula to simplify, we obtain the proposed

formula.

0

Example 3 is found in Hardy’s paper [l, Eq. (3.49)] and Bailey’s book [4, p. 961.The next example is equality (3.44) in Hardy’s paper [l].

Example 4. Zj Re x > 4, then

kzo(- lJk(2k+ llm(1c-

x>: Xk

=

l-(x + r(2x)

1) ’

PROOF.In Corollary 5, let n = 1, replace x by x - 1, and set y = x - 1. 0

Example 5. Zf Re x > ), then x-l
1+3- x+1+5

(x - 1)(x - 2) (x+ I)(x+2- )+“‘=x.

10. Hypergeometric Series, 1

21

PR~~E‘.Put n = 1 and replace x by x - 1in Coroilary 14.

0

Example 5 is given by both Hardy [l, Eq. (3.41)] and Bailey [4, p. 961. Examlple6 is also given by Hardy [ 1, Eq. (3.46)].

Example 6. Zf Re x > 0, then

x-l +(x-l)(x-2)+~~~=~-‘T~(x+l)~

l+-

~~

x + 1 (x + 1)(x + 2)

I-(2x + 1)

PRO~F’. In Corollary 13, put n = 1, replace x by x - 1, and use Legendre’s

duplication formula to simplify.

0

Examlple 7. If Re x > i, tken
l x-1 I (x-1)(x-2)-.,.x+ 1 (x+ 1)(x+2)

x 2x - 1.

PRO~F’.In Corollary 12,set n = 1 and replace x by x - 1.

0

Examples 7 and 8 are given by Hardy [l, Eqs. (3.47), (3.42)]. See also Bailey’s tract [4, p. 961 for Example 8.

Examlple 8. If Re x > 1, tken

1~3~YIL+s(x-l-

-...= 0.

(x + 1)(x + 2)

PROOF.Put n = 1 and replace x by x - 1 in Corollary 9.

0

Examlple 9. Zf Re x > 0, tken

c ---- m (- l)k(l - x)k 22”-2r2(x) 1 m

(7.6)

k=O (k + I)(l + x)k

Wx)

PROOF.Replace x by x - 1 in Kummer’s formula, Corollary 13. Then logarithnncally differentiate both sideswith respectto n and setn = 0. Using (0.2) and (0.3), we find that

Now,

1

---~. k+x-1

(7.7)

22

ccn (- l)k(l - x)k = x1 zl (- ‘“fcl-;x,flk

k=l

Wk

10. HypergeometricSeries,1 - 4

(7.8) by Example 6. Combining (7.7) and (7.8), WI: deduce the desired result.

Example 10. Zf Re x > 0, then

(l - x)k k=O (k + I)(l + x)k

PROOF.By Entry 9 below and (0.2),

(l - x)k

f (-x)k

k=,,(k + I)(l + x)k = -k=l k(x)k

= bw4 - $W

=--- 1 1 X 2x + ,gl (

and the proof is complete.

0

The next example is in Hardy% paper [Il, Eq. (3.48)] and Bailey’s book
C4,P. 961.

Example 11. Zf Re x > 0, then

1-p x-l

(x - 1)(x - 2)

24”l-4(x + 1)

3(x + 1) + 5(x + 1)(x + 2) -- ... = 4xr2(2x + 1)’

PROOF. In Corollary 11, put n = 1 and replace x by x - 1. After using the Legendre duplication formula, we easily obtain the proposed equality. 0

Example 12. If x is a positive integer, then
t1 - x)k
(k + 1)2(1 + x)k =

PROOF. Consider Dixon’s formula, Entry 7, and logarithmically differentiate both sideswith respect to y. Setting y = 0 and using (0.2) and (0.3), we find that

10. lH,ypergeometric Series, 1

23

1 k+n
Next, replace x by x -- 1 and differentiate both sides of (7.9) with respect to n. Setting n = 0 and using (0.3), we deduce that

(7.10)

On the other hand, by Example 10,

f (1 - X)k __ f (1 - X),-l@ - 4

k=l k’(X),

k=:, k*(l + x)k-1x

==: ,& $,;lxf

x)~ - j$ (k +:,1+

x)~

(7.11)

By combining (7.10) and (7.11) and using the fact that x is a positive integer,

we complete the proof.

0

Examlple 13. If Re x > 2, then

13+335+53

(x - 1)(x - 2) + . . . = x(4x - 3). (x + 1)(x + 2)

PROOF. We shall apply Entry 31 below with n = 1,y = -1,~ = u = -3,and

x replaced by x - 1. Accordingly, we lïnd that

,“3”,-x, 1 qx) 6F5 23,2+3 ,27$,231 + :Y,’1 .’ -,

=
l-(1 + 4 sF*[ -” ;;;”

‘1

=x , I (-1)(1-x) i a

= x(4x - 3).

0

Examlple 14

24

10. Hypergeometric Series, 1

PROOF. Let n = -x = -y = 4 in Corollary 5 to obtain

which is equivalent to the proposed formula.

0

Examples 14 and 15 were communicated by Ramanujan in his first letter to Hardy [16, pp. xxvi, xxv, respectively]. IHardy [2], [7, pp. 517, 5181 has observed the simple proofs that we offer hem. Evidently, Example 14 was lïrst established in 1859 by Bauer [l]. Examples 14 and 15 may also be found in Bailey’s tract [4, p. 961 and Hardy’s book [!), p. 71.

Example 15

PROOF. Set x = y = z = -n = -$ in Entry 5, and the proposed equality

follows forthwith.

q

Example 16

l+~(;>‘+;(~y+...=&.

PROOF. In Dixon’s theorem, Entry 7, let x =: -$, y = -$, and n = 4. [7

Example 17

1 +~(~)+~(~)+...=8~~~~~).

PROOF. In Dixon’s theorem, Entry 7, put n ==3 and x = y = -a.

0

Example 18

PROOF. Set - x = -y = n = 3 in Dixon’s thleorem, Entry 7.

q

Example 19

PROOF. In Kummer’s theorem, Corollary 13, set n = -x = f.

0

10. Hypergeometric Series, 1

25

Example 20. If Re n < 5, then

PROOF. In Dixon’s theorem, Entry 7, set x = y = -n. After several applica-

tions of the reflection principle and some simplificat:ion, we deduce the desired

formula.

Cl

Example 20 is due to Morley [l] in 1902. See Railey’s tract [4, p. 131 for further references.

Entry 8. If Re(x + y + n + 1) > 0, then

r(n + ~)I(X + y + n + 1)

pl(-x, -y; n + 1) = l-(x + n + l)r(y + n + 1).

(8.1)

As mentioned earlier, Entry 8 is Gauss’s theorem [l]. Following Entry 8, Ramanujan indicates, in one sentence, how he dedluced Entry 8. This is the only clue to the methods used by Ramanujan in his derivations of the several theomms in Chapter 10.
Assume that n and x are integers with n 2 0 and n + x 2 0. Expanding (1 + q+” and (1 + l/u)” in their forma1 binomial series and taking their product, we lïnd that, if a, is the coefficient of u”,

On the other hand, expanding (1 + u)x+y+” in its binomial series and dividing by uX, we find that

r(x + n + l)lQy + 1)

(8.3)

Comparing (8.2) and (8.3), we deduce (8.1).

Entry 9. If Re(a - p) > 0, then

(9.1)

PROOF. In Gauss’s theorem, Entry 8, put /I = -x and CI= n + 1. Take the

loganthmic derivative of both sides of (8.1) with respect to y and set y = 0.

Using (0.3), we complete the proof.

cl

Entry 10. If Re x > - 1, then
f (4 = r(n)r(x + :Il
k=O(n + k)k! r(n + x + :ïj

(10.1)

26

10. Hypergeometric Series, 1

PROOF. In Gauss’s theorem, Entry 8, let y = -n.

Cl

Example 1. If Re n > - 1, then

f (k - l)!
k=l k(n + l)k

PROOF. Differentiate both sides of (9.1) with respect to fi and set b = 0 and

a = n + 1. Using (0.3), we complete the proof.

0

Example 2. Zf Re n < 1, then

‘+

n(n + 1)

n (n+nl)l! +(n+2)2!+

7t =sin(nn)*

PROOF. Set x = -n in Entry 10.

0

Example 3. Zf n is arbitrary, then

PROOF. Let x = -2 and replace n by n + 1 in Entry 10.

0

Example 4. Zf Re n > - 1, then

1-L

n(n - 1)

3.1! +--- 5*2!

. . . __x/-n + 1) -- 2r(n + $) ’

PROOF. In Entry 10, replace x by n and n by 3.

0

Example 5. Zf Re x > - 1, then

‘f

(-x)k

k=O (n + k)2k! =

PROOF. Differentiate both sides of (10.1) with respect to n and use (0.2). 0

Example 6. Zf n is arbitrary, then
-(n++1 &1)2 (;)+&(g)+-
1 k + n ++ .

PROOF. Let x = -3 and replace n by n + 1 in Example 5.

0

10. Hypergeometric Series, 1

27

Example 7. IJ Re n < 1, then

71 $ ‘r+?

(n +Y)3!

+

n(n + 1) (n + 2)22!

+

. . . =-) sin(rrn) &

1 k+ n-

1

--

1 k >

.

PFUJF. Let x = -n in Example 5.

cl

Entry 11. Let n > 0 and suppose that Re(a - fi - 1) > 0. Then

kzo {(a + 4” - (P + 1 + k)“} {E$}”

= a”.

k

PROOIF. Observe that, for each positive integer m,

Thus, it suffices to show that lim (P + ‘), 0. m-m Mn
Since Re(fl + 1) < Re ~1t,he statement above is true by Stirling’s formula. 0
Corolllary 1. Zf Re(a - @- 1) > 0, then

ftg=

a-p-1p.

PROCIF. If n = 1, Entry 11 yields
(a-8-l)$y&a.

Multiplying both sides by /?/{N(U - fi - l)}, we obtain the desired formula.

Alternatively, in Entry 8, set n + 1 = 01,x = -- 1, and y = -8, and the

formula of Corollary 1 readily follows.

cl

Corollary 2. Zf Re(cc- fi - 1) > 0, then

kzl (a + fi + 2k - l)$ = -C8C2-fi-l’

PROOF. Apply Entry 11 with n = 2. Since (a + k)’ - (/? + 1 + k)’ = (a - fl - l)(a + j? + 2k + 1),
we find that

28

10. Hypergeometric Series, 1

Multiplying

(a - j3 - 1) $c (a + fi + 2k + l)w

= c?.
k

both sides by p2/{a2(a - /3 - l)), we complete the proof.

An alternate proof cari be obtained by letting x = y = -B and n = CI+ /? - 1 in Corollary 3 of Section 7.

Entry 12(a). Suppose that f(x) = C?=I (A,x”/k) in some neighborhood of the origin. Define Pk, 0 5 k < 00, by

efcX) = z. Pkxk.

(12.1)

Then P, = 1 and, for n 2 1,
np, = c AkP,,-,.
k=l
PROOF. It is clear that P0 = 1. Differentiating both sides of (12.1) with respect to x, we fmd that

Akxk-’ = 7 P,,nx”-l. “Z

Equating coefficients of x”-r on both sides, we deduce the required recursion

formula.

cl

Entry 12(b) is an instance of the inclusion-exclusion principle, but Ramanujan cleverly deduces Entry 12(b) from Entry 12(a). According to Macmahon [l, p. 61, Entry 12(b) is due to Newton.

Entry 12(b). For positive integers n and r, dgfine

and wherea,,a,,...,

9, = Pr(n) = C
1 <kiln klCk2C...<kr

aklak2”‘akrT

r I n,

a,, are arbitrary nonzero cornplex+zumbers. Then, if r 2 1,

where go = 1.

t-9, = i (-l)k+l&g,-k,
k=l

(12.2)

PROOF. In Entry 12(a), let Aj = (- ly’“S,,

j> 1.

10. H:ypergeometricSeries1,

29

Let r = max 1~” b,l. Then if 1x1-c l/a,

= exp t Log(1 + akx)

j=l

>

= kQ (1 + akx)

Hence, in the notation of Entry 12(a),P, = P,, and (12.2)follows immediately

from the conclusion of Entry 12(a).

0

In preparation for Entry 13, we need to make two delïnitions and prove one lemma. For each positive integer r, define

1

sr = SA% 4 = kgo & (

- ~(k+n+x+l)’

) ’

Let (pi(O=) 1, and delïne cp(n,x, r) = cp(r),r 2 1, recursively by

(13.1)

rcp(r) = c S,cp(r - k).
k=l
Lemma. Zf r is a positive integer, then

(13.2)

$‘P(r) = - c &+ldr - k). k=l

(13.3)

PROOF. We proceed by induction on r. If r = 1, equality (13.3)implies that

$<p(l) = $S’ = -s,,
which is easily verified from the definition (13.1). Now assumethat

$A.d

= -kil

sk+lv(j

- 4,

Hence:,by (13.2), (13.1),and (13.4),

lsljlr-1.

(13.4)

=b -k$l

kSk+l dr - 4 - kil sk ;gr sj+l dr - k -il)

30

10. Hypergeometric Series,1

which completes the proof.

q

Entry 13. Zf Re x > - 1 and r is any positive integer, then

f (-XIe
k=O (n + k)‘+‘k!

r(n)r(x + 1) = r~(n +~x + 1) cptr).

(13.5)

PROOF. Now by Example 5 in Section 10,
2 (-4k WWtx + Os = rtn)rtx + 1)
k=O (n + k)‘k! = r(n + x + 1) i r(n + x + 1) CPU).
Thus, (13.5) is valid for r = 1. Proceeding by induction, we assume that (13.5) holds for any lïxed positive
integer r and show that (13.5) is true with r replaced by r + 1. Differentiating both sides of (13.5) with respect to n and using the foregoing lemma, we find that

{4W - Il/@+ x + l)}cptr) + ~dr))

r(n)r(x + 1)

= r(n + x + 1)( -&cpW

- c Sk+,cptr - 4
k=l

r(n)r(x + 1)

=-T(n+x+l)

tr + l)cptr + 11,

from which (13.5), with r replaced by r + 1, follows.

cl

Corollary 1. Let S*(n, x) and cp(n,x, r) be defned by (13.1) and (13.2), respectively. Zj n = f and x = -3, then S, = 2 Log 2, S, = (2’ - 2)[(r), r 2 2, and

1 +&(~)+$(~)+...=~q(r),

r> 1. (13.6)

10. Hypergeometric Series, 1

31

PROOF. The proposed formulas for S,, r 2 1, are easily determined from (13.1) after brief calculations. Setting n = 3 and x = -i in Entry 13 yields

from which (13.6) trivially follows.

Cl

In the notebooks (p. 124), Ramanujan redelïnes S, for Corollary 1. We emphasize that his formulation of Corollary 1 is correct, however. Likewise, in Corollary 2, Ramanujan has redelïned S, in the notebooks. In fact, Ramanujan has proved Corollary 1 in his second published paper Cl], [ 16, pp. 15-171 by another method. Entry 13 and Example 1 below are also given in [l].

Corollary 2. Let S, and q(r) be defined by (13.1) a& (13.2), respectively, with n = 1 and x = -f. Then S, = 2 - 2 Log 2, S, = (2 - 2’)c(r) + 2’, r 2 2, and

1+~(~)+~(~)+...=2,1,,

r2 1.

PR~OF. Let n = 1 and x = -3 in Entry 13. The proof is completely analogous

to tha.t of Corollary 1.

0

Examlple 1

PROO’F. Letting S denote the infinite series above, we find from Corollary 1 and (13.2) that

s = 3P(2) = ;1w

+ S2Y~Ko~

=3s: +s,>

Examlple 2

=; 4Log2 2+1; .

I

1

a 0 8 cet 8 Log(sin 0) d0 = -: Log’ 2 - f .
s

PR~O:F. Letting u = sin 19and integrating by parts, we first find that

snl2 8 cet 8 Log(sin e) dtl = -i 0

l -~Log2 u s 0 ,/ÏTdU.

0 (13.7)

32

10. HypergeometricSeries1,

Next, for each nonnegative integer k, an elementary calculation showsthat

Lastly, recall that

1 l

1

UkLog2 u du = (k + 1)3

2 s0

(13.8)

(1 - uy/2

= 1 + 5u” + !&4

+ . ..)

lu1 < 1.

(13.9)

Now substitute (13.9) into (13.7) and integrate termwise with the help of (13.8) to obtain

ni2

-

8 cet 8 Log(sin 0) dB = 1 + -’ (21) + L (2E4)

s 0

33

s

+ ... .

Using Example 1, we complete the proof.

Cl

In preparation for Entry 14, define

and

where m and n are positive integers with m 2 2. Entry 14. Let n be an integer with n 2 2. Then

PROOF. Consider the decomposition from Nielsen’s book [ 1, p. 481

1

n-1

1

1

(k + x)“(k - j) = - ,=co (k + x)“-r(j +~ X)~+I + (j + x)“(k - j)’

Summing on j, 0 5 j I k - 1, we lïnd that

Next, sum on k, 1 I k < COt,o obtain

10. Hypergeometric Series, 1

33

1

n-2

- = - 2 L,r+1 j

- GI, l(X)

Observe that
zn(x)sn(x)= K?l+nc4+ G,.(x) + Cn,mM
Thus, (14.1)may be written in the form

> .
m, n 2 2.

(14.1)

-2k&$l jio]&+2s”+,(4

m k-l

+ “kg1 jgo,I:,,

This completes the proof.

Cl

Ra.manujan’sformulation of Entry 14 (p. 124)is somewhat imprecise. For several other results of this type, seeChapter 9 and the relevant references mentioned in Part 1 [9].

Entry 15. If CIand b are arbitrary complex numbers,then

1-l 1-(ct+ k + i=co T(M + j

l)l-(/? + k + + k + 2)k!

1) -

Log n -

$(a

+

1) -

$(p

+ 1) -

y,

as n tendsto co.

PROF. From a theorem in Luke’s book [l, p. 110,Eq. (35)],

WW)
r(U + b)

nc-1
k=O (U

(&@)k-
+ b)kk!

Logn-

$(a)-

W) -

Y,

asn t’endsto CO.Putting a = CI+ 1 and b = fi + 1,we deduce Entry 15. 0

Corollary. Let 0 < x < 1. Then usx tendsto 0, 7c2Fl($,3; 1; 1 - x) - Log x + 4 Log 2.

PROOF.From a general theorem in Luke’s text [ 1, p. 87, Eq. (1l)],

2Fl(a, b; a + b; 1 - x) - - r(a + b)(Log x + fi(u) + Il/(b) + 2y), (15.1) WW)

as x’ fends to 0, 0 < x < 1. The corollary now follows by putting a = b = f

and using the fact that +(i) = -y - 2 Log 2 (Luke [l, p. 131).

0

34

10. Hypergeometric Series, 1

It follows from Entry 15 that

“-l I-(cr + k + l)IY(B + k + 1)

c
k=O

I-(a+P+k+2)k!

~ - Log n,

as n tends to 00. This weaker result is due to Hi11 [l], [2]. See also Copson’s book [2, p. 2661. According to Copson [2, p. 2671, Gauss showed that

2Fl(a, b; a + b; x) T(a + b) j$-! Log{ l/(l - X)} =TOT(b)’

which is a consequence of (15.1). See also Whittaker and Watson’s text [l, p. 2991.

Entry 16. 1f A,, A,, . . . , A, are any complex numbers and

Pr = i Ak(-l)k ; , r 2 0,

k=O

0

then

A, = i Pk(-l)k
k=O

r 2 0.

A proof of this well-known inversion formula cari be found in Riordan’s book [l, pp. 43,443.

Entry 17. Suppose that

f(x) = f(r,-4= k$o$,

(17.1)

is analyticfor 1x1 > R. For 1x1> SU~(R, [hi), Write

(17.2)

Then

k 2 0.

PROOF. For Ix( > R, Ihl,
f(x) = k$ok!x’+c“r)(kBlk+ h/X)‘+k

Now equate coefficients of x-‘-” in (17.1) and (17.3) to deduce that

(17.3)

10. Hypergeometric Series, 1

35

After a straightforward calculation, the foregoing equality yields, for h # 0,

~ =k$o An(-

1)

h”

Bkt-h)k(;),

n 2 0.

Applying the inversion formula of Entry 16 and simplifying, we conclude that

B,,h-” = i Akh-k n

k=O

0k ’

n :2 0,

which implies the desired conclusion.

0

Entry 18(i). Supposethat (17.1) ho& Assumealso that

(18.1)

for 1x1> SU~(R,1). Furthermore, assumethut ~~=,,(Akxk/k!) is andytic for 1x1< R*. Then.for 1x1< R*,

(18.2)

PROOIF. Apply Entry 17 with h = - 1. Comparing (17.2) and (18.1) we find that

A, = i Aj(-1)j
j=O

k I> 0.

On the other hand, for 1x1< R*, by the Cauchy multiplication of power series,

(18.4)

where:

C,= $ (-l)‘Aj ; > k I> 0.

j=O

0

(18.5)

By (18.3)and (18.9, A, = C,, k 2 0. Thus, (18.4)becomesthe equality that we

sought to prove.

q

In Entry 18(ii), Ramanujan claims that if (17.1) arnd(18.1) hold, then
1 f o,A, cpc4 -$H} q?‘(X) k=O k! i is always an even function of x. This is clearly false. For example, letting q(x) := x and r = 1 provides a counterexample.

36

10. Hypergeometric Series, 1

Entry 18(iii). Supposethat (17.1) and (18.1) ,hold.Then, $ n is un eoeninteger,

*nA n-l =

(22k - 1)&‘&,-2k,

whereBj denotesthe jth Bernoulli number.

n 2 2,

(18.6)

PROOF. From the generating function (Abramowitz and Stegun [l, p. 804]),

X

m B,,x”

-e=cx-, - 1 “=o n.

1x1< 274

we fmd that, for 1x1< rc,

X

X

ex + 1 =-----e=Ixx - 1

2x
e2x - 1

Oo B,(l - 2k)Xk

k=l

k!

.

(18.7)

We now use the representation for ex given by (18.2) on the left side of (18.7). After some manipulation and simplification, we deduce that, for 1x1< min(rr, R*),

xc m $--+

j=O

.

= 2 2 Bk(1 ;,2k)Xk jf-o ‘%&.
k=l

If we equate coefficients of x”, with n even, on both sides above, we readily

deduce (18.6).

Cl

Entry 19. Supposethat 1x1,Ix - 11> 1. The,a .Y’ 2F1(r, m; n; 1/x) = (x - l)-’ 2F1(r, n - m; n; - 1/(x - 1)).

This transformation is well known (Bailey [4, p. 101) and is generally attributed to Gauss or Kummer. However, Askey [l] has indicated that it was originally discovered by Pfaff [l]. We shall give what was evidently Ramanujan’s argument.
PROOF. Apply Entry 17 with A, = (m)k/(n)k and h = 1. We then see that it suffices to show that

k 2 0.

(19.1)

But this is simply Vandermonde’s theorem (Bailey [4, p. 3]), which is a special

case of Gauss’s theorem, Entry 8.

cl

Entry 20. Let

OD q”‘(1)

cpG4 = rc=o -+x .

- 1)

(20.1)

be unulytic for Ix - 11< R, where R > 1. Supposethut m und n are complex

10. Hypergeometric Series, 1

31

pararneters suchthat the order of summationin

f (Mk .f <P(~)U)(- llY-dk

k=O (n)kk! r=k

r!

may be inverted. Then

.f (m)kdkW)

k=O

Wkk!

(20.2)

PRCMIFU. sing (20.1) to calculate C#~‘(O)0, I k < 00, and inverting the order of summation by hypothesis, we fit-rdthat

f tm)k <pck’to)

k=O

(n)kk!

= z. (- lYo(;;U$ -3,

r

by Vandermonde’s theorem (19.1).

cl

Note that if C~(X)= (x - l)‘, where r is a nonnegative integer, then (20.2) yields

i (m)kter)k k=O (n)kk!

=---- cn - m)p

(n),

’

Hence, in this case,(20.2) reducesto Vandermonde’s theorem.

Entry 21. For any complex numbersm, n, and x,

ex f (-l)k(n k=O

- m)kXk (“)kk!

PRO~F. Now,

ex f (- ljk(n - m)kXk

k=O

(n),k!

The coefficient of x’ on the right sideis

r (-l)k(n

- m)k

cm),

kc=o (n),k!(r - k)! = (n),rT’

by Vandermonde’s theorem (19.1). This completes the proof.

0

Entry 21 is due to Kummer Cl]. An alternate proof cari be obtained from Entry 19 by replacing x by r/x and letting r tend to CO.

38

10. Hypergeometric Series, 1

Entry 22. Supposethat (xl, (x + 1( > 1. Then (X + l)-’ 2Fl(r, m; 2m; 1/(x + 1)) = .x-l 2F1(r, m; 2m; - I/x).

PROOF.Set n = 2m and replace x by x + 1 in Entry 19.

0

Entry 23. Let m and x be any complex numbers.Then
m (- l)k(m)kxk 1: (m),xr ex k?o (2m),k! = ,& (2m),r! ’

PROOF.Set n = 2m in Entry 21.

q

Corollary 1. If x is any complex number,then

eX(I-(i);+(E)&...)=

1+k(~)~+(~)<+....

PROOF.Let m = f in Entry 23.

0

Corollary 2. If 1x1< 1 and Re x < t, then

PROOF.In Entry 22, let r = m = i and replace x by - 1/x.

Cl

The function zF1($, 9; 1; x) is a constant multiple of the complete elliptic integral of the lïrst kind and is central to the theory of elliptic functions. See
Part III of our account [ 1l] of Ramanujan’s notebooks. T. Matala-Aho and K. Vaananen [l] have studied the arithmetic properties
of LFl(i, i; 1; 0) when 13is algebraic.

Entry 24. Let (xl, Jx - 1) > 1andsupposethat mis arbitrary and that Re n > 0. Then

f

tm)k

k=O(n + k)k!x”+k = Jo ~;lt,lp-~;bk’

PROOF.In Entry 19,replace n by n + 1 and set r = n + 1 to obtain

-f (mL
k=Ok!x

= f (- ‘)“b

+ ’ - m)k

k=O k!(x -- l)“+k+’ ’

Integrate both sidesover [x, CO)to achieve the desired result.

Cl

Entry 25. Let Ix(, Ix - 11 > 1 and supposethat n is arbitrary. Then

= z. F&‘!

l)k+’ ’

10. Hypergeometric Series, 1

39

PRO~F. Put Y = m = 1 and replace n by n + 1in Entry 19,and multiply both

sidesby l/n.

0

Entry 26. If 1x1< 1 and M,/3,and y are arbitrary, then (1 - x)“+p ,F, (a, p; y; x) = (1 - x)’ *F1(y -- a, y - p; y; x).

Entry 26 is elementary and well known; seeBailey’s tract [4, p. 23.

Entry 27. If Re(n + 1) > - Re(x + y), - Re(p + q), then

I-(x + y + n + 1)

-p, -q,xi-y+n+

I-(x + n + l)r(y + n + 1) 3Fz[ x+n+l,y+n+l

rtp

r(p
+ n

++~q-i)r+tqn

+
+

1) n +

1) 3F2

-x,-y,p+q+n+l p+n+l,q+n+l

1 1
1 ’

Entry 27is a famous theorem of Thomae [l] and cari bederived from Entry 26. Hardy [l, p. 4991, [7, p. 5123 hasextensively discussedEntry 27 and has given references to other proofs. In Bailey’s bool< [4, p. 143, Entry 27 is equivalent to formula (1).

Entry 28. Zf Re(n + 1) > -Re(x + y), -Re(p - l), then

l- 3F2

[

-x, -y,p+n n,p+n+

1

(P + n)Wr(x

-t y + n + 1- )

r(x + n + i)ryy + n + 1)

-p,l,x+y+n+l

x 3F2 x+n+l,y+n+l

1

PROOF.Set q = - 1 in Entry 27.

0

Entry 29(a). If Re n > - 1, then

3F2[‘;;;;++2’=] FA3F,[ --;:;’ ‘1.

PR~OF. In Entry 28, put x = y = -3, n = 1, and p = n. Entry 29(b). If n is a nonnegative integer, then
A3F2 [‘;;;++21] = ::;; ; ;;.& $$

q (29.1)

This extremely interesting result was communicated in Ramanujan’s [ 16, p. 3511lïrst letter to Hardy and was lïrst establishsedin print by Watson [4] in 1929.A flurry of paperswaswritten on this formula and certain generalizations in the years 192991931.Referencesmay be found in Bailey’s book [4,
pp. 92.-951. Related results are given in Entry 32 and Section 35 below. A

40

10. Hypergeometric Series, 1

more recent proof of (29.1) has been given by Dutka Cl]. Further identities for partial sums of hypergeometric series have been established by Lamm and Szabo [l], [2] in their work on Coulomb alpproximations. The tïnite sum on the right side of (29.1) arises in the theory of Ifunctions of one complex variable and is called Landau’s constant. For details of this connection, see Watson%
paper C51.
Entry 29(c). If n is any complex number, then

PR~OF. InEntry27,letp=

-n-l,q=

replace n by n + 5.

Entry 29(d). If Re n > -3, then

--3,x= -n-$,andy=

-$,and
Cl

PROOF.In Entry 27, put p = -3, q = - 1, x = n, y = -3, and n = 1. Cl

Corollary 1. Zf G denotes Catalan’s constant, that is,

(29.2)
then (29.3)

PROOF.Putting n = -3 in Entry 29(a), we find that

1 ,F,($, 3, $; 1, $) = ,F&, 1, 1; +,+>.

On the other hand, from Example (i) in Section 32 of Chaptér 9 (seePart 1 [SI),

,F,(+, 1, 1; $3) == 2G.

Combining these two equalities, we deduce (29.3).

0

Corollary 2. As n tends to CO, n 3F2(+, 2, n; 1, n + 1) N Log n + 4 Log 2 + y.

Watson [S] has established an asymptotic formula for the fïnite sum on the right side of (29.1) as n tends to CO.Thus, Corollary 2 follows from Entry 29(b), Watson% theorem, and Stirling’s formula. We shall not relate any more details, because Entry 35(i) below gives a very closely related, fuller asymptotic expansion. R. J. Evans [l, Theorem 211 has generalized Corollary 2 by

10. HypergeometricSeries1,

41

showing that

I---W3F-f(ab) fhPc1lf]=Log c - y - $(a) - $(b) + 0

>

IT(tz + b)

as real c tends to 00.

Entry 30. Zj Re n > -Re x, -Re y, then

PROOF. Let y = - 1 and p = y in Entry 28.

Cl

Entry 31. Zf Re(x + y + n + 1) > 0 and Re(2x + 2y + 22 + 2u + 3n + 4) > 0, then

+n + 1, n, -x, -y, -2, -u
T(x + n + i)r(y + n + 1) -x,-y,z+u+n+l r(n + qr(x + y + n + 1) 3F2[ z+n+l,u+n+l

1 1 1 ’

Entry 31 is an immediate consequenceof Whipple’s transformation (6.1).

SeeBailey’s tract [4, p. 281 for details.

It is interesting to note that although Ramanujan did not discover Whipple’s

transfiarmation, hedid find this important specialcaseapproximately 20 years

before:Whipple’s proof [l] in 1926.An enlightening discussionof Whipple’s

theorem cari be found in Askey’s paper [3].

Supposethatweset -n=x=y=z=u=

-4inEntry31.Then

[“11 1

23 2> 2

= r($)r(*)3F2 1,i

2
=r4(t)3

(31.1)

by Example 18 in Section 7. This result may be found in Ramanujan’s [16, p. xxviii] lïrst letter to Hardy as well as in Hardy? book [9, p. 7, Eq. (1.4)]. Equality (31.1) wasestablishedby Watson [6], who gave the sameproof that we have given. Another proof wasgiven by Hardy 1121[,7, pp. 517,518].

Entry 32. If x + y + z = 0 and x is a positive integer, then

3F2p;;;,;y]

= r(~,n~~(:~)l)f~(n)~~~~!z)k.

(32.1)

PROOF.Consider the following result

42

10. Hypergeometric Series,1

“2 (-=aM&
rc=o (fhk!

T(a + n)T(b + n) a,b,f+n-1 r(n)IJa + b + n) ‘8” f,a+b+n

1 (32.2) ’

due to Bailey [2], [4, p. 933. Set a = n, b = y + z, f = z, and n = x + 1 in

(32.2) and use the fact that x + y + z = 0 to complete the proof.

Cl

In fac6 Entry 29(b) is not a special case of Entry 32. However, (32.2) does generalize Entry 29(b). The hypothesis x + y + z = 0 is not mentioned by Ramanujan. If x + y + z # 0, (32.1) is false in general. For example, if x = 2
and y = z = -5, then (32.1) is erroneous, as cari be seen by a comparison with the correct formula (32.2) with the proper parameters. For Entry 33 below, Ramanujan does provide the hypothesis x + y + z = 0.

Entry 33. If x + y + z = 0 and x + y + n is a positive integer, then

3F2[“,=;,,y]

=

r(n r(x

+ +

i)r(x + y n + i)r(y

+ + ~

n n

+ +

1) =+y 1) k=O

(-XM-Y)k (z),‘k!

.

PROOF. In (32.2), set a = -x, b = -y, and f = z, and replace n by x + y +

n + 1.

0

Entry 34. Zf x and y are arbitrary, then

Jk-($ + *y + )j

2Fl(x3Y; Hx + Y + ‘1; 3) = r(LX + +)r(ly + ‘1’

2

2

2

Entry 34 is due to Gauss [l]. In Bailey’s text [4, p. 111, Entry 34 is Eq. (2). The following result is due to Kummer [l] and cari be found in Bailey’s monograph [4, p. 11, Eq. (3)].

Corollary. Zf x and n are arbitrary, then
Jk2(1-“)‘2 r(&n + 3) ,F,(* - ix, + + 3x; +n + +; 4) = ~
r(+{n - x + 2})r(${n + x + 2))’

We refrain from explicitly stating Examples 1 and 2 which are merely the
special cases x = 0 and x = $, respectively, of the previous corollary. In Entry 35(i), Ramanujan delïnes

and then states an asymptotic formula for q({n + 1}/4) as n tends to 00. More

properly, q(n) should be delïned by (29.1). Thus, for a11complex n, define

,n r2(n + 3) dn=) [ii 1 r(n)r(n + 1) 3F2 l,;t+l .

(35.1)

10. HypergeometricSeries1,

43

Entry 35(i) is thus an extension of Corollary 2 in Section 29. Watson [S] and Dutka [l] have each derived asymptotic expansions for q(n). However, i.heexpansions of Watson, Dutka, and Ramanujan are a11of different forms. We shall employ Dutka’s asymptotic seriesto establish Ramanujan’s formulation.
Entry 35(i). Let q(n) be drfined by (35.1). Then asn tendsto CO,

PROOF. According to Dutka Cl], as n tends to CO, where

From Legendre’sduplication formula, it is easy to show that

(35.2)

Thus, as n tends to 00,

- Log 2 - u,. Using Stirling’s formula for Log Il/(x) (Luke [l, p. 33]),
$(x) N Log x - & - f B,,xT,
k=l 2k where x tends to COand B,, 0 < n < CO,denotes the nth Bernoulli number, we finsdthat, as n tends to 00,
n<p(q)-i(qJ)+3Log2+y+&
m &2=(2= - 1)
+ckEl (2k)(n + 1)2k - “.
Recalling that U, isdetïnedby (35.2),we now expressthe terms of l/(n + 1)-

44

10. Hypergeometric Series, 1

U, in terms of quotients of gamma functions. Thus, as n tends to CO,
.,(+(q+3Log2+g

a2 B,,22k(22k - 1) 1 _ - + 2k(2k + 1)

+c k=l

(2k)n2’

n

2n2

- 2k(2k + 1)(2k + 2) + 2k(2k + 1)(2k + 2)(2k + 3) _ . . .

6n3

24n4

r(+n + *) 12.32 + 2r(+n + 2) 242!2

12.32.52.72 -
284!4

r-(

n+3 4

>

r- n+19-( 4 >

12.32.52.72.92 21°5!5

12.32.52.72.92.112 2126!6

(35.3)

For each quotient of gamma functions displayed above, we use a general asymptotic formula for T(x + a)/T(x + b) due to Tricomi and Erdélyi [l] and reproduced in Luke’s book [l, p. 331. Omitting the numerical calculations,
we fïnd that, as n tends to 00,

r- n+l ( 2 > =;
r- n+5 ( 2>

l-

4 13 40 121 i + 2 - nj + n4 + Ofnm7),

i

i

r-(n+43 1 =$

l- 10 79 580 4141

r- n+ 11

i

( 4>

r-(-n+43 >

r-(n

+ 4

15
>

-2+1--n

310 n2

3990 n3 + O(n-‘), 1

(35.6)

10. Hypergeometric Series, 1 36 850 ; + 7 + O(n-‘),

45 (35.7)

(35.8) and
(35.9)

Substituting (35.4))(35.9) into (35.3), we now calculate the coefficients of npk,

2 I k I 6. After somelengthy calculations, we tïnd that a11the coefficients

agreewith what Ramanujan hasclaimed in Entry 35(i).

0

Entry 35(ii). Let q(n) bedefined by (35.1).Then for eachnonnegativeinteger n,

+ 2G,

(35.10)

where G is defined by (29.2).

Entry 35(iii). Let q(n) be defined by (35.1). Then for each nonnegative integer n,

(35.11)

Entry 35(iv). If q(n) is defined by (35.1) then

We shall lïrst prove Entry 35(iv) and then prove Entries 35(ii) and 35(iii) by induction.
PR~~ITOFENTRY35(iv). By (35.1) and Corollary 1 in Section 29,

46

10. Hypergeometric Series, 1

BY (35.1)and Dixon’s theorem, Entry 7, with n = 3, x = -4, and Y = -4, we lind that

cp($)= r($r”)(rt()s) 3F2(3,f, ii 192) = 3.

Cl

PROOF OF ENTRY 35(ii). We proceed by induction on n. For n = 0, formula (35.10)is valid by Entry 35(iv). Assume now that (35.10) holds for any fixed nonnegative integer n. Thus, it remains to prove (35.10) with n replaced by n + 1.
We lïrst establish the recursion formula

r2(n + 4) cp(n+ 1) = v(n) + 7rlY2(n+ 1)’

(35.12)

where n is any complex number, or, by (35.1),

(2n + 1)2(&+(;)‘&+(;y&+-*)

=4n’(~+(~~&+(~~&+*-*)+~.

(35.13)

In the courseof proving Entry 29(b), Darling [ 1,p. 9, line l] proved precisely the formula (35.13).
Hence, by (35.12) and (35.10),

%cp(n+f)=~g(n++)+

nr2(n + 1) 4r2(n + 3)

=;Y$

+2(3+02

(3):

=,9X

which completes the proof.

cl

PROOF OF ENTRY 35(iii). We induct on n. If n = 0, then (35.11)holds by Entry 35(iv). Assumenow that (35.11)holds for any lïxed nonnegative integer n, and SOit sufficesto prove (35.11)with n replaced by n + 1.
By (35.12) and (35.11),

10. Hypergeometric Series, 1

47

and the desired result follows.

0

In the first notebook (p. 239), Entry 35(iv) is listed before (35.10) and (35.11). Furthermore, prior to the latter two formulas, Ramanujan states the recursion formula (35.12). Thus, it seems clear that Ramanujan also used induction to establish (35.10) and (35.11).
At the beginning of Darling’s paper [l], in conjunction with Entry 29(b), he remarks, “His (Watson’s) own proof is by transformation of series, and it seems probable that Ramanujan obtained the theorem in a similar manner; but the following two proofs by induction, which will perhaps appeal more to the average analyst, may be of interest.” It appears that Darling’s specula-
tion is incorrect, and that he, in fact, had likely found Ramanujan’s proof. Dutka [l] has found a different proof of Entry 35(ii).

CHAPTER 11
Hypergeometric Series, II
Much of Chapter 11 is contained in Chapters 13 and 15 of the lïrst notebook, while some formulas from Chapter 11 may be found scattered among the “working pages” of the tïrst notebook.
In Chapter 11, Ramanujan gives many results on quadratic transformations of hypergeometric series. Several of these results cari be traced back to Kummer Cl], [2]. Ramanujan also offers many theorems on products of hypergeometric series. Although some of these results were established in the 19th Century, most are originally due to Ramanujan. Entry 34(iii) is a particularly elegant formula which combines a product formula and a quadratic transformation. Much of Bailey’s work in the 1930s on products of hypergeometric series was motivated by Ramanujan’s discoveries.
Corollary 2 in Section 24 offers a certain asymptotic formula for zerobalanced 3F2 series. Such formulas in the literature have previously been established only for zero-balanced ZF, series. It is interesting that this elegant formula had been overlooked for 60 years after Ramanujan’s death. We provide here an elegant proof of this asymptotic formula by R. J. Evans and D. Stanton [l]. However, their proof depends on knowing the formula in advance. It would be interesting to have a more direct proof that might shed some light on Ramanujan’s approach.
There are two additional formulas in Chapter 11 which are amazing indeed. The fïrst is Entry 22, which involves a remarkable recursively delïned sequence A, and which leads to two intriguing binomial coefficient identities (22.22) and (22.23). The second is Entry 31(ii), which we were only able to prove by using the theory of second-order inhomogeneous linear differential equations and equating coefficients in 15 power series. Unfortunately, we have no idea how Ramanujan discovered these two extraordinary formulas (as well as most of the results in this chapter). Our proofs of these two theorems are certainly

11. HypergeometricSeriesI,I

49

not those found by Ramanujan; he must have derived theseformulas more naturally. Although differential equations have traditionally played a strong role in the theory of hypergeometric series,there is nloevidence that Ramanujan significantly utilized this connection. The hypergeometric differential equation doesappear in somewhatdisguisedform in Entry 31(i). The formulas in Sections 30 and 31 of Chapter 11 are the only ones in Chapters 10 and 11 with links to differential equations.
A few formulas in Chapter 11 are apparently without meaning. Entry 24 is suchan example; we have not been able to lïnd any functions for which the proposed formula is valid.
We usethe notation that was setforth in the introduction to Chapter 10. In that chapter, we considered the casep = 4 + 1. Since in this chapter, we establishtheoremsfor p # q + 1,we offer further remarks about convergence. If p < q + 1, then PF4convergesfor a11Imite values a’fx; if p > q + 1, then ,,F4 convergesfor only x = 0 unlessthe seriesterminates. For most of the theorems and examplesin the sequel,we shall not state the region of validity because it cari readily be ascertained from the general remarks we have made about convergence.
In the sequel,we shall frequently appeal to the treatisesof Erdélyi [l] and Bailey [4].
Entry 1. Let <pbe any function. Then, provided the seriesconverges,

is an evenfunction of x.

PROOF. Consider the quadratic transformation found in Erdélyi’s work [l, p. 112,formula (26)] and due to Kummer [l, p. 781: [2, p. 1141,

22
ZFl(r, m; 2m; z) = (1 - z)-“’ zFl +r, m - $r; m + 4; ~ 4(z - 1) .

Setting z = 1 - <p(- X)/C~(X)w, e find after somesimplification that

1 --zFl cp’(4

r, m; 2m; 1 - “I,rx;‘>

which ciearly is an even function of x.

Entry 2 2x
2F1 r, m; 2m; ~ 1+x

= (1 + x)l ,F,($r, $(r + 1);$(2m + 1);x2).

50

11. HypergeometricSeriesI,I

Entry 2 is a well-known quadratic transformation (seeErdélyi’s book Cl, p. 111,Eq. (4)]) that is due to Kummer Cl, p. 781, [2, p. 1141.

Entry 3
( 4x
2Fl r, mi 2m; (1 + x)2 =(l+~)~‘~F~(r,r-rn+~;rn+~;x~). >
Entry 3 is precisely Eq. (5) of Erdélyi’s treatise Cl, p. 11l] and is due to Gauss Cl]. This formula is also mentioned by Hardy [l, p. 5021,[7, p. 5151.

Entry 4
( 4x
,F, +r, *(r + 1); 2(2m + 1); (1

= (1 + x)l ,F, (r, r - m + +; m + $; x). >

PROOF.In Entry 2, replace x by 2,,1%/(1 + x) to find that

4x

~FI ( 3r, t(r

+ 1); 3(2m

+

1); (1 + x)2 )

(1 +x) = (1 + fi)2r

2F1

r9m;2m;(1+4fi $J2
(

)

= (x + 1)’ Jl(r, r - m -t- 3; m + 4; x),

by Entry 3.

0

Entry >

r,i; I;(I~ > 4x

2F1

= (1 + x)~’ 2FI (r, r; 1; x2).

PROOF. Put m = 2 in Entry 3.

Ci

Entry 6
2F1(tr,f(r + 1);1;(14+x x)2>= (1 + x)l $,(r,r; 1;x).

PROOF. Put m = 4 in Entry 4.

n

Entry 7

,F,(m; 2m; 2x) = eX ,,FI(m + f; x2/4).

(7.1)

Entry 7 is due to Kummer [l, p. 1401, [2, p. 1341 and was recorded by Hardy [l, p. 5021,[7, p. 5151.Entry 7 follows from Entry 2 by replacing x by xjr there and then letting r tend to 00.

11. Hypergeometric Series, II

51

Corollary. IF,& 1; x) = e”‘2 ($,(l; (~/4)~).

PROOF. In Entry 7, put m = 4 and replace x by x/2.

0

Entry 8. Let C~(X)be anaZytic for Ix - 11< R, where R > 1. Supposethat m and cpare suchthat the order of summationin

may be inverted. Then
m cp’k’(0)2k(m) kc=O (2m),k! Ir = z. 22kc;;b’

PROOF. Since cpis analytic for (x - 11 < R, R > 1, we readily find that

m cp(“)(l)( - l)“( -n)k

cp’k’(0)= c

n=k

n!

’

k 2 0.

Hence, inverting the order of summation, by hypothesis, we find that

m q’k’(o)2k(m)k & (2m),k!

(8.1)

Now multiply both sides of (7.1) by emx and then equate coeffkients of x”, n 2 0, on both sides to obtain the evaluation

02,”(+mI>.,,(+iiffnnniisjs’oeddv. en(’8.2)

If we substitute (8.2) into (8.1), we complete the proof.

ci

Entry 9. If n is an integer, then

2”-‘T(n + 4)

J,(n + 3; (3x)‘) =

ex2Fo(n, 1 -n;&

J71X”

i

>

( + cos(n7c)e-x2F, n, 1 - n; -2x1 . )l

Observe that

$,(n + +; (jx)‘) = r(n + 3)(2/~)“-“‘Z,-~,~(x),

(9-l)

where Z, is the Bessel function of imaginary argument usually SOdenoted (see Watson3 treatise [9, p. 771). Thus, Entry 9 is a well-known result in the theory
of Bessel functions (ibid. [9, p. 80, formulas (lO), (1 l)]).

52

11. Hypergeometric Series, II

Corollary. As x tends to CO,

OFlu; (‘ix)‘)

l2

12.32 12.32.52

22(2x) + 242!(2x)2 + 263!(2x)3 + ‘*’ > ’ (9.2)

PROOF.Undoubtedly, Ramanujan formally deduced this formula from Entry 9 by setting n = 3 there.
However, by (9.1), which holds for any complex number n,

OFlu; (3x)2) = I,(x).

Remembering that x is positive, we observe that (9.2) is precisely the asymp-

totic expansion of I,(x) given by Watson [9, p. 2031.

ci

It is possible that Ramanujan did not restrict n to be an integer in Entry 9. In such a case,the right sideof Entry 9 is an asymptotic expansion for the left sideas 1x1tends to 00 when larg XI < $7~(Watson [9, p. 203]), provided that Cos(w) is replaced by exp(inn).

Entry 10. If n is an integer, then

oFl(n+ +;-(3x)2) = “7’ ‘){cos(+nn_ x) f q;f;il)z

n)2k

XX”

k=O

- sin(+nr - x) Cm (- 1)k(n)2k+l(1

- 11)2k+l

k=,, (2k + 1)!(2x)2k+’ ’

PROOF. Replace x by ix in Entry 9 and equate real parts on both sides.After

somesimplification, we achieve the desiredequality.

0

Corollary. Supposethat n is an integer. Let x0 be a root of
oF,(n + 3; -(ix)‘) = 0. Let p bean odd integer chosenSOthat Ix0 - $I(F + n)l is minimal. Then if x0 is “large,”

x0 -

4~ + 4 2

1, 41 - 4 + n(l 4~ + 4

4 PU - 4 37c3(p+ n)”

61 + . . . .

(lo 1)

PROOF.By Entry 10,we want to approximate large roots of

cos(& - x) 1Oo (- 1)k(n)2k(1

- n)2k

k=O

(2k)!(2x)2k

- sin(+ - x) Cm (- r)k(n)2k+l(1

- n)2k+l

k=,, (2k + 1)!(2X)2k+’

We shall usea method of successiveapproximations.

= o
’

(10.2)

11. Hypergeometric Series, II

53

From (10.2) it is clear that we should take as a lïrst approximation

x = $C(c( + n).

For our second approximation, consider &T(P + n) + y, where, by (10.2), y

should satisfy the equation

n(1 - n)

cos(+n - {+n(p + n) + y}) - sin(+ - {f7@ + n) + y})-

= 0.

After a short calculation, we find that

n(1 - n) tan y = ~
4P + 4’ Hence, as our second approximation, we shall take

For our third approximation,

4P

+ 2

n)

, I

n(1 0

- n) + 4’

consider

7.0 +
2

4

+

n(1 ~ 4P

+

n) 4

+

z,

where, by (10.2), z is to satisfy the equation

-sin(E+z){l

-n(n+2~z((l~~~~~“)}

+...l~+;l{~(~+n)~~~~~;:,

Hence,

- n(n + l)(n + 2)(1 - n)(2 - n)(3 - n) 67r3(~ + n)3

n(n + I)(n + 2)(1 - n)(2 - n)(3 - n)

!Y(rl + l)(l - n)(2 - n) 27c2(p + n)2

n(1 - n) n(1 - n) I

-2n(l

_ n) _ (n + l)(rl + W - W - 4

4~ + 4 7c3(p + n)”

6

+ n(n + l)(l - n)(2 - n) 2

(10.3)

Now,

54

11. Hypergeometric Series, II

Thus, from (10.3) and (10.4),

z ~ 41 - 4 -2n(l - n) - -(n + l)(n + 2)(2 - n)(3 - n)

7c3(p + ny

6

+ -

-

-

+ n(n l)(l 2 n)(2 n) n2(1 3 H)z 1

41-n)

7

ZZ7Zc3(p + n)3 i 3” z-Zne32 1 .

Hence, our third-order approximation is precisely that claimed by Ramanujan

in (10.1).

0

Entry 11. If

sx ~sin u du = ; - r cos(x - e) 0 u

(11.1)

x 1 - COSU

du = y + Log x - r sin(x - e),

s 0

?A

where y denotes Euler’s constant, then

m (- l)k(2k)!

r cas 9 - C

X2k+l

’

k=O

Oo(- l)k+1(2k - l)!

r sine - 1

>

k=l

XZk

and

as x tends to CO. PROOF. By (ll.l),
So’$?&(j)j-~)~&,

= -7--c 2

r cas x cas 8 -

r sin x sin 8.

By successively integrating by parts, we easily !ïnd that

(11.2) (11.3) (11.4) (11.5)
(11.6)

11. Hypergeometric Series, II

55

asx tends to 00. Thus, (11.3)and (11.4) follow from (11.6)and (11.7). We next show that (11.2) is consistent with (11.3) and (11.4). From (11.2)
and the tables of Gradshteyn and Ryzhik [ 1, p. 92811,

x 1 - COSu

du=y+Logx+

s 0

u

ccC~OSu du sx U

=y+Logx-rsinxcostl+rcosxsin8.

On the other hand, by successivelyintegrating by parts,

(11.8)

mCOSu ~ du - -sinx
sx u

m (- l)k(2k)!

g (- l)k+‘(2k - l)!

C X2k+l k=O

+ COS x ‘i k =l

X Zk

(11.9)

asx tends to CO.Using (11.8)and (11.9) we again deduce(11.3) and (11.4). From (11.3)and (11.4)

r2 - z. (-yfk)!j2

+ {zl (- l)‘ilJk

- ‘Y)‘,

i

asx tends to 00.The coefficient of X-‘“, n 2 1, above is equal to

n-1

n-l

(- l)“+’ 1 (2k)!(2n - 2 - 2k)! + (- 1)” 1 (2k - 1)!(2n - 2k -

k=O

k=l

zn-2
= (- 1)” C (- l)k+‘k!(2n - 2 - k)!.
k=O

l)! (11.10)

Comparing (11S) and (11.10) and replacing n by n -- 1, we seethat it suffices to show that

kio(-l)kk!(2n-k)!=(2;;;)!,

n>O.

(11.11)

Let S, denote the left side of (11.11). Using (32.2) in Chapter 10 with n replaced by 2n + 1, a = b = 1, and f = -2n - E,where E> 0, we fïnd that

=GM! ‘2 kto(-;;ky;)kk,

P(2n + 2)

1, 1, -E

= (2n)!

lim 3F2

l-(2n + l)I-(2n + 3) E-0

-2n-E,2n+3

1

r2(2n + 2)

1 + lim f

= r(2n + 3)

&+O k=Zn+l (-2n

(l)k(-E)k - &)k(h

+ 3); >

56

11. Hypergeometric Series, II

=PL~ -‘y-;’

(1)2”+1-4( 2”+1 ~ f Qn + 2),& + 1- 4k

&+O (- 2n - &n+1(2n + 3)2,+1 k=O c1 - E)k(4n + 4)k

(2n + 1)!(2n + 2)!

2n + 1,2n + 2

( 1+ 1) (4n + 3)!

2F1 [ 4n + 4 ;l

(2n + 1)!(2n + 2)! r(4n + 4)F(l)
( l+- (4n + 3)! r(2n + 3)r(2n + 2)

= ig?;(l

+ 1) = (2; 1 y! >

where we have employed Gauss’s theorem, which is Entry 8 of Chapter 10.

This completes the proof of (11.11).

Cl

For results similar to Entry 11, see the author’s [9] account of Chapter 4 of Ramanujan’s second notebook.

Example 1. S:I2 COS(~ sin2 0) df3 = 0.

PROOF. Letting

sin2 0 = $(l - cas 20)

and replacing 0 by 7112- 8, we find that

ni2
COS(~~
s 0

sin2 e)

w

dtl =

sin(& COS28) dB

s 0

ni2

=-

sin(+r COS28) dB,

s 0

from which the desired result follows.

(11.12) 0

Example 2. JO” COS(~~ sin2 0) de = -S;i” COS(~ sin 0) dB.

PROOF. As above, the proof is quite elementary. First, usethe identity (11.12)

and then replace 28 by 7r/2- 8. After simplifying, we obtain the desired

equality.

0

Example 3. ~~cos(2+sin20)dU=$SU/2cos(~sine)d&

PR~OF.The stepsare exactly the sameasin the previous proof.

0

Entry 12. Zf x + y + z = 3, then zF1(-X, -y; z; p) = $71(-2x, -2y; z; f(1 - .J1-p)).

11. Hypergeometric Series, II

57

With obvious changes in the parameters, Entry 12 is the same as equality (10) of Erdélyi’s book [l, p. 1111. A formula equivalent to Entry 12 was given by Hardy Cl, p. 5021, [7, p. 5153 in his overview. En-try 12 is due to Gauss [l].

Corollary
l2 + n
'+Fx+

Cl2 + W2 + njx2 + (12 + n)(5’! + n)(92 + n)X3 + ...

42.82

42. g2. 122

+ ....

PROOF. In Entry 12, set x = (- 1 + i&)/4, y = (- 1 - i&)/4, LT= 1, ad

p = x.

0

Example 1

l+$+Er k:2;fir( l-23 l)(l --+(l-

ji&)&

= l + kEl k:i (1+;)(1+;)-(1

+$)(‘-yy.

PROOF. In the corollary above set n = 3. For k 2 2, we are led to examine

(12 + 3)(52 + 3)(92 + 3)...((4k - 3)’ + 3) 42. 82.. . (4k - 4)2(4k)2

(12 + 3)(52 + 3)(92 + 3)...((4k - 3)2 + 3) 22.42...(2k - 2)2(2k)24k

(22 + 2 + 1) (42 + 4 + 1) ((2k - 2)2 + (2k - 2) + 1) 1

22

42 -...

(2k - :!)2

W2

2.4...(2k - 2) 1(22 + 2 + 1) 3(42 + 4 + l)...

1.3...(2k - 3) 23

43

(2k - 3)((2k - 2)2 + (2k - 2) + 1) 1

X

(2k - 2)3

G’d2

(12.1)

where in the middle expression above we used the equality 4((2n)2 + 2n + 1) = (4n + 1)2 + 3.
We are also led to examine, for k 2 1,

58

11. Hypergeometric Series, II

(12 + 3)(32 + 3)...((2k - 1)2 + 3) 22 . 42.. . (2k)2
= (22 - 2 + 1)...(k2 - k + 1) 12.22.. . k2

= 2(12-1+1)3(22-2+1)

l3

23

(k+l)(k2-k+l)

‘.‘-

k3

1 k+l

(12.2)

where in the second expression above we used the equality 4(n2 - n + 1) =

(2n - 1)2 + 3.

Using (12.1) and (12.2) in the previous corollary, we obtain the desired

result.

0

Example 2. Zf CI+ p = 1, then

(1+3

2F1(4(+aY)4, <P+ Y)Y; + 1;4

= ,F,(a, /3; y + 1; +(l - Jl - x)).

Example 2 is well-known (e.g., seeErdélyi’s compendium [l, p. 112, formula
WI 1.

Entry 13. Zf a + fi + y = 0, then
,Ff(--a, --fi; y + 3; x) = 3F2(-2a, -28, y; y + +, 2~; x).
Entry 13 is a famous result of Clausen [l]. Other results on products of hypergeometric series are given in the sequel. See also Bailey’s tract [4, Chapter 101. Entry 13 was mentioned by Hardy [l, p. 5033, [7, p. 5161.

Corollary 1

12+n x + u2 + n)(52 + njx2 + ... 2

I+F

42 *82

112+n = l + yFx

1.3 Cl2 + n)(32 + n)x2 + ... . + m-22.42

PROOF. Put a = (- 1 + i,/%)/4, /I = (- 1 - i&4, and y = 3 in Entry 13. cl
Corollary 2. Zf J, denotesthe ordinary Besselfunction of order 0, thenfor a11x, Jt(i&) = 1F,($; 1, 1; x).

11. Hypergeometric Series, II

59

PROOF.In Watson%text [9, formula (6), p. 1473,setv = 0 and z = iJx to find that

Since(2k)!/(k!22k) = (f),, the desired result follows.

cl

According to Watson [9, p. 1453,Corollary 2 is originally due to Schlafli
VI.

Entry 14. If c(+ fi + 1 = y + 6, then

2Fl(% P; y; 21 - JG))

,F,@, /3; 6; $1 - &a

This equality cari be found in Bailey’s monograph [4, p. 88, formula (3)], where x = 4z( 1 - z). The first published proof of Entry 14is due to Bailey [3] in 1935.

Entry 15. For any x,

OF,(y; x) OF1(S;x) = f (’ + ;r,;(&=

k=O

k k.

A short calculation shows that

(+(Y + s))k(+b (Y+s-

+ d - 1))k22k l)k

= (y + 6 + k - f)k.

Thus, Entry 15 cari be written in terms of hypergeometric series,

oF,(y; 4 OF,@; 4 = P&(Y + 4, +(Y + 6 - 1); Y, 6, Y + 6 - 1; 4.9

In fact, Entry 15 gives a formula for J,-1(2i,&)Ja-,(2i&),

where J,

denotes the ordinary Besselfunction of order v. This result is due to Schlafli

[l] and thus representsa generalization of Corollary 2 in the previous section.

Entry 15 is also given by Watson [9, p. 147, formula (5)], Hardy [l, p. 5031,

[7, p. 5161,and Erdélyi [l, p. 185,formula (2)]. Bailey [l] hasalso established

Entry 15aswell as generalizations.

Entry 16. If x is arbitrary, then

OF,(m + 1, n + 1; x) oF,(m + 1, n + 1; -x)

=f

(- l)k(m + n + 2k + l),xZk

k=O (m + l)k(n + l)k(m + l)2k(n+l),,k!’

(16.1)

60

11. HypergeometricSeriesI,I

A brief calculation shows that

~~

(+(m+ n + l))&(m + n + 2)),(3(m + n + 3)),33k

(+b + n + l))k(+(m

+ n + 2))k(+(m+

l))k(+trn

+ 2))k(+(n

+ l))k<+b

(m + n + 2k + l)k

tm + lhk@

+ l),k’

Thus, Entry 16may be written in the form

()F,(rn + 1, n + 1; x) J,(m + 1, n + 1; -x)

f(m+n+ l),)(m+n+2),&n+n+3)

+ 2))k26k
1 27 2
-Gïx .

The first published proof of Entry 16is evidently due to Hardy [ 1, p. 5033, [7, p. 5161 who stated Entry 16 in the latter form. Seealso Erdélyi’s treatise [l, p. 186,formula (7)].

Entry 17

$,(m + n + 1, n + 1;x) oF,(m + 1, 1 - n; -x)

= 1 + .f xk(@m

+ n + k + 2))k(2x)k

k=l (m + n + l)k(m + l),k! ’

where,for k 2 1,

(17.1)

Gc=k!

(n’_ lZ)(nZ

p:2)...(n2 1

-k2)’ ifkisodd9

(17.2)

(n’ _ 22)(n2 -;2)...($ _ k2)’ ifk iseuen’

PROOF. For r 2 0, the coefficient of x’ on the left sideof (17.1)is equal to

c,:= i,

(- lJk

k+ (m + n i- l),-,(n + l)r-k(r - k)!(m + l)k(l - n)kk!

1

c (-m-n - r)k(-n - r)k(-r)k

= (m + n + l),(n + l),r! kEO (m + l),(l - n),k!

’

where we have used the elementary relation

(- Uk(4,

(a),-k =

(-a

- r + l)k

(17.3)

with a = m + n + 1, n + 1, and 1. We now apply Dixon’s theorem, Entry 7 of Chapter 10,with x = m + n + r,
y = r, and n replaced by -n - r. Accordingly, we find that

11. Hypergeometric Series, II

61

l-($(2 - n - r))T(m + l)l-(1 - n)l-(4(2 + 2m + n + 34)
c, = (m + n + l),(n + I),r!r(l - n - r)r(+(2 + 2m + n + r))r()(2 - n + r))r(l + m + r)
(4(2m + n + r + 2))J(n + l)r(l - n)I(3(2 - n - r)) = (m + n + i),(m + qrr!r(i + II + r)r(l - n -- r)r(+(2 - n + I))’
(17.4)
after a considerable amount of simplification. Comparing (17.4) with (17.1), we lïnd that it suffrcesto show that

r(n + l)r(l r(1 + n + r)r(l

-

n)r(+(2 - n n - 1)r(+(2 -

r)) n +

r))

== 2*u,,

r 2 0. (17.5)

After using the functional equation of the gamma function, we readily establish

(17.5), and therefore Entry 17 is proved.

cl

Entry 18

,F,(-PG Y; -4 ,F,(-P; y; 4 = 2F3(-B> p + y; y, )y, %y + 1); x2/4).

Evidently, the lïrst published proof of Entry 18 was given by Hardy [l, p. 5033, [7, p. 5161. (There is a misprint in Hardy% formulation; read x2/4 instead of -x2/4.) See also Erdélyi’s book [l, p. 186, formula (5)]. For extensions and q-analogues of Entries 16 and 18, see the Ipaper by Srivastava [l].
Ramanujan (p. 133) has an extra factor of (y + 4) in the denominator of the coefficient of x4 on the right side above.
If we replace x by -X//I in Entry 18 and let p tend to 00, we find that

oF,b; - 4 oF,(y; 4 = &(Y, 3r, +(Y + 1); -x2/4).

(18.1)

Entry 19. If a or j3is a nonnegative integer, &,(-a, -B; 4 Ad-~, -B; -4 = j. <-cdk<-a>kc;; - B + k)kXZk

= 4F1(-CG -8, -+(a + /?), -$(a + jl- :L);-u - /l; 4x2).

Entry 19 may be proved by multiplying termwise the two series on the left side and applying Dixon’s theorem. Entry 19 may be found in Erdélyi’s treatise [l, p. 186, formula (4)].

Entry 20. If x is arbitrary and uk, k 2 1, is defined b,y(17.2), then

,F,(-m; n + 1; -x) ,F,(-m - n; I! - n; x)

=

1

+

z

uk(~(-

2m - n - k)h(--2x)k

k=l

k!

(20.1)

62

11. Hypergeometric Series, II

PROOF. Using (17.3), we find that the coefficient of x’, r 2 1, on the left side of (20.1) is equal to

d .= i (-m),-k(-l)l-k(-m

- n)k

” k=(, (n + l)r-k(l - k)!(l - n),k!

= (- l)‘(-4

r (-r)k(-m

(n + l),r! kg0 (m-r+

- n)k(-n - 7gk l),(l -n),k! ’

Apply Dixon’s theorem (17.4) with a = -n - r, b = -r, and c = -m - n and use (17.6) to get

d,=--

I(l - n)r(*(2 - n - r))

(- l)‘r(*(2 + 2m + n + r))

(n + l),r(l - n - r)r(+(2 - n + r)j r(+(2 + 2m + n - r))

x (-m),Vm + 1 - 4 r!T(m + 1)

= 2’cr,(+( -2m - n - r))rr!C. -1)

This completes t he proof.

q

Example 1

a, c--

X3k

cm -(=-X)3k

k=,, (3k)! k=O (3k)!

PR~OF. In Entry 16, let m = -4 and n = -3 and replace x by (~/3)~. Then (16.1) becomes

(- l)k3kXGk =l+f
k=l (2k - 1)!(3k)2.5...(6k - 1)1.4...(6k - 2)

=1+g2 m (- l)k33kX6k k 1 (6k)! ’

from which the desired result follows.

q

Example 2

11. Hypergeometric Series, II

63

PROOF.Putting m = n = 0 in Entry 16yields

,F,(l, 1; x) oF2(1, 1; -x) = f !s!Fk + 1)kx2k. k=,, (k!)“((2k)!)2

The desiredequality readily follows.

0

Example 2 is mentioned by Hardy in his book [9, p. 71 and is found in Ramanujan’s letters [16, p. xxvi] to Hardy.

Example 3

X3k+l

f (- l)kx=+i

k=O(3k + l)! k=O (3k + l)!

= 2 m (- l)k(3X2)3k+’ j&- (6k + 2)! .

PROOF.In Entry 16,set m = 3 and n = -s and replace x by (x/3)j. We then find that

(- l)k(2k + l),3kX6k k=O(3k + 1)!4.7...(6k + 1)2.5...(6k - 1)

=f

(- l)k3kX6k

k=O(3k + 1)(2k)!4.7...(6k + 1)2.5...(6k - 1)

= ; kzo ‘-“+;;!x”k

On multiplying both sidesabove by x2 we complete the proof.

0

Example 4. COSx cash x = Cm ( - l)k(2XZ)2k

k=O

(4k)! .

PROOF.In (18.1), set y = $ and replace x by x2/4. After somesimplification,

the desired result follows.

0

Example 5. sin x sinh x = 1‘X2(- l)k(2X2)2k+’ k=o (4k + 2)!

PROOF.In (18.l), let y = 3 and replace x by x2/4.

CI

kc=,,(k(-!)x22()2kk)! ’

PROOF.Set y = 1 in (18.1).

0

Example 7

,F,($; 1; x) ,F&; 1; -x) = tF,($; 1, 1; x2/4).

64

11. Hypergeometric Series, II

PROOF. Set fi = -i and y = 1 in Entry 18.

0

Example 8

,F,(l; 3;$2) ,F,(l; 5; -x/2) = kzo(2k(fy;;2; l)r.

PROOF. Apply Entry 18 with fi = - 1, y = 3, and x replaced by x/2 to obtain

,F,(l; 3; x/2) ,F,(l; 3; -x/2) = ,F& +; $,$, 2; x2/16).

An elementary calculation shows that

-ZZZ1
<4>k@k

26k(2k)!
(4k + l)! .

The proposed equality now follows.

Cl

Example 9
,~,(l; n + 1;x) IFlu; n + 1; -x) = kzo(n k;;-k+ 1)
2k
= &(l, n; n + 1, +(n + l), +(n + 2); x2/4).

PROOF. Set j = - 1 and y = n + 1 in Entry 18.

0

Example 10. If n is a nonnegative integer, then

2Fo(-n, 1; x) 2F,,(-n, 1; -x) = f (-n)k(-n + 1 + k),x2k.
k=O

PROOF. In Entry 19, let M = n and /I = - 1. The proposed equality easily

follows.

cl

Entry 21

,F,(m, n; +(VI + n + 1); +(l + x))

&r($(m + n + 1)) 2F1($m, +n; 4; x2) r(+(m + l))r(+(n + 1))

+ _2Jkr(+(m + n +
r(fm)r($n)

1))~ 2Fl(+(m

+ l), +(n +

1); 3; x2).

Entry 21 is originally due to Kummer [l, p. 821, [2, p. 1181. See also Erdélyi’s compendium [ 1, p. 111, formula (3)].

Entry 22. Let m be a nonpositive integer and put p=+m(m- 1).

(22.1)

11. Hypergeometric Series, II

65

For each nonnegative integer k, let

A, = pk - 3kr(k - lJpkml + k(k - 116 - ‘Wk - l)pk-2 5!

. . . + 2(k - 1)!(22k- l)B,,

+

(22.2)

1.3.5...(2k - 1) ”

where Bj, 0 I j < 00, denotes the jth Bernoulli number. Then, if 1x1 < 71,

emmxkg *(l

- e-2x)k = 1 + z1 $$$.

(22.3)

Before commencing the proof of Entry 22, we make one comment. We have stated Entry 22 exactly as Ramanujan gives it. Note that, by (22.2),A, is not well defined becausethere apparently is no general formula for the coefficient of pj, 1 < j I k. However, A, is well delïned by a recursion formula given by (22.13) below.
PROOF.Replacing m by -n and x by ix in (22.3), we rewrite (22.3)in the form

We show first that f,(x) = P,,(cos x), n 2 0, where P, (denotesthe nth Legendre polynomial. (This fact was lïrst kindly pointed out to us by R. J. Evans.) By Bailey’s book [4, p. 41,

f,(x) = einx 2F1(-n, i; 1; 1 - eë2iX)

r(n + 3) einx 2F1(-n, f; -YI + 4; eë2iX). = r(n + l)fi

(22.5)

By using (17.3), we may easily show that

(-n),-k(t)n-k

_ (-n)k(i)k

(-n + $)n-k(n - k)! (-n ++)kk!’

Hence, from (22.9,

f.(x) = 2 r(n + i)
r(n + l)fi

[n’21 (-n)k(&)k
c

> cLostn _ 2kjx

k=o (-n + +)kk!

(22.6)

wherethe prime on the summation signindicates that if k = n/2, this summand is to be multiplied by 3. From a representation for P”(C~S x) in Whittaker and Watson5 text [l, p. 3031, it follows that f.(x) = P”(C~S x). Hence, it remains to show that

m (- l)kAk~2k PJCOS x) = 1 + 1
k=l 2k(k!)2 ’

(XI < 71.

(22.7)

66

11. HypergeometricSeriesI,I

It is well known (e.g., seeCopson’s text [2, p. 2731) that P”(C~Sx) is a solution of Legendre’s differential equation

y” + (cet x)y’ + n(n + 1)y = 0.

(22.8)

Since P,(l) = 1 (Whittaker and Watson [l, p. 3021)and P,(cos x) is an even function of x, P,,(cosx) has a power seriesexpansion of the form

P,(cos x) = T a2kx2k,
k=O

a, = 1.

(22.9)

Our procedure Will be as follows. We shall actually assumethat (22.7) holds; that is, we assumethat

a2k = (-- lJkAk
2k(k!)2 ’

k2 1,

(22.10)

and then we show that A, has the properties evinced by the formula (22.2). Recall that

as (- l)kB2,22kx2k-’

cet x = ,y

k=O

(2k)!

’

1x1 < 7c.

(22.11)

Substituting (22.9) and (22.11)into (22.8),we fïnd that

‘f a2,2k(2k - 1)~~~~~+ f q$:““‘*-’

k=l

k=O

kg1a2k2kx2k-’

+ n(n + 1) f aZkxZk= 0.
k=O
Equating coefficients of x~‘-~, r 2 1, on both sides,we find that

(W2a2,

r-1 (-
+ 1 k=l

1)kB2k22k(2r

- 2kh-2,
(2k)!

+ +

+ lla2 _ _ o. r2-

t22 12J

Noting, from (22.1), that p = $(n + 1) and using (22.10), we find, after some simplification, that the recursion relation (22.12)takes the form

r-1 23k-’ {(r - 1)!)2&,&k
Ar + kgl (r - k)!(r _ k - l)l.(2k)r. - pAr-1 = O7

(22.13)

wherer2 1 andA, = 1. First, letting r = 1 in (22.13),we lïnd that

A, =p.

(22.14)

Second,letting r = 2, using (22.14),and recalling that B, = i, we find that

A, = p2 - $p.

(22.15)

Third, letting r = 3, using (22.14) and (22.15), and recalling that B4 = --A, we find that

A, = p3 - p2 + $p.

(22.16)

11. H ypergeometric Series, II

67

Observe that the formulas for A,, A,, and A, given by (22.14)-(22.16) respectively, are in complete agreement with the formula for A, given by (22.2). In particular, the coefficient of p in (22.2) is in corroboration with (22.14)-(22.16) for k = 1, 2, 3.
We now proceed by induction and assumethat, for k = 1, 2, . . . , r, the leading three coefficients and the last coefficient of A, are in agreement with those prescribed in the formula (22.2). Thus, from (22.13) and the inductive hypothesis,
23k-‘(r!)2&,A,+,-, A ,+1 = pA’ - k$l (y + 1 - k)!(r - k)‘.(2k)f .
= pA, - 3rA’ + &r’(r - l)A’-,

+ . . - ~3’-‘v)2B2’
(2r)! lJ

= (p - $r) p’ - 6r(r - l)p,-l + r(r - l)(r - 2)(3r - l)pr-2

i

5!

+ . . + 2(r - 1)!(22’ - l)B,’ 1.3...(2r - 1) ’ 1

+ $r2(r

2(r - 2)!(2’!‘-2 - l)B,‘-,

- 1) p’-’ + ... + i

1.3...(2r - 3) ’ i

+ ... _ ~3’-‘(r92B2r
(2r)! ‘.

The coefficient of p’+’ above is equal to 1 in agreement with (22.2). The coefficient of p’ above is equal to

r r(r - 1) (r + 1)r

3

6

-7

which also agreeswith (22.2).The coefficient of p’-’ above is found to be

r2(r - 1) rJ - l)(r - 2)(3r - 1) 2r2(r - 1) (r + l)r(r - 1)(3r + 2)

18 +

5!

+ 45 =

5!

>

which again is what we desireby (22.2). Lastly, the coefficient of p above is equal to

c’ 23k-‘(r!)2B2k(r - k)!2’+2-k(r + 1 - k)!(22’f2-2k - l)B,‘+,-,,

c,:= -

k=l

(r + 1 - k)!(r - k)!(2k)!(2r + 2 - 2k)!

’ 2’+2k(22’+2-2k- 1)B2kB2’+2-2k

= -2(r!)2 C

k=l

(2k)!(2r + 2 - 2k)!

.

(22.17)

Recalling the Laurent expansions for coth(2x) and tanh x, we have, for 1x1< 742,

68

11. HypergeometricSeriesI,I

m 24k-‘B

X2k-1

coth(2x) tanh x = c

k=O

(2k;!

Jl 22k(22k,$!kx2k-’

The coefficient of xzr, r 2 0, on the right sideis equal to

d, := 2’+l i; 2r+2k(22r+2-2k

- 1)bc&r+2-zk

k=O

(2k)!(2r + 2 - 2k)!

= 2r+l

c,

2*(22r+2 - lV32r+2

-2(r!)Z+

(2r + 2)!

’

by (22.17). On the other hand, for (xl < 742,

coth(2.x) tanh x = 1 - 3 sech2x

(22.18)

= 1 -i&tanhx

=1-C- m 22k-i(22k - 1)(2k - 1)B2,xZkP2

k=l

(2k)!

Hence, we have also found that, for r 2 1,

d,= - 22r+1(22r+2- 1)P + 1V32r+2

(2r + 2)!

’

(22.19)

Equating (22.18)and (22.19)and solving for c,, we find that

2’+2r!(r + 1)!(22’+2 - 1)B2,+2 c, =
(2r + 2)! -’

r2 1.

Examining the coefficient of p in (22.2) when k = r + 1, we find that this

coefficient is indeed equal to c,. This completesthe inductive proof.

0

Corollary. Zf p = 1, then

A,=A,(I)=~,

k>l;

(22.20)

if p = 3, then

A, = A,(3) = 3. 22k-2Ak(1), k> 1.

(22.21)

PROOF. In Entry 22, let m = - 1. Then by (22.1), p = 1. From (22.6), Pi(c~s x) = COSx. Thus, the coefficient of xZk, k 2 1, in Pi(c~s x) is equal to (- l)k/(2k)!. But from (22.7), the coefficient of xZk is also equal to (- l)kAk/{ (2k(k!)2}, k 2 1. Equating thesetwo coefficients, we deduce(22.20).
Second, let m = - 2 in Entry 22. Then p = 3. From (22.6),
P,(cos x) = 2 cos(2x) + a.
Thus, the coefficient of x2k, k > 1, in P,(cos x) is equal to 3( - l)k22k-2/(2k)!. Equating this with the coeflï&t of xZk given by (22.7), we deduce (22.21).
Cl

11. Hypergeometric Series, II

69

If we expand COS(~ - 2k)x, 0 I k I [n/2], in its Maclaurin series in (22.6) and, for P,,(cos x), equate the coefficient of x2! j 2 1, with the coefficient of x2j in (22.7) we obtain an elegant identity involving binomial coefficients. We shah further separate this identity into two cases. Replacing n by 2n and then n by 2n + 1, we find, respectively, that
(22.22)

where n, j 2 1 and p = n(2n + l), and that

‘; 1 ;v;)(2;

1 tk)(2k + 1j2’ = 2,“-‘+‘(2$R,(p),

(22.23)

where n 2 0, j 2 1, and p = (n + 1)(2n + 1). These identities are apparently new and cannot be found in the tables of Goulcl [l] or Hansen [l], for example.
Entry 23 is apparently meaningless. Ramanujan claims that if
q(x) = Cl + Jr = c2 + Jz = ... ==c, + &
and “if ci, c2, cg, . . ., c, appear to be similar,” thlen they are a11identically equal to c. He then concludes that
cp(x)=c+Ji+Jz+-$3.
The intent of this entry shall perhaps always remain a mystery.

Entry 24. Let

<p(r) =

Pkrn-k

and

QI = i. v,(r

+ k).

Then

kro cp(W

x)~ = z Qk(-~)k
k=O

. (24.1)

Corollary 1. Let q(r) be defined as above and let

Q: = r(m

r(m + 1)

f ---t-rncp+(ml)k

+ 1 - r)r(r

+ 1) k=O (m + 1 - r)k

Then

+ k).

70

11. Hypergeometric Series, II

kgo<p(m+ k)(l - x)~+~= k$oQk(-x)~

1-2 II+1
+ (LogL-)

k=

X

Entry 24 and Corollary 1 are enigmatic. It seemslikely that there are no functions <pfor which either of the proposed identities holds. For most choices of cp,the seriesfor Q, and Q: diverge. Employing the definition of Qj, we formally lïnd that

51Qj(-x)' = F $ ~p(j+ k)

j=O

j=O k=O

= z. d4 j. (;)ix)i

= nzo cp(n)(l

- x)“.

Comparing the formula above with (24.1) we fmd that the logarithmic series does not appear!

Corollary2. Letsr+fi+y+l=6+.swithy>

-l.ThenasxtendstoO+,

T(cY+ l)I-(p + l)l-(y + 1) lx + 1, fi + 1, y + 1

r(61)I-(E

+ 1) 3F2 6+1,s+1

;l-X

1

w -Log .y - l& + 1) - $(S + 1) - 2c + k$I (r+-l;$; where 4(z) = T’(z)/T(z) and C denotes Euler’s constant.

& 2
k

We cannot seehow Corollary 2 would follow from Entry 24. Corollary 2 should be compared with the more precise formula for ZFl in Entry 26 below. Corollary 2 is a very beautiful and signilïcant formula, for it is the only asymptotic formula for zero-balanced seriesbesidesthat which cari be obtained from Entry 26. R. J. Evans and D. Stanton [l] have recently found an elegant proof of Corollary 2 as well as of a q-analogue. They provide a complete proof of the q-analogue and sketch a proof of Corollary 2. In fact, they establish a slightly stronger version of Corollary 2. We follow Evans and Stanton in our development below. It Will be convenient to trivially alter the notation of Corollary 2 above.

Theorem 1. Zf a + b + c = d + e and Re c > 0, then

(24.2)

11. Hypergeometric Series, II

71

where

L = -3) - $@) - $(b) + f (d-C**,

k=l

(‘&(bhk

wherey denotesEuler’s constant. Furthermore, asm tends to CO,

(22.4)

wherethe implied constant dependson a, b, c, d, and e but not on m.

If c = e, then (24.4) reduces to the following asymptotic expansion for a partial sum of a zero-balance zFl series(e.g., see Luke’s book [l, p. 109, Eq. (34)l):

&{Log

m - y - $(a) -- $(b)} + o(k), (24.5)

asm tends to 00.A slightly lesspreciseversion of (24.5)is given by Ramanujan in Entry 15 of Chapter 10. It would be interesting if there existed a theorem for zero-balanced q+lFqseriesthat included (24.4) a:nd(24.5) asspecial cases.
Theorem 2 below is a slightly more precise thcorem than Ramanujan’s Corollary 2 given above.

Theorem 2. If a + b + c = d + e and Re c > 0, then as x tends to 1 with

WW3[2adbe,c,,;x1 O<x<l,
WI-(b)W

= -Log(l - x) + L + O((1 - x) Log(1 - x)),

(24.6)

where L is defined by (24.3).

In the sequal, we shah deduce Theorem 2 from Theorem 1. In order to establishTheorems 1 and 2, we shall needfour lemmas.

Lemma 1. If Re C > 0, S = D + E - A - B - C, and Re S > 0, then

AB,C, 3F2 [ 1 1 D, E

wvw-(s)

-D - C, E - C, S

= r(c)r(A + S)T(B + S) 3F2_ A + s, B + s .

Lemma 1 is a reformulation of Entry 27 in Chapter 10.

Lemma 2. Zf a and d are bounded,then asz tendsto 00 with Re z > 0,

T(a + Z)

-~

= z”-d(1 + O(l/z)).

T(d + z)

Lemma 2, of course, is an easy consequenceof Stirling’s formula for the gamma function, which cari be found in Entry 23 of Chapter 7.

72

11. Hypergeometric Series, II

Lemma 3. Let E > 0 befixed and let a complex number E befixed. Let Re z 2 E and suppose that k is any positive integer. Then there exists a constant N > 0 such that

(l+$boL),

(24.7)

where the implied constant is independent of z and k.

PROOF. Let F - Re E. If F 2 0, then, since Re ,Z2 E,

Hence, it suflïces to consider the case F 2 0. Let N = F + 1. First, suppose that k 5 (z(. Then

Thus, (24.7) easily follows. Finally, suppose that k > IzJ. Then

where the last series does indeed converge because F 2 0. This completes the

proof.

II!

Lemma 4. Let Re D be fixed, where D is not a nonpositive integer. Let k be any positive integer, and suppose that Re z 2 0. Then

t-D--= - ‘)k (D)k

0(~2nlz1/3) 2

where the implied constant is independent of z and k.

PROOF. For some constant N > 0 that is independent of z and k,

= Z ‘-D+j
k-l
<C(l + Izl)N n
j=O D+jZ 1
=(1+IzUNDn;;;l(1-2Re(Df;i) +I&l’>‘i

11. Hypergeometric Series, II

13

<< (1 + lzl)”

I-I D;l1

(1

+ I&i’>”

<<(l +(zl)N fi 1 +g lP

( m=l

>

(&;L?fi)‘.

= (1 + lzl)”

<< (1 + ~Z~)Ne~‘z”Z<< eZn’z”3.

PROOF OF THEOREM 1. By Lemma 2,
T(a + k)l-(b + k)l-(c + k) -~ 1 r(d + k)r(e + k)T(l + k) k + 1
as m tends to 00. Also, from Ayoub’s text [l, p. 433,
m-l CL=
k=o k + 1
as m tends to CO. Thus, it is readily seen that (24.4) follows from (24.2). It remains to prove (24.2).
We first prove (24.2) for c = 1. Then, inducting on c, we prove (24.2) for each positive integer c. Lastly, we establish (24.2) for a11c with Re c > 0.
For each E > 0, Write

= H, - H;t,

(24.8)

where

[ab,1,1 HI = 3F2 d, e + E

and
GMb)m r 1 l,b+m,a+m
H2 = (d),(e + E), 3F2 Ld + m, e + (s + m ’

upon a change of index of summation. By Lemma 1,

1 H = W)r(e + 434

d - 1, e -t E - 1, E

1 r(a + q-(b + E)3F2 a+&,b+&

and

H = JYd)r(e 2 r(U)r(b)r(l

+ m-)r(b + 4 + &)r(b + m + E) 3F2

1 d -- a, e - a + E, E

l+&,b+m+E

.

Thus, we may Write

H, - H, = G, + G, + G,,

(24.9)

where

74

11. Hypergeometric Series, II

G = lim fm-le + 4w

r(d)r(e

1 E+O ( r(a + E)r(b + E) - ~r(a)r(b)r(l

+ E)r(E)r(b + m) + E)r(b + m + E)>

= r(d)r(e) lim i+r’(l)+E+Oci

L-ii r(a)

Y(4 r2(J + “’ i

1 x T(b)

T’(b)

- .~~ 1

P(b)& + ...

umw

1 J-‘@+M+ T(b + m) ...

_ ww ww

r’(a) -‘-r(a)--

T’(b) + f’(b + m) l-(b) T(b + m) ’

(24.10)

G = ,im gW(e

+ 4W

.f Cd - lk(e + e - l)k(4k

2 E+O r(a + E)r(b + E) k=l (a + &(b + &k!

_ WW wr(b)
and

m Cd- l),(e - llk c k=l (de(b), k ’

(24.11)

G = lim _ Wm

+ 4wr(b

+ 4 _ f. Cd - u)k(e - U + &)k(&)k

3 Ed0 r(U)r(b)r(l

+ &)r(b i- PI -t E) kcl (1 + E)k(b + m + E)kk!

_ r(d)w

.f Cd - u)k(e - dk

r(U)r(b)

k=l (l)k(b + m)kk-'

Since(Luke Cl, p. 33, Eq. (8)]),

T’(b + m) T(b + m) asm tends to CO,we find from (24.10)and (24.12)that, respectively,

(24.12)

GI=%{--y-$(a)-$(b)+Logm}+O($) and

(24.13)

(24.14)

as m tends to CO. Putting (24.11),(24.13), and (24.14)in (24.9) and then (24.9) into (24.Q we
conclude that we have established(24.4)for c = 1. Assuming that (24.4)holds with c replaced by 1,2, . . . , c - 1,we examine

mg1~(u-)k(b)k(C)k k=~ (d)de),k!

_ (d - lb
(b -l)(c-

- 1) mi1 (4k(b

- l)k+l@

- l)k+l

k=~ (d - l),+r(e - l)k+lk!

11. HypergeometricSeriesI,I

75

(d - l)(e - 1) m VJ- l)k(C - l)k (4k = (b - l)(c - 1) c (d - l),(e - l),(k - l)! k
= (d - l)@ - 1) .f (4k@ - lJk@- llk
(b - l)(c - 1) k=O (d - l),(e - l),k!
(d - l)(e - 1) t (a - l),(b - l)k(~ - l)k - (b - l)(c - 1) k=e (d - l),(e - l),k!

b - l)k k

(d - l)(e - 1) Oo (a - l)k@ - l)k@ - ‘)k + o i

-(b- l)(C- l)k& (d- l)k(e-l)kk!

0 m'

as m tends to cc. Using again Lemma 1, we deduce that

mg1 (“)k(b)k(C)k
k=O (d)k(e)kk!
r(d) W = r(a)r(b)r(c)

Log m- Y- Il/(4 - W) + gq

+ f (d - dkte - dk -~

1

f @ - dkte - dk

k=l (a),@ - l)kk

b - 1 f=O (‘&(bh

= r”“c>

Log m - Y - $64 - W) {

+ -f (d - dkte

k=l

h)k

- dk

1

-~- 1

( (b - l),k (b - l)(b), 11

Lw in - Y - VW - W) + kzl (d juf’;b, ; ‘jk .
k
Thus, (24.4)hasbeen establishedfor each positive integer c. Letting m tend to cc in (24.4) and recalling the opening paragraph of this proof, we conclude that (24.2)holds for each positive integer c.
TO prove that (24.2)is valid for a11c with Re c > 0, it sufficesby Carlson’s theorem (Bailey [4, p. 391)to prove that, for a, b, d, and E > 0 Iïxed, both sides of (24.2)are analytic in c and equal to O(e2n’cti3f)or Re c 2 E.
Let D = Re(d - E),with d adjusted, if necessary,SOthat D is not a nonpositive integer. Let z = c + D - d. Thus,

where

s .= c (d - dkte

k=l

(d&ck

- dk

A = (a + b - dhc(Dh

k

(~h@hk

’

k;, 1.

16

11. HypergeometricSeriesI,I

By Lemma 2, A, = O(k-‘-“), while by Lemma 4, (D - .z)J(D)~ = O(e2nlzl’3). Thus, S is analytic in z and equals O(e2n1z1/f3o)r Re z 2 0. It follows that S is analytic in c and equal to O(e2n’c1’f3o)r Re c 2 E.
It remains to prove that

T:= f

I-(a + k)T(b + k)I-(c + k)

-~ 1

k=1 r(1 + k)T(d + k)T(a + b - d + c + k) k + 1

is analytic in c and equal to O(e2n1c113)foRret>s. Let E=d-a-b.

By

Lemma 2, since Re c > E,

1 T = $l k-E-l(c + k)E(l + k-‘O(1)) - ~k+l

= zl k-’ {( 1 + ;y - 1) (1 + k-‘O(1)) + O(l),

where the expressions O(1) are bounded analytic functions of c for Re c 2 E.

By Lemma 3, (1 + c/k)E - 1 = O(cN/k) for somepositive constant N. Thus, T

is analytic in c and equals O(C~) for Re c 2 E.This then completes the proof

of Theorem 1.

0

PROOF OF THEOREM 2. Delïne 1-(a + k)T(b + k)I-(c + k)
f(k) = I-(d + k)r(e + k)T(i + k)
and

V(x) = -f f(k)xk + Log(1 - x) - L,
k=O
where 0 < x < 1 and L is defined by (24.3).We must show that
V(x) = O((1 - x) Log(1 - x)),
as x tends to 1. By (24.2),

(24.15)

Now, by Lemma 2, f(k) - &

(24.16) = (1 - x) $i k-2 ‘2 x”
n=O

11. Hypergeometric Series, II

II

=(1- x)$, xn,=,+K, 2

C(l -x){;;+$l;}

« (1 - x) Log( 1 - x).

Using this in (24.16), we complete the proof of (24.15) and SOalso that of

Theorem 2.

0

The special casec = e of Theorem 2 gives an asymptotic expansion of a zero-balanced ,F, as x tends to 1-. This special case is also an easy consequenceof Entry 26 below. Moreover, it is equivalent to (24.5).
For further remarks on Theorems 1 and 2 aswell asq-analogues,cons& the paper of Evans and Stanton Cl], A generalization of Theorem 2 has recently been establishedby Bühring [1] who usesthe differential equation satisfiedby 3F2. His proof has the advantage that the form of the asymptotic formula doesnot have to be known in advance. BecauseRamanujan showed little interest in differential equations, he likely had yet a different proof.

Entry 25. Suppose that n is not an integer. Then

a+n+l,b+n+l

2 F 1

a+b+n+2

1 ;l-x

=- T(a + b + n + 2)l--n)

a+n+l,b+n+l

T(a + l)T(b + 1) 2F1

n+l

r(a + b + n + 2)r(n)x?

a+l,b+l

+l-(a

+ n+

l)r(b

+ n+

1) 2F1

-n+l

1;x 1;x

Entry 25 is a basicformula for the analytic continuation of hypergeometric seriesand cari be found in the treatises of Bailey 114p, . 43 and Erdélyi [1, p. 108,formula (l)].

Corollary 1. If n is a nonnegative integer, then

a+n+l,b+n+l

2F 1

a+b+n+2

1 ;l-x

T(a + b + n + 2)r(n)x-” ng (a + l),(b + l),xk r(a + n + l)r(b + n + 1) k=O (-n + l),k!

(- l)“E(a + b + n + 2) f (a + n + l),(b + n + 1)k

- T(a + l)T(b + l)T(n + 1) k=O

(n + l),k!

x {$(a + n + k + 1) + $(b + n + k + 1) - $(n + k + 1) - +(k + 1) + Log X}X~,

78

11. Hypergeometric Series, II

where $(z) = Y(z)/T(z). Zf n = 0, the first expression on the right side above is understood to be equal to 0.

Corollary 1 cari be found in Erdélyi’s synopsis [l, p. 110, formula (14)].

Corollary 2. If n is a nonpositive integer, then

a+n+l,b+n+l 2F 1 a+b+n+2

1 ;l-x

= F(a + b + n + 2)r(-n) -n-1 (a + n + l),(b + n + l)k~k

r(a + l)r(b + 1) & -

(n + l),k!

r(a + b + n + 2)(-x)-”

f (a + l)k@ + l)k

- T(a .+ n + l)I(b + n + i)r(i - n) k=O (1 - n),k!

x {$(a + k + 1) + $(b + k + 1)

- $(k - n + 1) - $(k + 1) + Log X}X~.

If n = 0, we employ the same convention as in Corollary 1. Corollary 2 is a reformulation of another formula in Erdélyi’s treatise
[l, p. 110, formula (12)].

Entry 26. We have

r(a + l)I(b + 1) 2Fl(a + 1, b + 1; a + b + 2; 1 - x)
T(a + b + 2)

+ Log x 2Fl(a + 1, b + 1; 1; x)

+ -f (a + l),(b + ljk

k=O

WI2

($(a + k + 1) + $(b + k + 1)

- 2$(k + 1))~~ = 0.

Entry 26 is simply the case n = 0 of either Corollary 1 or Corollary 2 above. Ramanujan has given a less precise version of Entry 26 in Chapter 10 (Section 15).
Corollary

ir2F1(+,+;1;1-x)=Log

; 2F1($,&1;x) 0

-4ce(+i

’ Xk

k=l (k!)2 j=I (2j - 1)(2j) ’

PR~OF. Putting a = b = -4 in Entry 26 and using familiar formulas for Il/(k + 1) and $(k + f) (Gradshteyn and Ryzhik [l, p. 945]), we find that

11. Hypergeometric Series, II

79

= - Log x 2~,($, 3; 1; x) - 2 k$ $$($(k + $1 - $6 + Wk

(26.1)

which completes the proof.

Cl

Example. Zf 0 < x < 1, then

42 7G tan(cp/2) d8 dq

=- 71 n/2

s 0 s 0 J 1 - x cos2 8 cos2 cp 4 s 0 J1--

+iLogx

dv (1 - x) sin2 cp
‘n/2
0 J&.

(26.2)

PROOF. First, for 1x1< 1,

sozopsin@Zlkq n’2 m (+>k k

= ~o~x&ki&!

xk = ; ;:F,(& 3; 1; x).

(26.3)

Second, for Il - XI < 1,

dv

=- ; 2F,(+, +; 1; 1 - x).

1 - (1 - x) sin2 cp

(26.4)

Third, using an integral evaluation in Gradshteyn and Ryzhik’s tables [ 1,p. 3761and the calculation (26.1), we lïnd that, for 1x1-C 1,

tan(cp/2) dtl dtp

1 - x COS20 COS2cp

= m &kxk
I<=- =Ok!

7G
X/2 tan(cp/2) cosZkq dcp cos2kede

s 0

s 0

80

11. Hypergeometric Series, II

2) $,(+, 3; 1; x). (26.5)

Using (26.3)-(26.5), we tïnd that (26.2) is equivalent to the identity
-zz.$$i Xk + ;(Log 2) 2F,(3, 3; 1; x) (2j -l1)(2j)

=~,F,(f,~;1;1-x)+~(L0gx),F,(i,i;1;x),

where 0 < x < 1. This last identity follows from the foregoing corollary, and

SOthe proof is complete.

0

The integral in (26.3) is the complete elliptic integral of the lïrst kind, and the formula (26.3) is a basic, well-known result in the theory of elliptic functions. For further ramifications, see Section 6 of Chapter 17 in Part III [l 11.

Entry27. For 1x1< 1,

-Xk

= -a 2Fl(), 3; 1; X) Log(1 - x).

(27.1)

PROOF. For n 2 1, the coefficient of x” on the right side of (27.1) is equal to

where we have employed (17.3). It thus suffices to show that

n2 1.

Let S, denote the left side of (27.2) and rewrite S, in the form

n-1 ( - 4k+l xl= c
k=O (k + 1)(+ - n)k+l

=---- n2 n: (1 - n):(l): (3 - PI)’k=O(2 - n);(2)kk!

(27.3)

The right side of the equality above is a balanced 4F3 and SOcari be trans-

formed by (6.3) in Chapter 10. Let y = z = 1, x = -n, u = u = 3 - n, w = 2,

and m = n. Then

=(-3-Ml),l,+, -n -+, -n 1 1 (+ - n),(2), 4F3 3 - n, +, -n

11. Hypergeometric Series, II

81

= (2n + 1Y n

1

n + 1 kc=-rJ(2k + 1)(2n + 1 - 24

1

1

2k + 1 + 2n + 1 - 2k

= (2n + 1Y n 1 (n + 1)2 kcz-O2k + 1’

Replacing n by n - 1 above and using the result in (27.3), we complete the

proof of (27.2).

cl

The expression on the left side below is fundamental in the theory of elliptic functions. See Section 6 of Chapter 17 in Part III [l 11.

( Example 1 exp -7~ 2F,($, f; 1; 1 - x) 1 =$ ( I+;x+gx2+...

)

2Ei(3,3; 1; 4

( PROOF.By the corollary in Section 26, 2Fl(+, +; 1; 1 - x) exp -rc 2F1(4, t; 1; 4 >

from which the sought result follows. Example 2
PROOF. Putting a = - f and b = -3 in Entry 26, wtt lïnd that -~2F1(+,$;1;1-~)
J3 = Log x ,F,(3,$; 1; x) + &ODo(32>kG>k {$(k + 3) + ti(k + 4) - 21C/(k+ l)}Xk

82

11. Hypergeometric Series, II

{3$(3k) - t++(k) - 2t,h(k + 1))~~

where we have used the facts (Gradshteyn and Ryzhik [l, p. 945]), t,b($) +

$(4)-211/(l)= -3 Log 3 and $(k+$)+$(k+3)=3$(3k)-$(k)-3

Log 3,

for k 2 1. Hence,

=exp(Log(x).g)=$(l+iX+...).

•!

Example 3

PROOF. Putting a = -$ and b = -2 in Entry 26, we find that -J%c pl($, 4; 1; 1 - x) = Log x ZFl(%, 2; 1; x)

+ -f <--i>(k$<(?k)k k=,, (k!)’

+ +) + t,h(k + a) - 2$(k + 1))~~

x {411/(4k) - Il/(k) - ij(k + +> - 2 Log 2 - 2$(k + 1))~~

= Log 6x4 zF,(& 2; 1; x) + ;x + ..., 0

where we have used the facts (Gradshteyn and Ryzhik [l, p. 945]),

$(a) + $($) - 2$(l) = -6 Log 2 and $(k + b) + t,h(k + 2) = 4$(4k) - $(k) -

Il/(k + 4) - 8 Log 2. The proposed formula now easily follows.

0

Example 4

ev ( -2*ZF1(&,2; 1;1- 4 =& 1+:x+... zF,(i, 2; 1;x) > (

. >

PROOF. In Entry 26, put a = -i and b = -2 to lïnd that

11. Hypergeometric Series, II

83

-27~ ,Fl(& 2; 1; 1 - x) = Log x 2Fl(& 2; 1; x)

-(l)(k

+ 5) + bl/(k + 2) - 2$(k t- l)}Xk

= Log s 2Fl(&;; 1; x) ( 1

<b)k@>k + f
k=l (k!)2

6$(6k) - 3$(3k) - 2 i L- + y - 2$(k + 1) xk j=l 25 - 1

As in Examples t-3, we have employed familiar properties of $(z) (Gradshteyn

and Ryzhik [l, p. 9451). We also have used the fact that $(i) + $(s) -

2$(l) = -4 Log 2 - 3 Log 3, which cari be deduced from results in Chapter

8 of the second notebook. (See the author’s book [9, Chap. 8, Eq. (5.2) and

Corollary 3 in Sec. 63. See also Gradshteyn and Ryzhik’s tables [l, p. 944,

formula (7)].) The desired formula now readily follows.

0

We do not know Ramanujan’s intention in giving Examples l-4.

Entry 28. Let q denote a polynomial of degreem. Supposethat n is not an integer and that Re(a + b + m + n + 1) < 0. Then

r(a + l)T(b + l)r(n) f (a + iffn:kk:<p(k)
k=O

+ r(a + n + l)T(b + n + l)r(-n) 2 (a+
k=O

‘)~nb~l~ k!l)k’(n + k,
k

= r(a + n + l)T(b + n + i)r(a + l)T(b + 1) F (a + lh(b + l),AkV(0)

T(a + b + n + 2)

k=(O (a + b + n + 2),k! ’

PROOF.Since 1, x, x(x - l), . . . , x(x - l)... (x - m + 1) form a basisfor the

setof a11polynomials of degreem over the lïeld of cumplex numbers,it suffices

to prove the result for C~(X)= C~,,,(X:=) x(x - 1)... (x - m + 1). We first

observe that

0,

k < m,

s,(k) = m!,

k := m,

Next, since

i (- l)“‘(-k),,

k :> m.

j~ow(;)ir=o>Osr<k,

84

11. Hypergeometric Series, II

where r is an integer, we find that
Akq,,JO) = i (- l)j
j=O

k < m, k = m, k > m.

Thus, for q(x) = q,(x), the proposed identity may be written as

r(a

+ l)I-(b

+ l)I-(n)

cm (a +
k=m

lh@ + lh(-l)“Y-k), (1 - n),k!

+ T(U+ n+ i)r(b + n+ w-(-n)

x ,f (U+ Il+ l),(b+ n+ l)k(- l)“(-n - 4,

k=O

(n + l),k!

= r(a + n+ ip-(b+ n+ qr(a + ip-(b+ i)(~ + i),(b + l),(-l)mm! r(a + b+ n+ 2)(a + b + n + 2),m!
(28.1)

Let S, denote the lïrst sum on the left side of (28.1). Replacing k by k + m, employing Gauss’s theorem, Entry 8 of Chapter 10, and simplifying, we find that

s, = r(a + 1)r(b+ i)r(n) 1m(a+ l),+,@+ Qk+J- l)“(- k - m),

k=O

(1 - h+&

+ m)!

= r(a + i)r(b + i)r(n)(a+ l),(b+ l), m(a+ m+ l),(b+ m+ l)k

(1- 4,

c
k=O

(1 - n + m),k!

=

r(q-(a

+ m

+

i)r(b + m + l)I(m (1- n),r(-a

- n + i)r(-a - n)r(-b - TI)

b -

m

-

n -

1)

=- T(a + m + 1)IJb + m + i)r(a + n+ i)r(b+ n + 1) sin n(a + n) sin ~l(b + n)
T(a + b + m + n + 2) sin(7rn)sin a(a + b + M + n + 1) (28.2)

Let S, denote the expression on the right side of (28.1). Then

s = (- ipr(a + n+ ip-(b+ n+ i)r(a + m + i)r(b + m + 1) (28 3)

3

T(a + b + m + n + 2)

If S, denotes the second series on the left side of (28.1), then, by (28.2) and (28.3), we must show that

s = r-tu+ n+ i)r(b + n+ i)r(a + m + i)r(b + m + 1)

2

r(0 + b + m + n + 2)

x (-l)“+

sin rr(n f a) sin 7c(n+ b) si.n(7zn) sin 7c(u+ b + m + n + 1) ’

(28.4)

We shah prove (28.4) by inducting on m. For m = 0, (28.4) is valid by Entry

11. HypergeometricSeriesI,I

85

25, sinceEntry 28 reducesto Entry 25 for x = 1 whemq(x) = 1. Assumethen that (28.4)holds with m replaced by 0, 1, 2, . . . , m - 1.Observe that

<p,(n+ k) = (n + k)cp,-,(n - 1 -t- k).

Thus, we may Write

s, = I-(a + n + l)T(b + n + l)I-( - n)

x f (a + n + l),(b + n + l),cp,-,(n - 1 + k-)

k=O

(n + l),-,k!

= -I-(U + 1 + (n - 1) + i)r(b + i + (TI - 1) +- i)r(-(n

- 1))

x-J-m- (a + 1 + (n - 1) + l),(b + 1 + (n - 1) + l),cp,-,(n - 1 + k)

k=O

(“),k!

We now apply the induction hypothesis, but with a, b, and n replaced by a + 1, b + 1, and n - 1, respectively. Hence,

s = -r(a + n + i)r(b + n + I)r(a + m + i)r(b+ m+ 1)

2

r(u + b + m + n + 2)

x (-l)“-’ i

sin z(n + a) sin n(n + b) +
sin n(n - 1) sin rc(u+ b + m + n + 1) ’

from which (28.4)follows. This completes the proof.

0

Corollary. Assumethe hypothesesof Entry 28. Then

-r(a + i)r(b + 1) cm (a + l),(b + l),Akd'D)

r(U + b + 2) k=O

(a + b + 2),k! -

+ f (a + l)d’ + l)&“(k) + -f (a + l),(b + l)&‘(k)

k=O

(k!)2

k=O

(k!)2

x {$(a + 1 + k) + $(b + 1 + k) - 2$(k + l)} = 0.

PROOF. After somemanipulation, we Write Entry 28 in the form

cm T(u + 1 + k)I-(b + 1 + k)cp(k)

k=O

I(l - n + k)k!

-&m r(u + n + 1 + k)T(b + n + 1 +- k)cp(n+ k) lY(n + 1 + k)k!

- ~sin(7cn)r(a + n + l)T(b + n + 1) n

m I-(a + 1 + k)T(b + 1 + k)Akq(0) _ o ' kg0 F(a + b + n + 2 + k)k! - - .

Differentiating both sideswith respectto n and then setting n = 0, we lïnd that

86

11. Hypergeometric Series, II

cm r(a + 1 + QI-(b + 1 + k)$(k + l)cp(k)

k=O

(k!)’

_ kzoI-‘(a + 1 + W-ipz+ 1 + k)cp(k)

_ z. W + 1+ kV-‘-‘;+1+ k)cp(k)

_ z. r(a + 1+ 4;;; 1+ kW(k)

m I-(a + 1 + k)r(b + 1 + k)$(k + l)cp(k)

+c k=O

(k!)’

- r(a + 1)1-p+ 1)1m, F(a + 1 + k)l-(b + 1 + k)Akq(0) = o

k=O

1-(a + b + 2 + k)k!

After somemanipulation and simplification, the formula above reducesto the

proposed formula.

0

Entry 29(i). Zf Re(cc + fl + y - 6 - E), Re(6 - y - 1) < 0, then
%PY?r(s)r(s - tl - b) 4 P,E- Y [ 1 1 3F2 6,E = r(6 - g(6 - fi) aF2[ cr+/?-6+1,s
+ r(qr(q-(~ + p - y)r(d+ E- a- p - 7) r(q-(p)r(&- y)r(6+ E- C-( p)

x F s-u,s-p,s+&-a-p-y
3 2 [ 6-a-P+1,6+&-a-fi

1 *

Entry 29(i) wascommunicated by Ramanujan in his secondletter to Hardy [16, p. xxviii]. For a proof of Entry 29(i) and an illuminating discussion of this formula, seeHardy’s paper [l, pp. 498, 4991, [7, pp. 511, 5121. Another proof cari be found in Bailey’s tract [4, p. 211.

1 [ -4-B,Y+61 Entry 29(ii). Zf ~1f,i, or y is a nonnegutiveinteger, 3 F2 --2C4l-/3+-a$,S -Y =4 F3 -u-p+~&,~(s+-Y,

1) . (29.1)

PROOF. R. Askey and J. Wilson [l] have recently given a short proof of Entry

29(ii) when either a or /3 is a nonnegative integer. Now supposethat y is a

nonnegative integer. If we multiply both sidesof (29.1) by (-c1 - /3+ s),,

then on each side we obtain a polynomial in CIof degree y. These two

polynomials agreefor each nonnegative integer 01H. ence, they must beidenti-

cally equal, and this completes the proof.

cl

11. Hypergeometric Series, Il

87

If n is a nonnegative integer, detïne

-n,n+a+y-$,y+ix,y-ix

P&) = Pznk 4 Y) = (- 1)” d-3 [

M+ :‘, y, Y + f

l-

Thesepolynomials in x arisefrom the right sideof (29.1) by a renaming of the parameters. Askey and Wilson [1] have shown that {P,,(x)}, 0 < n < CO, is an orthogonal set on (-00, 00) with respect to the weight function Ir(a + ix)I(y + ~X)I’. As we pointed out in Chapter 10, the integral over (-a, GO)of this weight function was lïrst evaluated by Ramanujan [S], [ 16,p. 571.There also exists a setof similarly defined polynomials Pz,+i (x) of odd degree2n + 1 SOthat {P,,(x)}, 0 I n < 00, forms a complete orthogonal set on (-CO, cc) with respect to the aforementioned measure [l].

Entry 30. Let M+ ,8 + 1 = y + 6, c = r(a)r(fi)/{r(y)I(8)),

and

Then

c ,F,(& P; 6; 1 - 4 Y= 2Flk P; y; 4 .
x-y(l - x)-” y’ = - &(a, p; y; x)

PROOF.From Entry 25,

CA, ZFl(C(, p; y; x) + cA,x’-y ,F,(d - CI,6 - p; 2 - y; x) y=---
2Fl@, 0; y; x)

where A, and A, are constants with CA, = l/(y - 1).Thus,

yl=- ’ d x1-y 2F1(6 - CY6, - /?; 2 - y; x)

y-ldx

2F1(4 B; y; 4

1 = i:Y- ‘1 *c%, B; Y;4

1 KFlb,
= (Y - 1) ,G(% B; Y; 4 1
=:(y - 1) ,F:(a, p; y; x) IVx),

B; y; x), cy ,F,(6 - CI,6 - 8; 2 - y; x))

where W(f, g) = W(x) denotesthe Wronskian off(x) and g(x). Now thesetwo functions are linearly independent solutions of the hypergeometric differential equation (Bailey [4, p. 11)

x(1 - x)yU + {y - (CY+ fl + l)x}y’ - apy = 0.

(30.2)

88

11. Hypergeometric Series, II

By Abel% formula (e.g., see the text of Coddington

y - (m + p + 1)x

W(x) = C exp -

(S

x(1 - x)

[l, p. 113]), 1

=cexp(J(-~+f&)dx):cx-YLx)-~,

where C is a particular constant. Suppose that we Write W(x) = xmYF(x). Then C = F(0). If we perform the differentiation in (30.1), we readily Iïnd that C = 1 - y. Thus,

W(x) = -(y - 1)x-7(1 - x)-d,

(30.3)

and, by (30.1), the proposed formula for y’ follows.

0

Corollary. Let Then

7c 2Fl(n, 1 - n; 1; 1 - x) Y = sin(xn) 2Fl(n, 1 - n; 1; x) ’
1 ” = -x(1 - x) ,Ff(n, 1 - n; 1; x)’

PROOF. Apply Entry 30 with LY= n, fi = 1 - n, and y = 6 = 1.

0

Entry 31(i). Let y = ,F,(a, 8; y; x). Then

(a - l)(b - 1) Xydx - x(1 - x)y’ = (y - l)(y - 1) - (e + j - 1)xy. s 0

PROOF. Upon differentiation, it is found that the proposed formula is equivalent to the formula

(c! - l)(P - 1)y - (1 - x)y’ + XJJ - x(1 - x)y”

= (y - 1)y’ - (c! + fi - 1)y - (a + fi - 1)xy’.

Upon simplification, this formula reduces to (30.2), the hypergeometric differ-

ential equation satisfted by $i(cz, j?; y; x).

0

Entry 31(ii). Let a + /i’ + 1 = y + 6. Assume that n > 1 and that n > Re y. 1f Y = ~(-4 = zFl(~, B; Y; 4, then

y@1) ; {j; t”-“y(t),,} uY(l-d;)6y2(u)

1 x”-Y(l_ x)1-d n - cI, n - /?, 1

(n - r)(n - 1) 3F2 [ n,n-y+1

‘x ’

(31.1)