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\vskip 1cm{\LARGE\bf Beukers' integrals and Ap\'{e}ry's recurrences}
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\large
Lalit Jain  \\
Faculty of Mathematics \\
University of Waterloo  \\
Waterloo, Ontario N2L 3G1 \\
CANADA   \\
\href{mailto:lkjain@uwaterloo.}{\tt lkjain@uwaterloo.ca} \\
\ \\
Pavlos Tzermias \\
Department of Mathematics  \\
University of Tennessee  \\
Knoxville,  TN 37996-1300 \\
USA  \\
\href{mailto:tzermias@math.utk.edu}{\tt tzermias@math.utk.edu} \\
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\begin{abstract}
We give a new and completely elementary proof of the fact that the rational 
approximations to $\pi^2$ obtained by Ap\'{e}ry in his famous proof of  
the irrationality of certain values of the Riemann zeta function  
are identical to those obtained by Beukers  
in one of his alternative proofs of Ap\'{e}ry's result.   
\end{abstract}

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\bigskip
 

\section{Introduction}

Ap\'{e}ry's famous proof (\cite{A}, \cite{P}) of the irrationality of 
$\zeta(3)$ makes ingenious use of certain identities and specific 
recurrences. The proof gives explicit rational approximations to $\zeta(3)$ 
which 
converge fast enough to prove its irrationality. Ap\'{e}ry's 
proof also produces analogous rational approximations to $\zeta(2)=\pi^2/6$, 
whose irrationality (in fact, transcendence) is well-known. Specifically, 
in the case of $\zeta(2)$,  
Ap\'{e}ry considers the recurrence relation 
$$ (n+1)^2 u_{n+1} - (11 n^2+11 n +3) u_n - n^2 u_{n-1} = 0 . $$ 
Let $\{a_n\}$ be the sequence solving the above recurrence with 
initial values $a_0=0$ and $a_1=5$. Also let $\{b_n\}$ be the sequence 
solving the recurrence with initial values $b_0=1$ and $b_1=3$.    
Then the sequence $\{a_n/b_n\}$ converges to $\zeta(2)$. Explicit formulas for 
$a_n$ and $b_n$ are given in \cite{A} and \cite{P}. We also note that 
$\{b_n\}$ is the sequence A005258 in the {\it On-Line Encyclopedia of 
Integer Sequences}. 
Ap\'{e}ry also 
gives analogous arguments for $\zeta(3)$.  

Shortly after Ap\'{e}ry announced his proof, Beukers (\cite{B3}) 
produced an elegant and 
entirely different proof of the irrationality of $\zeta(2)$ and $\zeta(3)$. 
In the case of $\zeta(2)$, Beukers considers the double integrals $I_n$ defined by   
$$I_n = \int_{0}^{1} \int_{0}^{1} \frac{(1-y)^n P_n(x)}{1-xy} dx dy , $$ 
where $n \in {\mathbb N}$ and $P_n(x)$ is the Legendre-type polynomial given by 
$$ P_n(x) = \frac{1}{n!} \frac{d^n}{dx^n} (x^n (1-x)^n) . $$ 
He then shows that 
$$I_n =  \beta_n \zeta(2) - \alpha_n , $$ 
where $\alpha_n$, $\beta_n \in {\mathbb Q}$, for all $n$.  
An estimation of the latter linear 
form shows that it tends to 0 (as $n$ approaches infinity) fast enough to yield 
the irrationality of $\zeta(2)$. Beukers also gives an analogous argument for 
the case of $\zeta(3)$ by using a triple integral instead. 
It is worth noting that an even more striking 
proof of Ap\'{e}ry's result was given by Beukers in \cite{B1} using 
modular forms. 

It is rather remarkable (and apparently known, as explained below) 
that the rational approximations to $\zeta(2)$ and 
$\zeta(3)$ obtained by Beukers are identical to those obtained 
by Ap\'{e}ry. We list below some places we were able to find 
in the literature where proofs of this fact (or of related facts) are 
given:     

\noindent For   
the 1-dimensional 
analogue of Beukers' method (i.e., appropriate 
single-variable integrals), a related fact was 
established by Alladi and Robinson in \cite{AR}, using properties of 
values of Legendre polynomials. The method is also discussed independently by 
Beukers in \cite{B2}.  
For the 2-dimensional and 3-dimensional analogues of Beukers' method (i.e.,  
Beukers' double and triple integrals), the 
fact is verified by Nesterenko in \cite{N}, by using rather advanced 
arguments involving, among other things, contour integrals of Barnes type and 
transformation properties of 
hypergeometric series (see also the recent preprint by 
Zudilin (\cite{Z}). The reader may also consult the article by Fischler 
(\cite{F}) for a complete survey of the subject.             

The purpose of this paper is to give a new, short and 
completely elementary proof 
of the fact mentioned above for $\zeta(2)$.     
  
\begin{theorem}
With notation as above, we have 
$$ a_n = \alpha_n , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  b_n = \beta_n , $$ 
for all $n \in {\mathbb N}$. 
\end{theorem}

As the referee of an earlier version of this paper pointed out, another 
elementary proof of the same fact can be given by combining clever 
manipulations with Zeilberger's powerful program Ekhad. Our proof is  
along different lines. 

Our attempts to 
apply similar elementary arguments 
to the case of $\zeta(3)$ have invariably (and not surprisingly) 
led us to certain 
expressions involving    
special values of generalized hypergeometric series, which are notoriously 
difficult to compute. We do not address the case of $\zeta(3)$ 
any further in 
this paper. 

The above theorem easily implies a recurrence relation 
between special values of certain generalized hypergeometric series (see the 
corollary below). It is 
not unlikely that this may also follow from the contiguous relations of Kummer 
(which were generalized by Wilson in \cite{W}); we have not 
attempted to verify this. The reader may also 
consult the books  
by 
Andrews, Askey and Roy (\cite{AAR}) or by Magnus, 
Oberhettinger and Soni (\cite{MOS})  
for a wealth of information regarding special 
functions of hypergeometric type. 

If $a$ is a positive integer, 
let $\sb 3 F \sb 2 (a, a , a ; 2a, 2a ; 1)$ denote
the value of the generalized hypergeometric series 
$$\sb 3 F \sb 2 (a, a, a ; 2a, 2a ; x) = 1 +\sum_{k=1}^{\infty} \frac{(a\ldots (a+k-1))^3}{((2a)\ldots (2a+k-1))^2} \frac{x^k}{k!} $$ 
at $x=1$. Then 
         
\begin{corollary}
For every integer $a$ such that $a \geq 2$, we have
$$ \sb 3 F \sb 2 (a+1, a+1, a+1 ; 2a+2, 2a+2 ; 1) $$ 
$$= - \frac{176 a^4-84 a^2+4a+12}{a^4} \ {\sb 3 F \sb 2 (a, a, a ; 2a, 2a ; 1)} + $$ 
$$ + \frac{256 a^4 - 128 a^2 +16}{a^4} \ {\sb 3 F \sb 2 (a-1, a-1, a-1 ; 2a-2, 2a-2 ; 1)} .  $$   
\end{corollary}

\bigskip
 
\section{The Proof}

We first point out that some of the integrals below are improper; 
their use can be justified by replacing $\int_{0}^{1}$ by 
$\int_{\epsilon}^{1-\epsilon}$ and letting $\epsilon$ tend to 0. Also, in what 
follows, our manipulations of the series involved are valid because of their 
absolute and/or uniform convergence. 

First note that 
$$I_0 = \int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} dx dy = \sum_{k=0}^{\infty} \int_{0}^{1} \int_{0}^{1} x^k y^k dx dy = \zeta(2), $$ 
$$I_1 = \int_{0}^{1} \int_{0}^{1} \frac{(1-y)(1-2x)}{1-xy} dx dy  $$ 
$$=\sum_{k=0}^{\infty} \int_{0}^{1} \int_{0}^{1} (x^k y^k -x^k y^{k+1} -2 x^{k+1} y^k +2 x^{k+1} y^{k+1}) dx dy = -5 +3 \zeta(2) , $$ 
so the theorem is true for $n=0$ and $n=1$. 
As Beukers shows in \cite{B3}, we have 
$$I_n = (-1)^n \int_{0}^{1} \int_{0}^{1} \frac{x^n y^n (1-x)^n (1-y)^n}{(1-xy)^{n+1}} dx dy ,  $$  
for all $n$. 
Now, $n$-fold differentiation of the geometric series identity 
$$\frac{1}{1-u} = \sum_{k=0}^{\infty} u^k $$ 
gives the following 
formal identity for $n \in {\mathbb N}$ and an indeterminate $u$:  
$$\frac{1}{(1-u)^{n+1}} = \sum_{k=0}^{\infty}  \frac{(n+k)!}{k! \ n!} u^k . $$  
Therefore, 
$$ I_n = (-1)^n \sum_{k=0}^{\infty} \frac{(n+k)!}{k! \ n!} \int_{0}^{1} \int_{0}^{1} x^{n+k} y^{n+k} (1-x)^n (1-y)^n dx dy = $$  
$$ (-1)^n \sum_{k=0}^{\infty} \frac{(n+k)!}{k! \ n!} (B(n+k+1,n+1))^2 , $$
where $B(\cdot,\cdot)$ denotes Euler's beta function. Therefore,  
$$I_n= (-1)^n \sum_{k=0}^{\infty} \frac{(n+k)!^3 \ n!}{(2n+k+1)!^2 \ k!} . $$  
Since $\zeta(2)$ is irrational and $\{a_n\}$, $\{b_n\}$ satisfy the same 
recurrence relation (with different initial conditions), it follows that, 
in order to prove the theorem, it suffices to show that the sequence 
$\{I_n\}$ satisfies Ap\'{e}ry's recurrence relation, i.e., we need to show 
that 
$$(n+1)^2 I_{n+1} - (11 n^2+11 n +3) I_n -n^2 I_{n-1} = 0 , $$ 
for all $n \geq 1$. Fix such an $n$. It suffices to show that 
$$ \sum_{k=0}^{\infty} ((n+1)^2 \frac{(n+k+1)!^3 \ (n+1)!}{(2n+k+3)!^2 \ k!} + (11n^2+11n+3)\frac{(n+k)!^3 \ n!}{(2n+k+1)!^2 \ k!} $$ 
$$ - n \frac{(n+k-1)!^3 \ n!}{(2n+k-1)!^2 \ k!} ) = 0 . $$  
Let $S_{k,n}$ denote the expression 
inside the infinite sum on the left-hand side 
of the above equality. A tedious calculation shows that 
$$S_{k,n} =\frac{(n+k-1)!^3 \ n!}{(2n+k+3)!^2 \ k!} (-n k^8 +(3-n -5 n^2) k^7 +(10n^3+66n^2+88n+31) k^6 +$$  
$$+(119 n^4 +563 n^3 +881 n^2 +548 n +114) k^5 + $$ 
$$+(314 n^5 +1749 n^4 +3479 n^3 +3128 n^2 +1268 n +183) k^4 +$$ 
$$+(340 n^6 +2412 n^5 +6121 n^4 +7310 n^3 +4330 n^2 +1179 n +109) k^3 + $$ 
$$+(71 n^7 +1119 n^6 +4172 n^5 +6737 n^4 +5364 n^3 +2043 n^2 +291 n) k^2 +$$ 
$$+ (-138 n^8- 503 n^7 -349 n^6 +782 n^5 +1468 n^4 +885 n^3 +183 n^2) k  $$ 
$$- (79 n^9 +474 n^8 +1141 n^7 +1400 n^6 +913 n^5 +294 n^4 +35 n^3)) . $$  
Let $T_{m,n}$ denote the $m$-th partial sum of the 
latter infinite series, i.e., 
$$T_{m,n} = \sum_{k=0}^{m} S_{k,n} . $$ 
We claim that $T_{m,n}$ is given by the following closed    
formula: 
$$T_{m,n}= \frac{(n+m)!^3 \ n!}{(2n+m+3)!^2 \ m!} (m^6 +(4n+9) m^5 -(13 n^2 -7 n-26) m^4  $$   
$$-(102 n^3+228 n^2+112 n-15) m^3 -(225 n^4+822 n^3 +1025 n^2 +479 n+52) m^2 $$ 
$$-(217 n^5 + 1057 n^4 +1957 n^3 + 1691 n^2 +658 n +84) m $$
$$-(79 n^6+474 n^5 +1141 n^4 +1400 n^3 +913 n^2 +294 n +35 )). $$        
Although this formula is difficult to guess, its proof 
is a tedious but straightforward induction argument 
(on $m$), 
using the 
explicit formula for $S_{k,n}$ given above. The authors suspected the 
existence of such a  
closed formula for $T_{m,n}$ after explicitly computing it for the 
first few values of $m$. It should be pointed out that there is an algorithm 
(due to Gosper, and lying at the heart of the Ekhad program) that, given a 
hypergeometric summand $S_k$, determines whether or not $\sum_{k=0}^{m} S_k$ 
has 
a hypergeometric closed form. Also, one may try 
to use the 
Wilf-Zeilberger algorithm of creative telescoping to compute $T_{m,n}$; we have not attempted to 
use any of these algorithms.  

It now remains to show that $T_{m,n}$ tends to 0 as $m$ approaches infinity. By 
Stirling's formula, we see that 
$$\lim_{m\to\infty} T_{m,n} =  
\lim_{m \to \infty} \frac{(\frac{n+m}{e})^{3n+3m}}{(\frac{m}{e})^m (\frac{2n+m+3}{e})^{4n+2m+6}} \frac{\sqrt{8 \pi^3 (n+m)^3}}{2\pi (2n+m+3) \sqrt{2\pi m}} m^6  $$ 
$$=\lim_{m \to \infty} m^{-n} e^{n+6} = 0 , $$
and this completes the proof of the theorem. 

\bigskip

Now note that the formula for $I_n$ given in our proof of the theorem may be 
restated as follows:  
$$I_n = (-1)^n \frac{n!^4}{(2n+1)!^2} \ {\sb 3 F \sb 2 (n+1, n+1, n+1 ; 2n+2, 2n+2 ; 1)} . $$
Since $I_n$ satisfies Ap\'{e}ry's recurrence, the corollary follows from an   
easy calculation. 

\section{Acknowledgments}

This work was done during the first author's 
participation in 
the {\it Research Experiences for Undergraduates} program at the University 
of Tennessee in the Summer of 2004. 
We thank Suzanne Lenhart for organizing the program and 
the National Science Foundation for providing financial 
support for it. We also thank David Dobbs and the referee for their 
suggestions on a previous version of the manuscript.    

  
\begin{thebibliography}{12}

\bibitem[1]{AR}
K. Alladi and M. Robinson, On certain irrational values of the logarithm, {\it Number Theory, 
Carbondale 1979, Lecture Notes in Math.} {\bf 751}, Spinger, Berlin 1979, 1--9.

\bibitem[2]{AAR}
G. Andrews, R. Askey and R. Roy, {\it Special Functions}, Encyclopedia of 
Mathematics and its Applications, {\bf 71}, Cambridge University Press, 1999. 
  
\bibitem[3]{A}
R. Ap\'{e}ry, Irrationalit\'{e} de $\zeta(2)$ et $\zeta(3)$, {\it Ast\'{e}risque} {\bf 61} (1979), 11--13.  

\bibitem[4]{B1}
F. Beukers, Irrationality proofs using modular forms, {\it Ast\'{e}risque} {\bf 147-148} (1987), 271-283. 

\bibitem[5]{B2}
F. Beukers, Legendre polynomials in irrationality proofs, {\it Bull. Austral. Math. Soc.} {\bf 22} (1980), 431--438. 
 
\bibitem[6]{B3} 
F. Beukers, 
A note on the irrationality of $\zeta(2)$ and $\zeta(3)$, {\it Bull. London 
Math. Soc.} {\bf 11} (1979), 268--272. 

\bibitem[7]{F}
S. Fischler, {\it Irrationalit\'{e} de valeurs de z\^{e}ta (d'apr\`{e}s Ap\'{e}ry, Rivoal, \ldots)}, S\'{e}minaire Bourbaki 2002-2003, expos\'{e} no. 910, to 
appear in {\it Ast\'{e}risque}.  
 
\bibitem[8]{MOS} 
W. Magnus, F. Oberhettinger and R. P. Soni, {\it Formulas and Theorems for the 
Special Functions of Mathematical Physics}, Third enlarged edition, Grund. der 
math. Wissen. {\bf 52}, Springer-Verlag, New York, 1966. 
 
\bibitem[9]{N} 
Y. Nesterenko, Integral identities and constructions of approximations to 
zeta-values, {\it J. Th\'{e}or. Nombres Bordeaux}, {\bf 15} (2003), 535-550.  

\bibitem[10]{P} 
A. van der Poorten, 
A proof that Euler missed... Ap\'{e}ry's proof of the irrationality of $\zeta(3)$, {\it Math. Intelligencer} {\bf 1} (1978/79), 195--203. 

\bibitem[11]{W} 
J. Wilson, {\it Hypergeometric Series, Recurrence relations and Some New 
Orthogonal polynomials}, Ph.D. Thesis, University of Wisconsin, Madison, 1978.
  
\bibitem[12]{Z} 
W. Zudilin, Approximations to -, di- and tri- logarithms, {\it preprint}, 
posted at
\href{http://front.math.ucdavis.edu/math.CA/0409023}{\tt http://front.math.ucdavis.edu/math.CA/0409023},
September 2, 2004.   
 
\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11J72; Secondary 11B83, 11Y55, 33C20 .

\noindent \emph{Keywords: }
Beukers' integrals, Ap\'{e}ry's recurrences.  

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A005258}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received September 6 2004;
revised version received  January 3 2005.
Published in {\it Journal of Integer Sequences}, January 10 2005.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.math.uwaterloo.ca/JIS/}.
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