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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}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section]
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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.
\bigskip
\hrule
\bigskip
\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.math.uwaterloo.ca/JIS/}.
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