\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf A Generalized Ap\'{e}ry Series} \vskip 1.0cm {\large M. L. Glasser \\ Department of Physics\\ Clarkson University \\ Postdam, NY 13699-5820\\ USA\\ \href{mailto:laryg@clarkson.edu}{\tt laryg@clarkson.edu}\\ } \end{center} \vskip .2 in \begin{abstract} The inverse central binomial series \begin{equation}S_k(z)=\sum_{n=1}^{\infty} \frac{n^k z^n}{\binom{2n}{n}},\nonumber \end{equation} popularized by Ap\'ery and Lehmer, is evaluated for positive integers $k$ along with the asymptotic behavior for large $k$. We show that the value $z=2$, as commented on by D. H. Lehmer, provides a unique relation to $\pi$. \end{abstract} \section{Introduction} Since the appearance of $S_{-3}(1)$ in Ap\'ery's famous proof \cite{aper} in 1979 that $\zeta(3)$ is irrational, an extensive literature has been devoted to the series \begin{equation} S_k(z)=\sum_{n=1}^{\infty}\frac{n^k z^n}{\binom{2n}{n}} \label{1} \end{equation} For example, in 1985 Lehmer \cite{lehm} presented a number of special cases which could be obtained from the Taylor series for $f(x)=x^{-1/2}(1-x)^{-1/2}\sin^{-1}x$ using only elementary calculus. In passing, he noted that when $k$ is a positive integer, $S_k(2)$ had the form $a_k-b_k\pi$ and that the rational number $a_k/b_k$ ``is a close approximation to $\pi$. This remark was recently taken up by Dyson et al. \cite{dyson}, who proved that $|a_k/b_k-\pi|=O(Q^{-k})$ as $k\rightarrow\infty$ where $Q=\sqrt{1+(2\pi/\ln2)^2}$. Lehmer also showed that for positive integer $k$ \begin{equation} S_k(z)=\frac{2^{k+z^{5/2}}z^{1/2}}{(4-z)^{k+{3/2}}}( A_k(z/4) \sin^{-1}(\sqrt{z/4})+\sqrt{z(4-z)}B_k(z/4)) \label{2} \end{equation} where $A_k$ and $B_k$ are recursively defined polynomials. It was apparently not until 2005 that (\ref{2}) was evaluated explicitly, for $z=1$, by J. Borwein and P. Girgensohn \cite{bor} who showed \begin{equation} S_k(1)= \frac{1}{2}(-1)^{k+1}\sum_{j=1}^{k+1} (-1)^j j!S(j+1,j) 3^{-j} \binom{2j}{j}\left(\sum_{i=1}^{j-1}\frac{3^i}{(2i+1) \binom{2i}{i}} + \frac{2}{3 \sqrt{3}} \pi \right). \label{3} \end{equation} where the {\it Stirling numbers of the second kind} are defined by \begin{equation} S(k,j)=\frac{(-1)^j}{j!}\sum_{m=0}^j(-1)^m m^k\binom{j}{m}. \label{4} \end{equation} The aim of the present note is to extend (\ref{3}) to complex $z$ and thus to continue (\ref{1}) analytically beyond its circle of convergence $|z|=4$. \section{Calculation} We begin with the observation that $\left(m\binom{2m}{m}\right)^{-1} =B(m,m+1)$, where B denotes Euler's beta integral. Hence, \begin{equation} S_k(z)=\int_0^1\frac{dt}{t}\sum_{m=1}^{\infty}m^{k+1}(zt(1-t))^m. \label{5} \end{equation} Next, equation (21) of Girgensohn and Borwein \cite{bor}, \begin{equation} \sum_{m=1}^{\infty}m^pX^m= \sum_{n=1}^p\sum_{m=1}^n(-1)^{m+n}\binom{n}{m} m^pX^n(1-X)^{-n-1}, \label{6} \end{equation} gives \begin{equation} S_k(z)=\sum_{n=1}^{k+1}\sum_{m=1}^n(-1)^{m+n}\binom{n}{m} m^{k+1}\int_0^1\frac{dt}{t}\frac{(zt(1-t))^n}{(1-zt(1-t))^{n+1}}. \label{7} \end{equation} In the appendix it is shown that \begin{equation} \int_0^1\frac{dt}{t}\frac{(zt(1-t))^n}{(1-zt(1-t))^{n+1}}= \frac{\sqrt{\pi}\Gamma(n)}{\Gamma(n+1/2)}X^n\;_2F_1(-1/2,n;n+1/2;-X) \label{8} \end{equation} \noindent where $X=z/(4-z)$, so \begin{equation}S_k(z)= \sum_{n=1}^{k+1}n!B(n,1/2)S(k+1,n)X^n\;_2F_1(-1/2,n;n+1/2;-X). \label{9} \end{equation} By induction, starting with the tabulated value for $n=1$ and using Gauss' contiguity relations we find (some details are given in the appendix) \begin{equation} _2F_1(-1/2,n;n+1/2;-X) = \nonumber \end{equation} \begin{eqnarray} \left( \frac{1}{2} \right)_n \left(\frac{1}{n!}+ \frac{1}{\sqrt{\pi}\Gamma(n)}\sum_{k=0}^{n-1}\frac{(-1)^k\Gamma(k+1/2)}{(k+1)!} \binom{n-1}{k} \left(\frac{X+1}{X}\right)^{k+1}\times \right. \nonumber \\ \left.\left[\sqrt{X}\sin^{-1}\sqrt{\frac{X}{X+1}}- \frac{1}{2}\sum_{l=1}^k\frac{(l-1)!}{(1/2)_l} \left(\frac{X}{X+1} \right)^l \right] \right). \label{10} \end{eqnarray} (We have used the ascending factorial notation $(a)_n=\Gamma(a+n)/\Gamma(a)$). Therefore we have the principal result \begin{equation} S_k(z)= \sum_{n=1}^{k+1}n!\left(\frac{z}{4-z}\right)^nS(k+1,n)\times\nonumber\end{equation} \begin{equation} \left(\frac{1}{n}+ \sum_{p=0}^{n-1}(-1)^p\frac{(1/2)_p}{(p+1)!}\binom{n-1}{p} \left(\frac{4}{z}\right)^{p+1} \left(\sqrt{\frac{z}{4-z}} \sin^{-1}\frac{\sqrt{z}}{2}- \frac{1}{2}\sum_{l=1}^p\frac{\Gamma(l)}{(1/2)_l}\left(\frac{z}{4}\right)^l \right) \right) \label{11} \end{equation} Equation (\ref{11}) is rather condensed; in unpacking it, sums with upper limit less than the lower limit are to be interpreted as 0. It is clear from (\ref{11}) that for {\it rational} $z$ \begin{equation}\sum_{m=1}^{\infty}\frac{n^kz^n}{\binom{2n}{n}}= R_1(z,k)+R_2(z,k)\sqrt{\frac{z}{4-z}}\sin^{-1}\frac{\sqrt{z}}{2}, \label{12} \end{equation} where $R_j$ is a rational number. One sees from (\ref{11}) that $ S_k(z)$ is analytic on the two-sheeted Riemann surface formed by two planes cut and rejoined along the real half-line $x>4$. The numbers in (\ref{12}) have the explicit expressions \begin{equation} R_1(z,k)=\label{13} \end{equation} \begin{equation}\sum_{n=1}^{k+1}n!S(k+1,n)\left(\frac{z}{4-z}\right)^n \left(\frac{1}{n}-\frac{1}{2}\sum_{p=1}^{n-1} \sum_{l=1}^p\frac{(-1)^p(1/2)_p}{(p+1)!(1/2)_l}\binom{n-1}{p} \Gamma(l)\left(\frac{4}{z}\right)^{p-l+1}\right), \nonumber \end{equation} \begin{equation} R_2(z,k)= \sum_{n=1}^{k+1}n!S(k+1,n)\sum_{p=0}^{n-1}\frac{(-1)^p}{(p+1)!}\binom{n-1}{p} \left(\frac{4}{z}\right)^{p+1}. \label{14} \end{equation} \section{Asymptotics} It is convenient to work in terms of the exponential generating function \begin{equation} G(z,t):=\sum_{k=0}^{\infty}S_k(z)\frac{t^k}{k!}=S_0(ze^t)= \frac{z}{4-ze^t}+\frac{4\sqrt{z}e^{t/2}} {(4-ze^t)^{3/2}}\sin^{-1}\frac{\sqrt{z}e^{t/2}}{2} \label{15} \end{equation} To find the generating functions $\rho_j(z,t):=\sum R_j(z,k)t^k/k!$, it would be simplest to start with a series $D_k(z)=R_1(z,k)-R_2(z,k)\sqrt{\frac{z}{4-z}}\sin^{-1}\sqrt{z}/2$, work out its generating function $D(z,t)$ and by taking the sum and difference identify $\rho_1$ and $\rho_2$. However, this series has not been found and there is nothing to guarantee its existence in tractable form. Therefore, the $\rho_j$ were evaluated directly from (\ref{13}) and (\ref{14}). The details are omitted as the results \begin{eqnarray} \rho_1(z,t) & = & \frac{ze^t}{4-ze^t}+ \nonumber \\ & & \frac{8}{\pi}\sqrt{\frac{ze^t} {(4-ze^t)^{3/2}}}\left(\sin^{-1}\frac{\sqrt{z}e^{t/2}}{2} \cos^{-1}\frac{\sqrt{z}}{2} -\cos^{-1}\frac{\sqrt{z}e^{t/2}}{2}\sin^{-1}\frac{\sqrt{z}}{2}\right), \label{16} \end{eqnarray} \begin{equation} \rho_2(z,t)=4\sqrt{\frac{(4-z)e^t}{(4-ze^t)^3}} \label{17} \end{equation} are easily verified. In the case $z=2$, (\ref{15}) and (\ref{16}) are identical to Dyson's formulas \cite{dyson, dysonc} obtained empirically. In view of the prominent role that the ratio $R_1(z,t)/R_2(z,t)$ plays in Dyson et al. \cite{dyson} for $z=2$ it is interesting to examine it for general $z$. From (\ref{17}) we have \begin{equation} R_2(z,k)= \frac{2k!\sqrt{4-z}}{\pi i} \oint\frac{ds}{s^{k+1}}\frac{e^{s/2}}{(4-ze^s)^{3/2}}. \label{18} \end{equation} The non-zero singularity closest to $s=0$ is $s_0=\ln(4/z)$ and it dominates the asymptotic behavior. Ignoring the other singularities, distorting the contour to a small circle about $s_0$ and translating back to the origin by $t=s-s_0$, we have \begin{equation} R_2(z,k)\sim-\frac{k!\sqrt{4-z}}{zs_0^{k+1}}\oint\frac{dt}{2\pi i} \frac{e^{t/2}}{(1-e^t))^{3/2}}. \label{19} \end{equation} The exact value of the integral in (\ref{19}) is $-(2/\pi)\sqrt{e/(e-1)}$, and so \begin{equation} R_2(z,k)\sim\frac{k!}{(\ln(4/z))^{k+1}} \frac{2}{\pi}\sqrt{\frac{e(4-z)}{z(e-1)}}. \label{20} \end{equation} In the same way we obtain \begin{equation} R_1(z,k)\sim \frac{k!}{(\ln(4/z))^{k+1}}\left(\sqrt{2}+\frac{2}{\pi}\left(\sqrt\frac{e}{e-1} -\sqrt{2}\right)\cos^{-1}\frac{\sqrt{z}}{2}-\frac{2^{3/2}}{\pi} \sin^{-1}\frac{\sqrt{z}}{2}\right). \label{21} \end{equation} \section{Discussion} From (\ref{20}) and (\ref{21}) we find \begin{equation} \lim_{k\to\infty}\left(\frac{R_1(z,k)}{R_2(z,k)}- \sqrt{\frac{z}{4-z}}\sin^{-1}\frac{\sqrt{z}}{2}\right)= \sqrt{\frac{z}{4-z}}\left(\cos^{-1}\frac{\sqrt{z}}{2}- \sin^{-1}\frac{\sqrt{z}}{2}\right). \label{22} \end{equation} It thus appears that Lehmer's choice, $z=2$, is the unique permissible case for which the limit (\ref{22}) vanishes. (Also the {\it Lehmer limit}, as defined by Dyson et al. \cite{dyson}, relates to $\pi/4$ here rather than $\pi$). Finally, for negative integer indices, since \begin{equation} 2 S_{-k}(z)=\;_{k+1}F_k(1,\dots,1;\tfrac{3}{2},2,\dots,2; \tfrac{1}{4}z), \label{23} \end{equation} the fact that $S_{-k}(z)$ can be obtained from $S_0(z)$ by successive integrations with respect to $z$ and the explicit evaluations by Lehmer \cite{lehm}, Borwein and Girgensohn \cite{bor} and others \cite{6,7,8,9, 10} it should be possible to obtain explicit values for sundry generalized hypergeometric functions. \section{ Appendix: Derivation of Equations (\ref{8}) and (\ref{10})} Let us consider, for any integrable function $F$, \begin{equation}I=\int_0^1\frac{dt}{t}F(t(1-t))\nonumber\end{equation} Let $u=t(1-t)$, so $u(0)=u(1)=0;\;\; u(1/2)=1/4$. Then there are two expressions for $t$: \begin{equation} t_+=\frac{1}{2}(1+\sqrt{1-4t})\quad\mbox { for }\frac{1}{2} \le t\le1, \, \text{ with } \frac{dt_+}{t_+}=\left(1-\frac{1}{\sqrt{1-4u}}\right)\frac{du}{u} \nonumber\end{equation} \noindent and \begin{equation} t_-=\frac{1}{2}(1-\sqrt{1-4t})\quad\mbox { for }0\le t\le\frac{1}{2}, \, \text{ with } \frac{dt_+}{t_+}=\left(1+\frac{1}{\sqrt{1-4u}}\right)\frac{du}{u}. \nonumber \end{equation} Consequently, \begin{eqnarray} I & = & \int_0^{1/2}\frac{dt_-}{t_-}F(u)+\int_{1/2}^1\frac{dt_+}{t_+}F(u) =2\int_0^{1/4}\frac{du}{u\sqrt{1-4u}}F(u) \nonumber \\ & = & 2\int_0^1\frac{dx}{x\sqrt{1-x}}F \left(\frac{1}{4}x \right)= 2 \int_0^1\frac{dt}{(1-t)\sqrt{t}}F\left(\frac{1-t}{4}\right) \nonumber \end{eqnarray} and, with $t=x^2$, \begin{equation} I=4\int_0^1\frac{dx}{1-x^2}F\left(\frac{1-x^2}{4}\right). \nonumber \end{equation} \vskip .1in Therefore, \begin{equation} L=\int_0^1\frac{dt}{t}\frac{(zt(1-t))^{\alpha}} {(1-zt(1-t))^{\beta}}= 2\left(\frac{z}{4}\right)^{\alpha-\beta} \int_0^1dx\frac{(1-x^2)^{\alpha-1}}{(a^2+x^2)^{\beta}}, \nonumber \end{equation} where $a^2=1/X=(4-z)/z$. From standard references \begin{equation} \int_0^1 dx \cos(xy)(1-x^2)^{\alpha-1}= \sqrt{\frac{\pi y}{8}}\left(\frac{2}{y}\right)^{\alpha} \Gamma(\alpha)J_{\alpha-1/2}(y), \nonumber \end{equation} \begin{equation} \int_0^{\infty}dx \cos(xy)(a^2+x^2)^{-\beta} =\frac{\sqrt{\pi}}{\Gamma(\beta)}\left(\frac{y}{2a}\right)^{\beta-1/2} K_{\beta-1/2}(ay) \nonumber \end{equation} so, by the Parseval relation for the Fourier transform \begin{equation} \int_0^1dx\frac{(1-x^2)^{\alpha-1}}{(a^2+x^2)^{\beta}}= \frac{2^{\alpha-\beta}}{a^{\beta-1/2}}\frac{\Gamma(\alpha)}{\Gamma(\beta)} \int_0^{\infty}dy y^{\beta-\alpha}J_{\alpha-1/2}(y)K_{\beta-1/2}(ay). \nonumber \end{equation} This is a tabulated Hankel Transform and yields \begin{equation}L= \sqrt{\pi}\left(\frac{4}{z}\right)^{\beta-\alpha}X^{\beta} \frac{\Gamma(\alpha)}{\Gamma(\alpha+1)}\; _2F_1 \left(\frac{1}{2},\beta;\alpha+\frac{1}{2};-X \right). \nonumber \end{equation} Consequently \begin{equation} \int_0^1\frac{dx}{x}\frac{(zt(1-t))^n}{(1-zt(1-t))^{n+1}} =\sqrt{\pi}\left(\frac{4}{z}\right)X^{n+1}\; _2F_1 \left(\frac{1}{2},n+1;n+\frac{1}{2};-X \right) \nonumber \end{equation} However, since $\;_2F_1(a,b;c;z)=(1-z)^{c-a-b}\;_2F_1(c-a,c-b;c;z)$, \begin{equation} \;_2F_1 \left(\tfrac{1}{2},n+1;n+\tfrac{1}{2};-X \right)= (1+X)^{-1}\;_2F_1 \left(-\tfrac{1}{2};n;n+\tfrac{1}{2};-X \right) \nonumber \end{equation} \vskip .1in Next, we note that \cite[p .\ 590]{prud} \begin{equation} \;_2F_1(-1/2,1;3/2;z)= \frac{1}{2} \left(1+(1-z)\frac{\tanh^{-1}\sqrt{z}}{\sqrt{z}} \right). \nonumber \end{equation} With $z\rightarrow-z$, noting that $-i \tanh^{-1} iw=\sin^{-1}\sqrt{\frac{w}{1+w}}$ one has \begin{equation} \;_2F_1(-1/2,1;3/2;-z)= \frac{1}{2}(1+(1+z)\frac{\sin^{-1}\sqrt{\frac{z}{1+z}}}{\sqrt{z}}). \label{24} \end{equation} We next apply Gauss' differentiation formula \begin{equation} \frac{d}{dz}((1+z)^k\;_2F_1(-1/2,k;k+1/2;-z))= \nonumber \end{equation} \begin{equation} \frac{2k(k+1)}{2k+1}(1+z)\;_2F_1(-1/2,k+1;k+3/2;-z). \label{25} \end{equation} Iteration of (\ref{25}) starting with (\ref{24}), after a great deal of tedious algebra, aided by Mathematica, results in (\ref{10}). \section{Acknowledgements} The author is grateful to Profs. N. E. Frankel for suggesting this problem and F. J. Dyson for enlightening correspondence. \vskip .2in \begin{thebibliography}{9} \bibliographystyle{plain} \bibitem {aper}R. Ap\'ery, Irrationalit\'e de $\zeta(2)$ et $\zeta(3)$, {\it Asterique} {\bf 61} (1979), 11--13. \bibitem{lehm} D. H. Lehmer, Interesting series involving the central binomial coefficient, {\it Amer. Math. Monthly} {\bf92} (1985), 449--457. \bibitem{dyson} F. J. Dyson, N. E. Frankel, and M. L. Glasser, Lehmer's second interesting series, preprint. Available at \url{http//:arxiv.org/abs/1009.4274}. \bibitem{bor} J. M. Borwein and R. Girgensohn, Evaluations of binomial series, {\it Aequ. Math.} {\bf 70} (2005), 25--36. \bibitem{dysonc} F. J. Dyson, personal correspondence. \bibitem {6} I. J. Zucker, On the series $\sum\binom{2n}{n} ^{-1}n^{-k}$ and related series, {\it J. Number Theory} {\bf 20} (1981), 92--102. \bibitem{7} N. Batir, On the series $\sum\binom{2k}{k}^{-1}k^{-a}x^k$, {\it Proc. Indian Acad. Sci.} {\bf 115} (2005), 371--382. \bibitem{8} J. -H. Yang and F. -Z. Zhao, Certain sums involving inverses of binomial coefficients and some integrals, {\it J. Integer Sequences} {\bf 10} (2007), 1--11. \bibitem{9} R. Sprugnoli, Sums of the reciprocals of the central binomial coefficients, {\it INTEGERS} {\bf 6} (2006), 1--18. \bibitem{10} R. J. Mathar, Corrigenda to ``Interesting series involving the central binomial coefficient by D. H. Lehmer, preprint. Available at \url{http//: arxiv.org/abs/0905.0215v1}. \bibitem{prud} A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, { \it Tables of Integrals and Series} Vol.\ 3, Gordon and Breach, 1998. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2000 {\it Mathematics Subject Classification}: Primary 11B65; Secondary 33B05. \noindent {\it Keywords}: Binomial coefficient, infinite series, generating function. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A008277} and \seqnum{A145557}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received October 8 2011; revised version received April 2 2012. Published in {\it Journal of Integer Sequences}, April 6 2012. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}