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\begin{center}
\vskip 1cm{\LARGE\bf 
Certain Sums Involving Inverses of Binomial \\
\vskip .12in
Coefficients and Some Integrals
}
\vskip 1cm
\large
Jin-Hua Yang\\
Zhoukou Normal University\\
Zhoukou, 466001\\
China\\
\ \\
Feng-Zhen Zhao \\
Department of Applied Mathematics\\
Dalian University of Technology\\
Dalian, 116024\\
China \\
\href{mailto:fengzhenzhao@yahoo.com.cn}{\tt fengzhenzhao@yahoo.com.cn} \\
\end{center}

\vskip .2 in


\begin{abstract}
In this paper, we are concerned with sums
involving inverses of binomial coefficients. We study certain sums
involving reciprocals of binomial coefficients by using some
integrals. Some recurrence relations related to inverses of binomial
coefficients are obtained. In addition, we give the approximate
values of certain sums involving the inverses of binomial
coefficients.
\end{abstract}

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}


\section{Introduction}

It is well known that binomial coefficients play an important role
in various subjects such as combinatorics, number theory, and
probability. There are many results for sums related to binomial
coefficients. Sums involving inverses of binomial coefficients have
been receiving much attention. For example, see
\cite{ref1}-\cite{ref2} or \cite{ref4}-\cite{ref14}. In this paper,
we are still interested in sums involving inverses of binomial
coefficients, and we investigate these kinds of sums by using some
integrals. For convenience, we first give the definition of binomial
coefficients. For nonnegative integers $m$ and $n$, the binomial
coefficient ${n\choose m}$ is defined by
$$
{n\choose m}=\begin{cases}
\displaystyle\frac{n!}{m!(n-m)!}, & \text{if } n\geq m,\\
                  0, & \text{if } n<m.
\end{cases}
$$
We know that integral is an effective method for computing sums
involving inverses of binomial coefficients (see \cite{ref8,ref11}).
It is based on Euler's well-known Beta function
defined by (see \cite{ref11})
$$
B(n, m)=\int^1_0t^{n-1}(1-t)^{m-1}dt
$$
for all positive integers $n$ and $m$. Since $B(n,
m)=\displaystyle\frac{\Gamma(n)\Gamma(m)}{\Gamma(n+m)}=\frac{(n-1)!(m-1)!}{(n+m-1)!}$,
the binomial coefficient ${n\choose m}$ satisfies the equation
\begin{eqnarray}
\mbox{}\hspace{1cm}{n\choose
m}^{-1}&=&(n+1)\int_0^1t^m(1-t)^{n-m}dt. \label{bc-1}
\end{eqnarray}the equation
It is clear that integrals have connections with inverses of
binomial coefficients. The purpose of this paper is to study 
sums of the following forms:
$$
\sum_{n=0}^{\infty}\frac{{n+k\choose n}f_n}{{2n\choose n}} \ \ \
{\rm and} \ \ \  \sum_{n=0}^{\infty}\frac{f_n}{{2n\choose
n}{n+k\choose n}},
$$
where $f_n$ is a rational function of $n$. In Section 2, we derive
some relations for
$\displaystyle\sum_{n=0}^{\infty}\frac{{n+k\choose
n}f_n}{{2n\choose n}}$ by means of (1) and other integrals. In the
meantime, we express the series
$\displaystyle\sum_{n=0}^{\infty}\frac{1}{{2n\choose n}^2}$ by
some double integrations. In addition, we discuss the approximate
value of the series
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s{2n\choose n}}$,
where $s$ is a positive integer.

\section{Main Results}

\begin{lemma}
For the integrals
$\displaystyle\int_0^1\frac{dx}{[1-x(1-x)]^{k+1}}$ and
$\displaystyle\int_0^1\frac{dx}{[1+x(1-x)]^{k+1}}$, we have
\begin{eqnarray}
\int_0^1\frac{dx}{[1-x(1-x)]^{k+2}}=\frac{2}{3(k+1)}+\frac{2(2k+1)}{3(k+1)}\int_0^1\frac{dx}{[1-x(1-x)]^{k+1}},
\label{bc-2}
\end{eqnarray}
\begin{eqnarray}
\int_0^1\frac{dx}{[1+x(1-x)]^{k+2}}=\frac{2}{5(k+1)}+\frac{2(2k+1)}{5(k+1)}\int_0^1\frac{dx}{[1+x(1-x)]^{k+1}}.
\label{bc-3}\
\end{eqnarray}
\end{lemma}

\begin{proof}
The proofs of (\ref{bc-2})-(\ref{bc-3}) are simple and are omitted here.
\end{proof}

\begin{theorem}
Assume that
$$
A_k=\sum_{n=0}^{\infty}\frac{{n+k\choose n}}{{2n\choose n}} \ \ \
{\rm and} \ \ \ B_k=\sum_{n=0}^{\infty}\frac{(-1)^n{n+k\choose
n}}{{2n\choose n}},
$$
where $k$ is a nonnegative integer. Then we have the following
recurrence relations:
\begin{eqnarray}
A_{k+1}=-\frac{2}{3(k+1)}+\frac{2(2k+3)}{3(k+1)}A_k, \label{bc-4}
\end{eqnarray}
\begin{eqnarray}
B_{k+1}=-\frac{2}{5(k+1)}+\frac{2(2k+3)}{5(k+1)}B_k. \label{bc-5}
\end{eqnarray}
\end{theorem}

\begin{proof}
It follows from (\ref{bc-1}) that
$$
A_k=\sum_{n=0}^{\infty}{n+k\choose n}(2n+1)\int_0^1x^n(1-x)^ndx.
$$
Since $\displaystyle\sum_{n=0}^{\infty}{n+k\choose
n}(2n+1)\int_0^1x^n(1-x)^ndx$ converges uniformly for $x\in[0, 1]$,
we have
\begin{eqnarray*}
A_k=\int_0^1\bigg[\sum_{n=0}^{\infty}{n+k\choose
n}(2n+1)x^n(1-x)^n\bigg]dx.
\end{eqnarray*}
It is well known that
\begin{eqnarray}
\sum_{n=0}^{\infty}{n+k\choose n}u^n=\frac{1}{(1-u)^{k+1}},  \ \
{\rm for} \ \ |u|<1,\label{bc-6}
\end{eqnarray}
and
\begin{eqnarray}
\sum_{n=0}^{\infty}n{n+k\choose n}u^n=\frac{(k+1)u}{(1-u)^{k+2}},\ \
{\rm for} \ \ |u|<1. \ \ \label{bc-7}
\end{eqnarray}
From (\ref{bc-6}) and (\ref{bc-7}) we have
\begin{eqnarray*}
A_k&=&\int_0^1\frac{dx}{[1-x(1-x)]^{k+1}}+2(k+1)\int_0^1\frac{x(1-x)dx}{[1-x(1-x)]^{k+2}}\\
&=&-(2k+1)\int_0^1\frac{dx}{[1-x(1-x)]^{k+1}}+2(k+1)\int_0^1\frac{dx}{[1-x(1-x)]^{k+2}}.
\end{eqnarray*}
Using the same method, we have
\begin{eqnarray*}
B_k&=&-(2k+1)\int_0^1\frac{dx}{[1+x(1-x)]^{k+1}}+2(k+1)\int_0^1\frac{dx}{[1+x(1-x)]^{k+2}}.
\end{eqnarray*}
It follows from (\ref{bc-2}) and (\ref{bc-3}) that
\begin{eqnarray}
A_k&=&\frac{4}{3}+\frac{2k+1}{3}\int_0^1\frac{dx}{[1-x(1-x)]^{k+1}}
\ \ \  \label{bc-8} \\
B_k&=&\frac{4}{5}-\frac{2k+1}{5}\int_0^1\frac{dx}{[1+x(1-x)]^{k+1}}.
\ \  \label{bc-9}
\end{eqnarray}
From (\ref{bc-8}) and (\ref{bc-9}) we have recurrence relations
(\ref{bc-4}) and (\ref{bc-5}). 
\end{proof}

\begin{corollary}
Let $m$ be a positive integer. Define
$$
A_{k, m}=\sum_{n=0}^{\infty}\frac{{n+k\choose n}}{{2mn\choose mn}} \
\ \ {\rm and} \ \ \ B_{k,
m}=\sum_{n=0}^{\infty}\frac{(-1)^n{n+k\choose n}}{{2mn\choose mn}}.
$$
Then
\begin{eqnarray*}
A_{k,
m}&=&-(2mk+2m-1)\int_0^1\frac{dx}{[1-x^m(1-x)^m]^{k+1}}\\
&&+2m(k+1)\int_0^1\frac{dx}{[1-x^m(1-x)^m]^{k+2}}\\
B_{k,
m}&=&-(2mk+2m-1)\int_0^1\frac{dx}{[1+x^m(1-x)^m]^{k+1}}\\
&&+2m(k+1)\int_0^1\frac{dx}{[1+x^m(1-x)^m]^{k+2}}.
\end{eqnarray*}
\end{corollary}

\begin{theorem}
Suppose that
$$
C_k=\sum_{n=1}^{\infty}\frac{{n+k\choose n}}{n{2n\choose n}} \ \ \
{\rm and} \ \ \ D_k=\sum_{n=1}^{\infty}\frac{{n+k-1\choose
k}}{n^2{2n\choose n}}.
$$
Then we have
\begin{eqnarray}
C_k&=&\frac{1}{2}\sum_{i=1}^{k+1}\int_0^1\frac{dx}{[1-x(1-x)]^i}, \ \  \label{bc-10} \\
D_k&=&\frac{1}{2k}\sum_{i=1}^k\int_0^1\frac{dx}{[1-x(1-x)]^i}, \ \ \label{bc-11}\\
C_{k+1}&=&\frac{1}{3(k+1)}+\frac{7k+5}{3(k+1)}C_k-\frac{4k+2}{3(k+1)}C_{k-1},
\ \ \  \label{bc-12}\\
D_{k+1}&=&\frac{1}{3k(k+1)}+\frac{7k-2}{3(k+1)}D_k-\frac{2(2k-1)(k-1)}{3k(k+1)}D_{k-1}.
\ \ \ \label{bc-13}
\end{eqnarray}
\end{theorem}

\begin{proof}
It is evident that
\begin{eqnarray*}
C_k&=&\frac{1}{2}\sum_{n=1}^{\infty}\frac{{n+k\choose
n}}{(2n-1){2n-2\choose n-1}}\\
D_k&=&\frac{1}{2}\sum_{n=1}^{\infty}\frac{{n+k-1\choose
k}}{n(2n-1){2n-2\choose n-1}}.
\end{eqnarray*}
It follows from (\ref{bc-1}) that
\begin{eqnarray*}
C_k&=&\frac{1}{2}\sum_{n=1}^{\infty}{n+k\choose
n}\int_0^1x^{n-1}(1-x)^{n-1}dx
=\frac{1}{2}\int_0^1\sum_{n=1}^{\infty}{n+k\choose
n}x^{n-1}(1-x)^{n-1}dx. \\
D_k&=&\frac{1}{2}\int_0^1\sum_{n=1}^{\infty}\frac{{n+k-1\choose
k}}{n}x^{n-1}(1-x)^{n-1}dx.
\end{eqnarray*}
Owing to (\ref{bc-6}), (\ref{bc-10}) holds.

We note that
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{{n+k-1\choose
k}}{n}u^n=\frac{1}{k}\bigg[\frac{1}{(1-u)^k}-1\bigg], \ \ {\rm for}
\ \ |u|<1. \ \ \ \label{bc-14}
\end{eqnarray}
It follows from (\ref{bc-14}) that
$$
D_k=\frac{1}{2k}\int_0^1\frac{1}{x(1-x)}\bigg\{\frac{1}{[1-x(1-x)]^k}-1\bigg\}dx.
$$
Hence, (\ref{bc-11}) holds.

It is clear that
\begin{eqnarray*}
C_{k+1}-C_k&=&\frac{1}{2}\int_0^1\frac{dx}{[1-x(1-x)]^{k+2}}\\
2(k+1)D_{k+1}-2kD_k&=&\int_0^1\frac{1}{[1-x(1-x)]^{k+1}}dx
\end{eqnarray*}
Using (\ref{bc-2}), we can obtain (\ref{bc-12})-(\ref{bc-13}).
\end{proof}

\noindent {\bf Remark:} By using the same method,
the reader can consider the sums
$$
\sum_{n=1}^{\infty}\frac{(-1)^n{n+k\choose n}}{n{2n\choose n}} \ \ \
{\rm and} \ \  \sum_{n=1}^{\infty}\frac{(-1)^n{n+k-1\choose
k}}{n^2{2n\choose n}}.
$$

\begin{theorem}
Let
\begin{eqnarray*}
E_k=\sum_{n=0}^{\infty}\frac{1}{{2n\choose n}{n+k\choose n}},\ \ \
F_k=\sum_{n=1}^{\infty}\frac{1}{n{2n\choose n}{n+k\choose n}}, \ \
G_k=\sum_{n=1}^{\infty}\frac{1}{n^2{2n\choose n}{n+k\choose n}}.
\end{eqnarray*}
Then\\
\begin{eqnarray}
E_k&=&12\sum_{i=0}^k{k\choose i}(-3)^id_{k-1,
i}+4(2k-3)\sum_{i=0}^k{k\choose i}(-3)^id_{k,
i}\nonumber\\
&&-8k\sum_{i=0}^k{k\choose i}(-3)^id_{k+1,
i}+24\int_0^1\frac{(1-y)^kdy}{(4-y)^2}\nonumber\\
&&+(4k-2)\int_0^1\frac{(1-y)^kdy}{4-y},\label{bc-15}\\
F_k&=&-8\sum_{i=0}^k{k\choose i}(-3)^id_{k+1,
i}+8\sum_{i=0}^k{k\choose i}(-3)^id_{k,
i}+4\int_0^1\frac{(1-y)^k}{4-y}dy\nonumber\\
&&-\frac{1}{k+1}+16(k+1)\sum_{i=0}^k{k\choose i}(-3)^id_{k+2, i+1},
\label{bc-16}\\
G_k&=&16(k+1)\sum_{i=0}^k{k\choose i}(-3)^ig_{k+2,
i}+16\sum_{i=0}^k{k\choose i}(-3)^id_{k+2, i+1},\label{bc-17}
\end{eqnarray}
where
\begin{eqnarray*}
d_{m, i}=\int_0^{\frac{1}{\sqrt 3}}\frac{u^{2i}\arctan
u}{(1+u^2)^m}du,  \ \ g_{m, i}=\int_0^{\frac{1}{\sqrt
3}}\frac{u^{2i+1}(\arctan u)^2}{(1+u^2)^m}du,
\end{eqnarray*}
and they satisfy the equations
\begin{eqnarray}
&&d_{m, i}+d_{m, i+1}=d_{m-1, i},\label{bc-18}\\
&&g_{m, i}+g_{m, i+1}=g_{m-1, i}. \label{bc-19}
\end{eqnarray}
\end{theorem}

\begin{proof}
It follows from (\ref{bc-1}) that
\begin{eqnarray*}
E_k&=&\sum_{n=0}^{\infty}(2n+1)(n+k+1)\int_0^1x^n(1-x)^ndx\int_0^1y^n(1-y)^kdy\\
&=&\nonumber\sum_{n=0}^{\infty}(2n+1)(n+1+k)\int_0^1\int_0^1
x^n(1-x)^ny^n(1-y)^kdxdy\\
&=&2\sum_{n=0}^{\infty}n^2\int_0^1\int_0^1
x^n(1-x)^ny^n(1-y)^kdxdy\\
&&+(2k+3)\sum_{n=0}^{\infty}n\int_0^1\int_0^1
x^n(1-x)^ny^n(1-y)^kdxdy\\
&&+(k+1)\sum_{n=0}^{\infty}\int_0^1\int_0^1
x^n(1-x)^ny^n(1-y)^kdxdy.
\end{eqnarray*}
By using (\ref{bc-6})-(\ref{bc-7}) and
\begin{eqnarray}
\sum_{n=0}^{\infty}n^2u^n=\frac{1}{1-u}-\frac{3}{(1-u)^2}+\frac{2}{(1-u)^3},
\label{bc-20}
\end{eqnarray}
 we obtain
\begin{eqnarray}
E_k&=&4\int_0^1\int_0^1\frac{(1-y)^k}{[1-x(1-x)y]^3}dxdy+(2k-3)\int_0^1\int_0^1\frac{(1-y)^k}{[1-x(1-x)y]^2}dxdy
\nonumber\\
&&-k\int_0^1\int_0^1\frac{(1-y)^k}{1-x(1-x)y}dxdy.\nonumber
\end{eqnarray}
When $|y|\leq 1$, put
$I_n(y)=\displaystyle\int_0^1\frac{dx}{[1-x(1-x)y]^n}$. One can
verify that
\begin{eqnarray}
I_{n+1}(y)=\frac{2(2n-1)}{n(4-y)}I_n(y)+\frac{2}{n(4-y)}.\label{bc-21}
\end{eqnarray}
From (\ref{bc-21}), we have
\begin{eqnarray*}
E_k&=&4\int_0^1(1-y)^k\bigg[\frac{6}{(4-y)^2}I_1(y)+\frac{6}{(4-y)^2}+\frac{1}{4-y}\bigg]dy\\
&&+(2k-3)\int_0^1(1-y)^k\bigg[\frac{2}{4-y}I_1(y)+\frac{2}{4-y}\bigg]dy-k\int_0^1(1-y)^kI_1(y)dy.
\end{eqnarray*}
Noting that
$$
I_1(y)=\frac{4}{\sqrt{y(4-y)}}\arctan\sqrt{\frac{y}{4-y}},
$$
\begin{eqnarray*}
E_k&=&96\int_0^1\frac{(1-y)^k}{(4-y)^2\sqrt{y(4-y)}}\arctan\sqrt{\frac{y}{4-y}}dy+24\int_0^1\frac{(1-y)^k}{(4-y)^2}dy
\\
&&+(4k-2)\int_0^1\frac{(1-y)^k}{4-y}dy+8(2k-3)\int_0^1\frac{(1-y)^k}{(4-y)\sqrt{y(4-y)}}\arctan\sqrt{\frac{y}{4-y}}dy
\\
&&-4k\int_0^1\frac{(1-y)^k}{\sqrt{y(4-y)}}\arctan\sqrt{\frac{y}{4-y}}dy.
\end{eqnarray*}
Put $\displaystyle\sqrt{\frac{y}{4-y}}=u$. Then we get
$y=\displaystyle\frac{4u^2}{1+u^2}$,
$dy=\displaystyle\frac{8udu}{(1+u^2)^2}$, and
\begin{eqnarray*}
E_k&=&12\int_0^{\frac{1}{\sqrt 3}}\frac{(1-3u^2)^k\arctan
u}{(1+u^2)^{k-1}}du+4(2k-3)\int_0^{\frac{1}{\sqrt
3}}\frac{(1-3u^2)^k\arctan u}{(1+u^2)^k}du\\
&&-8k\int_0^{\frac{1}{\sqrt 3}}\frac{(1-3u^2)^k\arctan
u}{(1+u^2)^{k+1}}du+24\int_0^1\frac{(1-y)^kdy}{(4-y)^2}+(4k-2)\int_0^1\frac{(1-y)^kdy}{4-y}.
\end{eqnarray*}
Hence (\ref{bc-15}) holds.


Using a similar approach, we have
\begin{eqnarray*}
F_k&=&\sum_{n=1}^{\infty}2n\int_0^1\int_0^1(xy)^n(1-x)^n(1-y)^kdxdy\\
&&+(2k+3)\sum_{n=1}^{\infty}\int_0^1\int_0^1(xy)^n(1-x)^n
(1-y)^kdxdy\\
&&+(k+1)\sum_{n=1}^{\infty}\frac{1}{n}\int_0^1\int_0^1(xy)^n(1-x)^n(1-y)^kdxdy\\
&=&2\int_0^1\int_0^1\frac{(1-y)^kxy(1-x)}{[1-x(1-x)y]^2}dxdy+(2k+3)\int_0^1\int_0^1\frac{(1-y)^kxy(1-x)}{1-xy(1-x)}dxdy\\
&&-(k+1)\int_0^1\int_0^1(1-y)^k\ln[1-xy(1-x)]dxdy\\
&=&-\int_0^1(1-y)^kI_1(y)dy+4\int_0^1\frac{(1-y)^k}{4-y}I_1(y)dy+4\int_0^1\frac{(1-y)^k}{4-y}dy-\frac{1}{k+1}\\
&&+\frac{k+1}{2}\int_0^1(1-y)^kyI_1(y)dy\\
&=&-8\int_0^{\frac{1}{\sqrt 3}}\frac{(1-3u^2)^k\arctan
u}{(1+u^2)^{k+1}}du+8\int_0^{\frac{1}{\sqrt
3}}\frac{(1-3u^2)^k\arctan u}{(1+u^2)^k}du\\
&&+4\int_0^1\frac{(1-y)^k}{4-y}dy-\frac{1}{k+1}+16(k+1)\int_0^{\frac{1}{\sqrt
3}}\frac{u^2(1-3u^2)^k\arctan u}{(1+u^2)^{k+2}}du.
\end{eqnarray*}
Then (\ref{bc-16}) holds.

Now we prove (\ref{bc-17}). After calculus, we have
\begin{eqnarray*}
G_k&=&\frac{k+1}{2}\int_0^1(1-y)^kdy\int_0^1\frac{\ln[1-x(1-x)y]}{-x(1-x)}dx+\frac{1}{2}\int_0^1y(1-y)^kI_1(y)dy.
\end{eqnarray*}
Let
$$
h(y)=\int_0^1\frac{\ln[1-x(1-x)y]}{-x(1-x)}dx, \ \ 0\leq y\leq 1.
$$
We can verify that
\begin{eqnarray*}
h^{\prime}(y)&=&\int_0^1\frac{1}{1-x(1-x)y}dx\\
&=&\frac{4}{y}\sqrt{\frac{y}{4-y}}\arctan\sqrt{\frac{y}{4-y}},\\
h(y)&=&4\bigg(\arctan\sqrt{\frac{y}{4-y}}\bigg)^2.
\end{eqnarray*}
Then
\begin{eqnarray*}
G_k&=&2(k+1)\int_0^1(1-y)^k\bigg(\arctan\sqrt{\frac{y}{4-y}}\bigg)^2dy+\frac{1}{2}\int_0^1y(1-y)^kI_1(y)dy.
\end{eqnarray*}
By the proof of (\ref{bc-15}), we get
\begin{eqnarray*}
G_k&=&16(k+1)\int_0^{\frac{1}{\sqrt 3}}\frac{(1-3u^2)^ku(\arctan
u)^2}{(1+u^2)^{k+2}}du+16\int_0^{\frac{1}{\sqrt
3}}\frac{(1-3u^2)^ku^2\arctan u}{(1+u^2)^{k+2}}du.
\end{eqnarray*}
Then (\ref{bc-17}) holds. The proofs of (\ref{bc-18})-(\ref{bc-19})
are omitted here.
\end{proof}

Now, we express the series
$\displaystyle\sum_{n=0}^{\infty}\frac{1}{{2n\choose n}^2}$ by means
of some double integrations. 

\begin{theorem}
For the series
$\displaystyle\sum_{n=0}^{\infty}\frac{1}{{2n\choose n}^2}$, we have
\begin{eqnarray}
\sum_{n=0}^{\infty}\frac{1}{{2n\choose
n}^2}&=&\int_0^1\int_0^1\frac{dxdy}{1-x(1-x)y(1-y)}-8\int_0^1\int_0^1\frac{1}{[1-x(1-x)y(1-y)]^2}dxdy
\nonumber\\
&&+8\int_0^1\int_0^1\frac{1}{[1-x(1-x)y(1-y)]^3}dxdy. \label{bc-22}
\end{eqnarray}
\end{theorem}

\begin{proof}
It follows from (\ref{bc-1}) that
\begin{eqnarray*}
\sum_{n=0}^{\infty}\frac{1}{{2n\choose
n}^2}&=&\sum_{n=0}^{\infty}(4n^2+4n+1)\int_0^1x^n(1-x)^ndx\int_0^1y^n(1-y)^ndy\\
&=&\sum_{n=0}^{\infty}(4n^2+4n+1)\int_0^1\int_0^1x^n(1-x)^ny^n(1-y)^ndxdy.
\end{eqnarray*}
From (\ref{bc-6})-(\ref{bc-7}) and (\ref{bc-20}) we can prove that
(\ref{bc-23}) holds.
\end{proof}

Finally, we discuss the computation of the series
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s{2n\choose n}}$.

Let
$$
\psi(s)=\sum_{n=1}^{\infty}\frac{1}{n^s{2n\choose n}}.
$$
One knows that (see \cite{ref3})
$$
\psi(1)=\frac{\sqrt 3\pi}{9}, \ \ \psi(2)=\frac{\pi^2}{18}, \ \
{\rm and} \ \ \psi(4)=\frac{17\pi^4}{3240}.
$$
But we do not know the accurate value of $\psi(s)$ ($s=3$ or
$s\geq 5$). Now we give the approximate value of $\psi(s)$. It is
clear that
\begin{eqnarray*}
\psi(s)-\frac{1}{2}-\frac{1}{6\times
2^s}&=&\sum_{n=3}^{\infty}\frac{1}{n^s{2n\choose
n}}<\sum_{n=3}^{\infty}\frac{1}{n^s}=3^{-s}\bigg(1+\frac{3^s}{4^s}+\frac{3^s}{5^s}+\cdots\bigg)\\
&<&3^{-s}\bigg(1+\int_3^{\infty}\frac{3^s}{x^s}dx\bigg)=3^{-s}\bigg(1+\frac{3}{s-1}\bigg).
\end{eqnarray*}
Then we have
\begin{eqnarray}
\lim_{s\rightarrow+\infty}s^3\bigg(\psi(s)-\frac{1}{2}-\frac{1}{6\times
2^s}\bigg)=0. \label{bc-23}
\end{eqnarray}


Using a similar approach, we have
\begin{eqnarray}
\lim_{s\rightarrow+\infty}s^3\bigg(\psi(s-\psi(s+1)-\frac{1}{6\times
2^{s+1}}\bigg)=0. \label{bc-24}
\end{eqnarray}
(\ref{bc-23})-(\ref{bc-24}) provide the following simple 
approximate
\begin{eqnarray}
&&\psi(s)\approx\frac{1}{2}+\frac{1}{6\times 2^s}, \label{bc-25}\\
&&\psi(s+1)\approx\psi(s)-\frac{1}{6\times 2^{s+1}}. \label{bc-26}
\end{eqnarray}
By (\ref{bc-25})-(\ref{bc-26}), we obtain
\begin{eqnarray*}
\psi(6)\approx\frac{193}{384}, \ \
\psi(3)\approx\frac{\pi^2}{18}-\frac{1}{48}, \ \
\psi(5)\approx\frac{17\pi^4}{3240}-\frac{1}{192}.
\end{eqnarray*}

Similarly, we get
\begin{eqnarray*}
\sum_{n=1}^{\infty}\frac{1}{n^s{2n\choose
n}^2}\approx\frac{1}{4}+\frac{1}{36\times 2^s},\ \ \
\sum_{n=1}^{\infty}\frac{1}{n^s{2n\choose
n}^r}\approx\frac{1}{2^r}+\frac{1}{2^s\times 6^r}, \ \ r\geq 3.
\end{eqnarray*}


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\end{thebibliography}


\bigskip
\hrule
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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B65.

\noindent \emph{Keywords: } binomial coefficient, integral, recurrence
relation, generating function.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received March 15 2007;
revised version received  August 16 2007.
Published in {\it Journal of Integer Sequences}, August 16 2007.

\bigskip
\hrule
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\noindent
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