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\begin{center}
\vskip 1cm{\LARGE\bf  Sums Involving the Inverses of Binomial \\
\vskip .05in
Coefficients}
\vskip 1cm \large
Jin-Hua Yang \\
Department of Applied Mathematics \\
Dalian University of Technology \\
Dalian, 116024 \\
China\\

and \\

 Zhoukou Normal University\\
 Zhoukou, 466001 \\
 China\\

 \bigskip
 
 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 compute certain sums involving
the inverses of binomial coefficients. We derive the recurrence
formulas for certain infinite sums related to the inverses of
binomial coefficients.
\end{abstract}





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\section{Introduction}

As usual, the binomial coefficient ${n\choose m}$ is defined by
$$
{n\choose m}=\begin{cases}
\frac{n!}{m!(n-m)!}, &amp; \text{if  $n\geq m$;}\\
                  0, &amp;  \text{if  $n&lt;m$;} 
\end{cases}                 
$$
where $n$ and $m$ are nonnegative integers.

There are many identities involving binomial coefficients. However,
computations related to the inverses of binomial coefficients are
difficult. For some results involving the inverses of binomial
coefficients, see \cite{ref2,ref3,ref4,ref5,ref6,ref7,ref8,ref9}. In order
to compute
sums involving the inverses of binomial coefficients, using integrals is an
effective approach. This idea is based on Euler's well-known Beta function
defined by (see \cite{ref6})
$$
B(n, m)=\int^1_0t^{n-1}(1-t)^{m-1}dt
$$
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 inverse binomial coefficient ${n\choose m}^{-1}$ satisfies the identity
\begin{eqnarray}
{n\choose m}^{-1}&amp;=&amp;(n+1)\int_0^1t^m(1-t)^{n-m}dt. \label{bc-1}
\end{eqnarray}
with this method, a series of identities related to the inverses of
binomial coefficients is obtained (see \cite{ref6,ref7,ref8,ref9}).
In this paper, we also use Eq.\ (\ref{bc-1}) to
evaluate the sums
$$
\sum_{n=1}^{\infty}\frac{\varepsilon^n}{n(n+k){2n\choose n}}, \ \ \
\  \sum_{n=1}^{\infty}\frac{\varepsilon^n}{n^2(n+k){2n\choose n}},
$$
$$
\sum_{n=1}^{\infty}\frac{\varepsilon^n}{n(n+k){2n+k\choose n}}, \ \
{\rm and} \ \
\sum_{n=1}^{\infty}\frac{\varepsilon^n}{n(n+k){2n+2k\choose n+k}},
$$
where $|\varepsilon|=1$, and $k$ is an arbitrary positive integer
with $k&gt;1$. For convenience, we put
\begin{eqnarray*}
S_1(k)&amp;=&amp;\sum_{n=1}^{\infty}\frac{1}{n(n+k){2n\choose n}}, \ \ \
S_2(k)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+k){2n\choose
n}},\nonumber\\
T_1(k)&amp;=&amp;\sum_{n=1}^{\infty}\frac{1}{n^2(n+k){2n\choose n}}, \ \ \
T_2(k)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2(n+k){2n\choose
n}},\nonumber\\
Q_1(k)&amp;=&amp;\sum_{n=1}^{\infty}\frac{1}{n(n+k){2n+k\choose n}}, \ \ \
Q_2(k)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+k){2n+k\choose
n}},\nonumber\\
R_1(k)&amp;=&amp;\sum_{n=1}^{\infty}\frac{1}{n(n+k){2n+2k\choose n+k}}, \ \
\ R_2(k)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+k){2n+2k\choose
n+k}}.\nonumber
\end{eqnarray*}
In the next section, we evaluate the sums above. In the third section,
we define
$W_k=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^k{2n\choose n}}$ and
\ \ $X_r=\displaystyle\frac{1}{r}\sum_{k=1}^rW_k$; our aim is to
compute $\displaystyle\lim_{r\rightarrow+\infty}X_r$.



\section{Some Results For $S_i(k)$ And $T_i(k) \ (1\leq i\leq 2)$}

\begin{theorem}
Let $k$ be a positive integer with $k\geq 2$. Then
\begin{eqnarray}
S_1(k)&amp;=&amp;\frac{1-2k}{k}\int_0^1\frac{\ln[1-t(1-t)]+\displaystyle\sum_{i=1}^kt^i(1-t)^i/i}{t^k(1-t)^k}dt
+\frac{1}{k}\bigg(2-\frac{\sqrt 3\pi}{3}\bigg),\label{bc-2}\\
S_2(k)&amp;=&amp;(-1)^k\frac{(1-2k)}{k}\int_0^1\frac{\ln[1+t(1-t)]+\displaystyle\sum_{i=1}^k(-1)^it^i(1-t)^i/i}{t^k(1-t)^k}dt\nonumber\\
&amp;&amp;-\frac{2}{k}\bigg(\sqrt 5\ln\frac{\sqrt 5+1}{2}-1\bigg),\label{bc-3}\\
T_1(k)&amp;=&amp;\frac{\pi^2}{18k}-\frac{S_1(k)}{k},
\label{bc-4}\\
T_2(k)&amp;=&amp;-\frac{2}{k}\bigg(\ln\frac{\sqrt
5-1}{2}\bigg)^2-\frac{S_2(k)}{k}.\label{bc-5}
\end{eqnarray}
\end{theorem}

\begin{proof}
It follows from Eq.\ (\ref{bc-1}) that
\begin{eqnarray*}
\mbox{}\hspace{-1cm}S_1(k)&amp;=&amp;\frac{1}{k}\sum_{n=1}^{\infty}\frac{1}{n}\int_0^1t^n(1-t)^ndt+\bigg(2-\frac{1}{k}\bigg)\sum_{n=1}^{\infty}
\frac{\int_0^1t^n(1-t)^ndt}{n+k}\nonumber\\
&amp;=&amp;\frac{1}{k}\sum_{n=1}^{\infty}\frac{1}{n}\int_0^1t^n(1-t)^ndt+\bigg(2-\frac{1}{k}\bigg)\sum_{n=k+1}^{\infty}
\frac{\int_0^1t^{n-k}(1-t)^{n-k}dt}{n},\nonumber
\end{eqnarray*}
\begin{eqnarray*}
\mbox{}\hspace{-1cm}S_2(k)&amp;=&amp;\frac{1}{k}\sum_{n=1}^{\infty}\frac{(-1)^n\int_0^1t^n(1-t)^ndt}{n}+\bigg(2-\frac{1}{k}
\bigg)\sum_{n=k+1}^{\infty}\frac{(-1)^{n-k}\int_0^1t^{n-k}(1-t)^{n-k}dt}{n}.\nonumber
\end{eqnarray*}
It is well known that
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{u^n}{n}=-\ln(1-u), \ \ {\rm for} \ \
u\in[-1, 1). \label{bc-6}
\end{eqnarray}
On the other hand,
\begin{eqnarray}
&amp;&amp;\int_0^1\ln(1-t+t^2)dt=-2+\frac{\sqrt 3\pi}{3},
\label{bc-7}\\
&amp;&amp;\int_0^1\ln(1+t-t^2)dt=2\bigg(\sqrt 5\ln\frac{\sqrt
5+1}{2}-1\bigg). \label{bc-8}
\end{eqnarray}
From Eqs.\ (\ref{bc-6}-\ref{bc-8}) we get
Eqs.\ (\ref{bc-2}) and (\ref{bc-3}).

Now we give the proofs of Eqs.\ (\ref{bc-4}-\ref{bc-5}).\\
One can verify that
\begin{eqnarray*}
\mbox{}\hspace{-2cm}&amp;&amp;T_1(k)=\frac{1}{k}\sum_{n=1}^{\infty}\frac{1}{n^2{2n\choose
n}}-\frac{1}{k}\sum_{n=1}^{\infty}\frac{1}{n(n+k){2n\choose n}}
=\frac{1}{k}\sum_{n=1}^{\infty}\frac{1}{n^2{2n\choose
n}}-\frac{1}{k}S_1(k),\nonumber\\
\mbox{}\hspace{-1cm}&amp;&amp;T_2(k)=\frac{1}{k}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2{2n\choose
n}}-\frac{1}{k}S_2(k).\nonumber
\end{eqnarray*}
It follows from \cite{ref1,ref9} that
$$
\sum_{n=1}^{\infty}\frac{1}{n^2{2n\choose n}}=\frac{\pi^2}{18}, \ \
\  \ \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2{2n\choose
n}}=-2\bigg(\ln\frac{\sqrt 5-1}{2}\bigg)^2 \ \ .
$$
Hence Eqs.\ (\ref{bc-4}-\ref{bc-5}) are valid.
% \ \ \ \ $\Box$
\end{proof}

Next, we extend $S_i(k)$ and $T_i(k) (i=1,2)$ to the following forms:
\begin{eqnarray*}
S_1(k, m)&amp;=&amp;\sum_{n=1}^{\infty}\frac{1}{n(n+k){2mn\choose mn}}, \ \
\ S_2(k, m)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+k){2mn\choose
mn}},\nonumber\\
T_1(k, m)&amp;=&amp;\sum_{n=1}^{\infty}\frac{1}{n^2(n+k){2mn\choose mn}}, \
\ {\rm and} \ \ T_2(k,
m)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2(n+k){2mn\choose
mn}}.\nonumber
\end{eqnarray*}
we give the corresponding results as a corollary:

\begin{corollary}
Let $k$ and $m$ be positive integers. Then
\begin{eqnarray}
S_1(k,
m)&amp;=&amp;\bigg(\frac{1}{k}-2m\bigg)\displaystyle\int_0^1\frac{\ln[1-t^m(1-t)^m]+\displaystyle\sum_{i=1}^k
t^{mi}(1-t)^{mi}/i}{t^{mk}(1-t)^{mk}}dt\nonumber\\
&amp;&amp;-\frac{1}{k}\int_0^1\ln[1-t^m(1-t)^m]dt,\label{bc-9}\\
S_2(k,
m)&amp;=&amp;(-1)^k\bigg(\frac{1}{k}-2m\bigg)\int_0^1\displaystyle\frac{\ln[1+t^m(1-t)^m]+\displaystyle
\sum_{i=1}^k(-1)^it^{mi}(1-t)^{mi}/i}{t^{mk}(1-t)^{mk}}dt\nonumber\\
&amp;&amp;
-\frac{1}{k}\int_0^1\ln[1+t^m(1-t)^m]dt,\label{bc-10}\\
T_1(k, m)&amp;=&amp;-\frac{m}{2k}\int^1_0\frac{\ln[1-t^m(1-t)^m]dt}{t(1-t)}-\frac{1}{k}S_1(k,m),\label{bc-11}\\
T_2(k,
m)&amp;=&amp;-\frac{m}{2k}\int^1_0\frac{\ln[1+t^m(1-t)^m]dt}{t(1-t)}-\frac{1}{k}S_2(k,m).\label{bc-12}
\end{eqnarray}
\end{corollary}

\begin{proof}
We only give the proofs of Eqs.\ (\ref{bc-11}-\ref{bc-12}),
and leave the proofs of Eqs.\ (\ref{bc-9}-\ref{bc-10}) to the reader.  We
can immediately obtain that
\begin{eqnarray*}
&amp;&amp;T_1(k,m)=\frac{1}{k}\sum_{n=1}^{\infty}\frac{1}{n^2{2mn\choose
 mn}}-\frac{1}{k}\sum_{n=1}^{\infty}\frac{1}{n(n+k){2mn\choose mn}}
=\frac{1}{k}\sum_{n=1}^{\infty}\frac{1}{n^2{2mn\choose
 mn}}-\frac{1}{k}S_1(k,m),\\
&amp;&amp;T_2(k,
m)=\frac{1}{k}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2{2mn\choose
mn}}-\frac{1}{k}S_2(k,m).
\end{eqnarray*}
Owing to the  conclusions (see \cite{ref9}):
$$
\sum_{n=1}^{\infty}\frac{1}{n^2{2mn\choose
mn}}=-\frac{m}{2}\int^1_0\frac{\ln[1-t^m(1-t)^m]dt}{t(1-t)}
$$
$$
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2{2mn\choose
mn}}=-\frac{m}{2}\int^1_0\frac{\ln[1+t^m(1-t)^m]dt}{t(1-t)},
$$
we can show that Eqs.\ (\ref{bc-11}-\ref{bc-12}) hold.
% \ \ \ $\Box$
\end{proof}

It is evident that Eqs.\ (\ref{bc-9}-\ref{bc-12}) are the generalizations
of Eqs.\ (\ref{bc-2}-\ref{bc-5}),
respectively. Now we give the recurrence relation for $S_i(k)$ and $T_i(k)$.\\

\begin{theorem}
Let $k$ be a positive integer with $k\geq 2$. Then
\begin{eqnarray}
S_1(k+1)&amp;=&amp;\frac{2(2k+1)}{k+1}S_1(k)+\frac{1}{(k+1)^2}-\frac{\sqrt 3\pi}{3(k+1)},\label{bc-13}\\
S_2(k+1)&amp;=&amp;-\frac{2(2k+1)}{k+1}S_2(k)+\frac{1}{(k+1)^2}-\frac{2\sqrt 5}{k+1}\ln\frac{\sqrt 5+1}{2},\label{bc-14}\\
T_1(k+1)&amp;=&amp;-\frac{(3k+1)\pi^2}{18(k+1)^2}+\frac{2k(2k+1)}{(k+1)^2}T_1(k)-\frac{1}{(k+1)^3}+\frac{\sqrt
3\pi}{3(k+1)^2},
 \label{bc-15}\\
T_2(k+1)&amp;=&amp;-\frac{2(5k+3)}{(k+1)^2}\bigg(\ln\frac{\sqrt
5-1}{2}\bigg)^2-\frac{2k(2k+1)}{(k+1)^2}T_2(k)-\frac{1}{(k+1)^3}\nonumber\\
&amp;&amp;+\frac{2\sqrt 5}{(k+1)^2}\ln\frac{\sqrt 5+1}{2}.\label{bc-16}
\end{eqnarray}
\end{theorem}

\begin{proof}
For $0&lt;a\leq 1$, we consider the integrals:
$$
I_k(a)=\bigg(\frac{1}{k}-2\bigg)\displaystyle\int_0^1\frac{\ln[1-at(1-t)]+\displaystyle\sum_{i=1}^ka^it^i(1-t)^i/i}{t^k(1-t)^k}dt
$$
\begin{eqnarray*}
J_k(a)&amp;=&amp;(-1)^k\bigg(\frac{1}{k}-2\bigg)\displaystyle\int_0^1\frac{\ln[1+at(1-t)]+\displaystyle\sum_{i=1}^k(-1)^ia^it^i
(1-t)^i/i}{t^k(1-t)^k}dt,\nonumber\\
\end{eqnarray*}
where $I_k(0)=0$ and $J_k(0)=0.$ It is clear that
$$
S_1(k)=I_k(1)+\frac{1}{k}\bigg(2-\frac{\sqrt 3\pi}{3}\bigg) \ \ {\rm
and} \ \ S_2(k)=J_k(1)-\frac{2}{k}\bigg(\sqrt 5\ln\frac{\sqrt
5+1}{2}-1\bigg).
$$
When $0&lt;a\leq 1$, we have
\begin{eqnarray*}
I_k^{\prime}(a)&amp;=&amp;\bigg(\frac{1}{k}-2\bigg)a^{k-1}\bigg(1-\int_0^1\frac{dt}{1-at+at^2}\bigg).\nonumber\\
&amp;=&amp;\bigg(\frac{1}{k}-2\bigg)\bigg(a^{k-1}-4a^{k-2}\sqrt{\frac{a}{4-a}}\arctan\sqrt{\frac{a}{4-a}}\bigg),\nonumber\\
J_k^{\prime}(a)&amp;=&amp;\bigg(\frac{1}{k}-2\bigg)\int_0^1\frac{a^kt(1-t)dt}{1+at(1-t)}\nonumber\\
&amp;=&amp;\bigg(\frac{1}{k}-2\bigg)\bigg(a^{k-1}-\frac{2a^{k-2}\sqrt
a}{\sqrt{a+4}}\ln\frac{\sqrt{\displaystyle\frac{a+4}{a}}+1}{\sqrt{\displaystyle\frac{a+4}{a}}-1}\bigg).\nonumber
\end{eqnarray*}
Hence
\begin{eqnarray*}
I_k(a)&amp;=&amp;-\frac{(2k-1)a^k}{k^2}+\frac{4(2k-1)}{k}\int
a^{k-2}\sqrt{\frac{a}{4-a}}\arctan\sqrt{\frac{a}{4-a}}da.\nonumber\\
J_k(a)&amp;=&amp;-\frac{(2k-1)a^k}{k^2}+\frac{2(2k-1)}{k}\int
a^{k-2}\sqrt{\frac{a}{a+4}}
\ln\frac{\sqrt{\displaystyle\frac{a+4}{a}}+1}{\sqrt{\displaystyle\frac{a+4}{a}}-1}da.\nonumber
\end{eqnarray*}
Let $u=\displaystyle\sqrt{\frac{a}{4-a}}$ and
$v=\displaystyle\sqrt{\frac{a+4}{a}}.$ Then we get
\begin{eqnarray}
I_k(a)&amp;=&amp;-\frac{(2k-1)a^k}{k^2}+\frac{(2k-1)2^{2k+1}}{k}\int\frac{u^{2k-2}\arctan
u}{(1+u^2)^k}du,\label{bc-17}\\
J_k(a)&amp;=&amp;-\frac{(2k-1)a^k}{k^2}-\frac{(2k-1)4^k}{k}\int\frac{1}{(v^2-1)^k}\ln\frac{v+1}{v-1}dv.\label{bc-18}
\end{eqnarray}
It is well known that
\begin{eqnarray*}
\int\frac{u^{2k}\arctan
u}{(1+u^2)^{k+1}}du&amp;=&amp;\frac{2k-1}{2k}\int\frac{u^{2k-2}\arctan
u}{(1+u^2)^k}du-\frac{u^{2k-1}\arctan u}{2k(1+u^2)^k}\nonumber\\
&amp;&amp;+\frac{1}{2k}\int\frac{u^{2k-1}}{(1+u^2)^{k+1}}du,\nonumber\\
\int\frac{1}{(v^2-1)^{k+1}}\ln\frac{v+1}{v-1}dv&amp;=&amp;-\frac{2k-1}{2k}\int\frac{1}{(v^2-1)^k}\ln\frac{v+1}{v-1}dv
+\frac{1}{2k^2(v^2-1)^k}\nonumber\\
&amp;&amp;-\frac{v}{2k(v^2-1)^k}\ln\frac{v+1}{v-1}.\nonumber
\end{eqnarray*}
In the meantime, we note that
$$
\int\frac{u^{2k-1}}{(1+u^2)^{k+1}}du=\frac{2}{4^{k+1}}\int a^{k-1}da
\ \ \ {\rm and} \ \ \ I_k(0)=J_k(0)=0.
$$
Therefore $I_k(a)$ and $J_k(a)$ satisfy that
\begin{eqnarray}
\frac{k+1}{2k+1}I_{k+1}(a)&amp;=&amp;2I_k(a)-\frac{a^{k+1}}{k+1}+\frac{4a^k}{k}-
\frac{4\sqrt{4-a}a^{k-1/2}}{k}\arctan\sqrt{\frac{a}{4-a}},\label{bc-19}\\
\mbox{}\hspace{-1cm}\frac{k+1}{2k+1}J_{k+1}(a)&amp;=&amp;-2J_k(a)-\frac{a^{k+1}}{k+1}-\frac{4a^k}{k}
+\frac{2a^{k-1/2}\sqrt{a+4}}{k}\ln\frac{\sqrt{a+4}+\sqrt
a}{\sqrt{a+4}-\sqrt a}.\nonumber\\
&amp;&amp; \label{bc-20}
\end{eqnarray}
From Eqs.\ (\ref{bc-19}-\ref{bc-20}) we can derive
Eqs.\ (\ref{bc-13}-\ref{bc-14}). According to Eqs.\ (\ref{bc-4}-\ref{bc-5}) we
can obtain Eqs.\ (\ref{bc-15}-\ref{bc-16}).
% \ \ \ \ \ $\Box$
\end{proof}

We note that the recurrences given in
Eqs.\ (\ref{bc-13}-\ref{bc-14}) are similar to \cite[Eq.\ (28)]{ref10}.\\

\begin{theorem}
Let $k$ be a positive integer with $k\geq 2$. Then
\begin{eqnarray}
Q_1(k)&amp;=&amp;\bigg(\frac{1}{k}-1\bigg)\displaystyle\int_0^1\frac{\ln[1-t(1-t)]+\displaystyle\sum_{i=1}^kt^i(1-t)^i/i
}{t^k}dt\nonumber\\
&amp;&amp;-\bigg(1+\frac{1}{k}\bigg)\int_0^1(1-t)^k\ln[1-t(1-t)]dt,
\label{bc-21}\\
Q_2(k)&amp;=&amp;(-1)^k\bigg(\frac{1}{k}-1\bigg)\displaystyle\int_0^1\frac{\ln[1+t(1-t)]+\displaystyle\sum_{i=1}^k(-1)^i
t^i(1-t)^i/i}{t^k}dt\nonumber\\
&amp;&amp;-\bigg(1+\frac{1}{k}\bigg)\int_0^1(1-t)^k\ln[1+t(1-t)]dt,
\label{bc-22}\\
R_1(k)&amp;=&amp;\frac{1}{k}\int_0^1\bigg\{\ln[1-t(1-t)]+\sum_{i=1}^k\frac{t^i(1-t)^i}{i}\bigg\}dt\nonumber\\
&amp;&amp;-\bigg(2+\frac{1}{k}\bigg)\int_0^1t^k(1-t)^k\ln[1-t(1-t)]dt,\label{bc-23}
\end{eqnarray}
\begin{eqnarray}
R_2(k)&amp;=&amp;\frac{(-1)^k}{k}\int_0^1\bigg\{\ln[1+t(1-t)]+\sum_{i=1}^k\frac{(-1)^it^i(1-t)^i}
{i}\bigg\}dt\nonumber\\
&amp;&amp;-\bigg(2+\frac{1}{k}\bigg)\int_0^1t^k(1-t)^k\ln[1+t(1-t)]dt.
\label{bc-24}
\end{eqnarray}
\end{theorem}

\begin{proof}
We only give the proof of Eq.\ (\ref{bc-21}). The proofs of
Eqs.\ (\ref{bc-22}-\ref{bc-24}) follow the same pattern and are omitted
here. It follows from Eq.\ (\ref{bc-1}) and Eq.\ (\ref{bc-6}) that
\begin{eqnarray*}
Q_1(k)&amp;=&amp;\sum_{n=1}^{\infty}\frac{2n+k+1}{n(n+k)}\int_0^1t^n(1-t)^{n+k}dt\nonumber\\
&amp;=&amp;\bigg(1-\frac{1}{k}\bigg)\sum_{n=k+1}^{\infty}\int_0^1\frac{t^{n-k}(1-t)^n}{n}dt+\bigg(1+\frac{1}{k}\bigg)
\sum_{n=1}^{\infty}\int_0^1\frac{t^n(1-t)^{n+k}}{n}dt.\nonumber\\
&amp;=&amp;\bigg(\frac{1}{k}-1\bigg)\displaystyle\int_0^1\frac{\ln[1-t(1-t)]+\displaystyle\sum_{i=1}^kt^i(1-t)^i/i}{t^k}dt\nonumber\\
&amp;&amp;-\bigg(1+\frac{1}{k}\bigg)\int_0^1(1-t)^k\ln[1-t(1-t)]dt.\nonumber
\end{eqnarray*}
Hence Eq.\ (\ref{bc-21}) holds.
% \ \ \  $\Box$
\end{proof}

By computing integrals of Eqs.\ (\ref{bc-17}-\ref{bc-18}) and
Eqs.\ (\ref{bc-21}-\ref{bc-24}), we can establish a series of identities
involving inverses of binomial coefficients. In the final part of this
section, we evaluate special cases of $S_i(k)$, $T_i(k)$, and
$R_i(k)(1\leq i \leq 2)$ according to the particular choice of $k$.
For example, when $k=2$ in Eqs.\ (\ref{bc-17}-\ref{bc-18}), we have
\begin{eqnarray*}
I_2(a)&amp;=&amp;-\frac{3a^2}{4}+48\int\frac{u^2\arctan
udu}{(1+u^2)^2}\nonumber\\
&amp;=&amp;-\frac{3a^2}{4}-\frac{24u\arctan u}{1+u^2}+12(\arctan
u)^2-\frac{12}{1+u^2}+c_2\nonumber\\
&amp;=&amp;-\frac{3a^2}{4}-6\sqrt{a(4-a)}\arctan\sqrt{\frac{a}{4-a}}+12\bigg(\arctan\sqrt{\frac{a}{4-a}}\bigg)^2-12+3a
+c_2.\nonumber\\
J_2(a)&amp;=&amp;-\frac{3a^2}{4}-24\int\frac{1}{(v^2-1)^2}\ln\frac{v+1}{v-1}dv\nonumber\\
&amp;=&amp;-\frac{3a^2}{4}+\frac{12v}{v^2-1}\ln\frac{v+1}{v-1}-\frac{12}{v^2-1}-3\ln^2\bigg(\frac{v+1}{v-1}\bigg)
+c_2^{\prime}\nonumber\\
&amp;=&amp;-\frac{3a^2}{4}+3\sqrt{a(a+4)}\ln\frac{\sqrt{a+4}+\sqrt
a}{\sqrt{a+4}-\sqrt a}-3a-3\ln^2\bigg(\frac{\sqrt{a+4}+\sqrt
a}{\sqrt{a+4}-\sqrt a}\bigg)+c_2^{\prime}.\nonumber
\end{eqnarray*}
Since $I_2(0)=0$ and $J_2(0)=0$, then $c_2=12$, $c_2^{\prime}=0$,
$$
I_2(1)=\frac{9}{4}-\sqrt 3\pi+\frac{\pi^2}{3}, \ \
J_2(1)=-\frac{15}{4}+6\sqrt 5\ln\frac{\sqrt
5+1}{2}-12\ln^2\bigg(\frac{\sqrt 5+1}{2}\bigg).
$$
Hence, we get
\begin{eqnarray*}
S_1(2)&amp;=&amp;\frac{13}{4}-\frac{7\sqrt 3\pi}{6}+\frac{\pi^2}{3},\ \
T_1(2)=\frac{7\sqrt 3\pi}{12}-\frac{13}{8}-\frac{5\pi^2}{36},\nonumber\\
S_2(2)&amp;=&amp;-\frac{11}{4}+5\sqrt 5\ln\frac{\sqrt
5+1}{2}-12\ln^2\bigg(\frac{\sqrt 5+1}{2}\bigg),\nonumber\\
T_2(2)&amp;=&amp;\frac{11}{8}-\frac{5\sqrt 5}{2}\ln\frac{\sqrt
5+1}{2}+5\ln^2\bigg(\frac{\sqrt 5+1}{2}\bigg).\nonumber
\end{eqnarray*}
By means of $S_i(2)$, $T_i(2)$, and Eqs.\ (\ref{bc-13}-\ref{bc-16}), we
can compute other values of $S_i(k)$ and $T_i(k)(1\leq i\leq 2,
k&gt;2)$.

If $k=2$ in Eqs.\ (\ref{bc-23}-\ref{bc-24}), we can obtain
\begin{eqnarray*}
R_1(2)&amp;=&amp;\frac{1}{2}\int_0^1\bigg[\ln(1-t+t^2)+t(1-t)+\frac{t^2(1-t)^2}{2}\bigg]dt\nonumber\\
&amp;&amp;-\frac{5}{2}\int_0^1t^2(1-t)^2 \ln(1-t+t^2)dt\nonumber\\
&amp;=&amp;\frac{17}{36}-\frac{\sqrt 3\pi}{12},\nonumber\\
R_2(2)&amp;=&amp;\frac{1}{2}\int_0^1\bigg[\ln(1+t-t^2)-t(1-t)+\frac{t^2(1-t)^2}{2}\bigg]dt\nonumber\\
&amp;&amp;-\frac{5}{2}\int_0^1t^2(1-t)^2\ln(1+t-t^2)dt\nonumber\\
&amp;=&amp;\frac{1}{36}+\frac{5\sqrt 5}{6}\ln\frac{\sqrt 5-1}{2}.\nonumber
\end{eqnarray*}

\section{The Value of $\displaystyle\lim_{r\rightarrow+\infty}X_r$ }

We know that $ W_1=\displaystyle\frac{\sqrt 3\pi}{9}, \ \
W_2=\displaystyle\frac{\pi^2}{18}, \ \ {\rm and } \ \
W_4=\displaystyle\frac{17\pi^4}{3240}$ (see \cite{ref1}).
However, we do not know how to evaluate $W_k$ in closed form when $k\geq 5$.
In this section, we are interested in the average $X_r$ of $W_k$. We
compute
$\displaystyle\lim_{r\rightarrow+\infty}X_r$ by Eq.\ (\ref{bc-1}). \\

\begin{theorem}
Let $r$ be a positive integer with $r&gt;4$. Then
$\displaystyle\lim_{r\rightarrow+\infty}X_r=\displaystyle\frac{1}{2}.$\\
\end{theorem}

\begin{proof}
One can verify that
$$
X_r=\frac{1}{2}+\frac{1}{r}\sum_{n=2}^{\infty}\frac{1-\frac{1}{n^r}}{(n-1){2n\choose
n}}.
$$
Since $1-\displaystyle\frac{1}{n^{r+1}}&lt;1$, the series
$\displaystyle\sum_{n=2}^{\infty}\frac{1-\displaystyle\frac{1}{n^r}}{(n-1){2n\choose
n}}$ satisfies that
$$
\sum_{n=2}^{\infty}\frac{1-\frac{1}{n^r}}{(n-1){2n\choose
n}}\leq\sum_{n=2}^{\infty}\frac{1}{(n-1){2n\choose n}}.
$$
It follows from Eq.\ (\ref{bc-1}) that
$$
\sum_{n=2}^{\infty}\frac{1}{(n-1){2n\choose
n}}=\sum_{n=2}^{\infty}\frac{(2n+1)\int_0^1t^n(1-t)^ndt}{n-1}.
$$
Then
\begin{eqnarray*}
\sum_{n=2}^{\infty}\frac{1}{(n-1){2n\choose
n}}&amp;=&amp;\sum_{n=1}^{\infty}\frac{(2n+3)\int_0^1t^{n+1}(1-t)^{n+1}dt}{n}\nonumber\\
&amp;=&amp;2\int_0^1\frac{t^2(1-t)^2dt}{[1-t(1-t)]^2}-3\int_0^1t(1-t)\ln[1-t(1-t)]dt.\nonumber
\end{eqnarray*}
Hence $\displaystyle\lim_{r\rightarrow+\infty}X_r=\frac{1}{2}. $
%  \ \ \ $\Box$
\end{proof}


\section{Acknowledgments}

The authors are grateful to the referee for a careful reading and
for numerous suggestions, all of which have helped improve the
presentation of this article.


\begin{thebibliography}{9}
\bibliographystyle{plain}
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 \bibitem{ref1} L. Comtet, {\it Advanced Combinatorics}, Reidel,
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\bibitem{ref10} C. Elsner, On recurrence formulae for sums involving binomial
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 \bibitem{ref3} P. Juan, The sum of inverses of binomial coefficients
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 \bibitem{ref2} P. Nicolae, Problem C:1280, {\it Gaz. Mat.} {\bf
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\bibitem{ref8} B. Sury, Tianming Wang, and Feng-Zhen Zhao, 
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\bibitem{ref6} T. Tiberiu, Combinatorial sums and series involving
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\bibitem{ref9} Feng-Zhen Zhao and Tianming Wang,
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\end{thebibliography}

\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: } binomial coefficients, integral,
recurrence relation.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received June 26 2006;
revised version received  September 3 2006.
Published in {\it Journal of Integer Sequences}, September 3 2006.

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\noindent
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