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\begin{center}
{\large Notes on Sums Involving the Inverses of Binomial
Coefficients}
\end{center}

\begin{center}
Jin-Hua Yang \\
Department of Applied Mathematics, Dalian University
of Technology, Dalian, 116024, China \\
Zhoukou Normal University, Zhoukou, 466001, China
\mbox{}\\[4pt]
\parskip=6pt
 Feng-Zhen Zhao\\
Department of Applied Mathematics, Dalian
University of Technology, \\
Dalian 116024, China.
\end{center}
\mbox{}\\[4pt]
\parskip=6pt
{\bf 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.
\mbox{}\\[4pt]
\parskip=6pt
{\bf keywords:} binomial coefficients, integral, recurrence
relation.
\mbox{}\\[4pt]
\parskip=6pt
{\bf AMS Classification Number:} 11B65.

\begin{flushleft}
{\bf 1. Introduction}
\end{flushleft}

  As usual, the binomial coefficient ${n\choose m}$ is defined by
$$
{n\choose m}=\cases{\displaystyle\frac{n!}{m!(n-m)!}, \ \ n\geq
m,\cr
                  0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n<m,
                 }
$$
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 references
\cite{ref2}--\cite{ref9}. Computing sums involving the inverses of
binomial coefficients, integral is an effective approach. It is
based on Euler's well-known Beta function defined by (see
reference \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 binomial coefficient ${n\choose m}$ satisfies that
\begin{eqnarray}
{n\choose m}^{-1}&=&(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 are obtained (see references
\cite{ref6}-\cite{ref9}). In this paper, we also use (\ref{bc-1})
to discuss the computation of 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 $k>1$. For convenience, we put
\begin{eqnarray*}
S_1(k)&=&\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)&=&\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)&=&\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)&=&\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 calculate above sums. In the third section,
we denote
$W_k=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^k{2n\choose n}}$ and
\ \ $X_r=\displaystyle\frac{1}{r}\sum_{k=1}^rW_k,$ our aim is to
compute $\displaystyle\lim_{r\rightarrow+\infty}X_r$.
\mbox{}\\[4pt]
\parskip=6pt
\begin{flushleft}
{\bf 2. Some Results For $S_i(k)$ And $T_i(k) (1\leq i\leq 2)$}
\end{flushleft}
{\bf Theorem 1:} Let $k$ be a positive integer with $k\geq 2$. Then
\begin{eqnarray}
S_1(k)&=&\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)&=&(-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\\
&&-\frac{2}{k}\bigg(\sqrt 5\ln\frac{\sqrt 5+1}{2}-1\bigg),\label{bc-3}\\
T_1(k)&=&\frac{\pi^2}{18k}-\frac{S_1(k)}{k},
\label{bc-4}\\
T_2(k)&=&-\frac{2}{k}\bigg(\ln\frac{\sqrt
5-1}{2}\bigg)^2-\frac{S_2(k)}{k}.\label{bc-5}
\end{eqnarray}
{\bf Proof:} It follows from (\ref{bc-1}) that
\begin{eqnarray*}
\mbox{}\hspace{-1cm}S_1(k)&=&\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\\
&=&\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)&=&\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}
&&\int_0^1\ln(1-t+t^2)dt=-2+\frac{\sqrt 3\pi}{3},
\label{bc-7}\\
&&\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 (\ref{bc-6}-\ref{bc-8}) we have (\ref{bc-2}) and (\ref{bc-3}). Now we give the proofs of (\ref{bc-4}-\ref{bc-5}).\\

One can verify that
\begin{eqnarray*}
\mbox{}\hspace{-2cm}&&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}&&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 references \cite{ref1} and \cite{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 (\ref{bc-4}-\ref{bc-5}) are valid. \ \ \ \ $\Box$

Next,we extend $S_i(k)$ and $T_i(k) (i=1,2)$ to the following forms:
\begin{eqnarray*}
S_1(k, m)&=&\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)&=&\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:\\
{\bf Corollary :} Let $k$ and $m$ be positive integers. Then
\begin{eqnarray}
S_1(k,
m)&=&\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\\
&&-\frac{1}{k}\int_0^1\ln[1-t^m(1-t)^m]dt,\label{bc-9}\\
S_2(k,
m)&=&(-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\\
&&
-\frac{1}{k}\int_0^1\ln[1+t^m(1-t)^m]dt,\label{bc-10}\\
T_1(k, m)&=&-\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)&=&-\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}
{\bf Proof:} We only give the proofs of (\ref{bc-11}-\ref{bc-12}),
and leave the proofs of (\ref{bc-9}-\ref{bc-10}) to the readers.
We can immediately obtain that
\begin{eqnarray*}
&&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),\\
&&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 reference \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 (\ref{bc-11}-\ref{bc-12}) hold. \ \ \ $\Box$

It is evident that (\ref{bc-9}-\ref{bc-12}) are the
generalizations of (\ref{bc-2}-\ref{bc-5}),
respectively. Now we give the recurrence relation for $S_i(k)$ and $T_i(k)$.\\
{\bf Theorem 2:} Let $k$ be a positive integer with $k\geq 2$. Then
\begin{eqnarray}
S_1(k+1)&=&\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)&=&-\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)&=&-\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)&=&-\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\\
&&+\frac{2\sqrt 5}{(k+1)^2}\ln\frac{\sqrt 5+1}{2}.\label{bc-16}
\end{eqnarray}
{\bf Proof:} For $0<a\leq 1$, 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)&=&(-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 $0<a\leq 1$, we have
\begin{eqnarray*}
I_k^{\prime}(a)&=&\bigg(\frac{1}{k}-2\bigg)a^{k-1}\bigg(1-\int_0^1\frac{dt}{1-at+at^2}\bigg).\nonumber\\
&=&\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)&=&\bigg(\frac{1}{k}-2\bigg)\int_0^1\frac{a^kt(1-t)dt}{1+at(1-t)}\nonumber\\
&=&\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)&=&-\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)&=&-\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)&=&-\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)&=&-\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&=&\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\\
&&+\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&=&-\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\\
&&-\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)&=&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)&=&-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\\
&& \label{bc-20}
\end{eqnarray}
From (\ref{bc-19}-\ref{bc-20}) we can derive
(\ref{bc-13}-\ref{bc-14}). According to (\ref{bc-4}-\ref{bc-5}) we
can obtain (\ref{bc-15}-\ref{bc-16}). \ \ \ \ \ $\Box$

We note that the recurrences (\ref{bc-13}-\ref{bc-14}) are similar to (28) of reference \cite{ref10}.\\
{\bf Theorem 3:} Let $k$ be a positive integer with $k\geq 2$. Then
\begin{eqnarray}
Q_1(k)&=&\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\\
&&-\bigg(1+\frac{1}{k}\bigg)\int_0^1(1-t)^k\ln[1-t(1-t)]dt,
\label{bc-21}\\
Q_2(k)&=&(-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\\
&&-\bigg(1+\frac{1}{k}\bigg)\int_0^1(1-t)^k\ln[1+t(1-t)]dt,
\label{bc-22}\\
R_1(k)&=&\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\\
&&-\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)&=&\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\\
&&-\bigg(2+\frac{1}{k}\bigg)\int_0^1t^k(1-t)^k\ln[1+t(1-t)]dt.
\label{bc-24}
\end{eqnarray}
{\bf Proof:} We only give the proof of (\ref{bc-21}). The proofs
of (\ref{bc-22}-\ref{bc-24}) follow the same pattern and are
omitted here. It follows from (\ref{bc-1}) and (\ref{bc-6}) that
\begin{eqnarray*}
Q_1(k)&=&\sum_{n=1}^{\infty}\frac{2n+k+1}{n(n+k)}\int_0^1t^n(1-t)^{n+k}dt\nonumber\\
&=&\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\\
&=&\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\\
&&-\bigg(1+\frac{1}{k}\bigg)\int_0^1(1-t)^k\ln[1-t(1-t)]dt.\nonumber
\end{eqnarray*}
Hence (\ref{bc-21}) holds. \ \ \  $\Box$

By computing integrals of (\ref{bc-17}-\ref{bc-18}) and
(\ref{bc-21}-\ref{bc-24}), we can establish a series of identities
involving inverses of binomial coefficients. In the final of this
section, we give 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 (\ref{bc-17}-\ref{bc-18}), we have
\begin{eqnarray*}
I_2(a)&=&-\frac{3a^2}{4}+48\int\frac{u^2\arctan
udu}{(1+u^2)^2}\nonumber\\
&=&-\frac{3a^2}{4}-\frac{24u\arctan u}{1+u^2}+12(\arctan
u)^2-\frac{12}{1+u^2}+c_2\nonumber\\
&=&-\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)&=&-\frac{3a^2}{4}-24\int\frac{1}{(v^2-1)^2}\ln\frac{v+1}{v-1}dv\nonumber\\
&=&-\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\\
&=&-\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)&=&\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)&=&-\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)&=&\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 (\ref{bc-13}-\ref{bc-16}), we
can compute other values of $S_i(k)$ and $T_i(k)(1\leq i\leq 2,
k>2)$.

If $k=2$ in (\ref{bc-23}-\ref{bc-24}), we can obtain
\begin{eqnarray*}
R_1(2)&=&\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\\
&&-\frac{5}{2}\int_0^1t^2(1-t)^2 \ln(1-t+t^2)dt\nonumber\\
&=&\frac{17}{36}-\frac{\sqrt 3\pi}{12},\nonumber\\
R_2(2)&=&\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\\
&&-\frac{5}{2}\int_0^1t^2(1-t)^2\ln(1+t-t^2)dt\nonumber\\
&=&\frac{1}{36}+\frac{5\sqrt 5}{6}\ln\frac{\sqrt 5-1}{2}.\nonumber
\end{eqnarray*}
\mbox{}\\[4pt]
\parskip=6pt
\begin{flushleft}
{\bf 3. The Value of $\displaystyle\lim_{r\rightarrow+\infty}X_r$ }
\end{flushleft}

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 reference
\cite{ref1}). However we cannot give the accurate value of $W_k$
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 (\ref{bc-1}). \\
{\bf Theorem 4:} Let $r$ be a positive integer with $r>4$. Then
$\displaystyle\lim_{r\rightarrow+\infty}X_r=\displaystyle\frac{1}{2}.$\\
{\bf 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 $1-\displaystyle\frac{1}{n^{r+1}}<1$, 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 (\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}}&=&\sum_{n=1}^{\infty}\frac{(2n+3)\int_0^1t^{n+1}(1-t)^{n+1}dt}{n}\nonumber\\
&=&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 $\displaystyle\lim_{r\rightarrow+\infty}X_r=\frac{1}{2}.$ \ \
\ $\Box$
\mbox{}\\[4pt]
\parskip=6pt

{\bf Acknowledgment} The authors wish to thank the anonymous referee
for his/her valuable suggestions for this paper.
\mbox{}\\[4pt]
\parskip=6pt
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\end{document}
