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\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5(1) 
(2005), \#A22} 
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\begin{center}
 \uppercase{\bf Some Results for Sums of the Inverses of Binomial
Coefficients}
 \vskip 20pt
 {\bf Feng-Zhen Zhao }\\
 {\smallit Department of Applied Mathematics, Dalian University of Technology,
Dalian, 116024, China }\\
 \vskip 10pt
 {\bf Tianming Wang}\\
 {\smallit Department of Applied Mathematics, Dalian University of Technology,
Dalian, 116024, China }\\
 \end{center}
 \vskip 30pt
 \centerline{\smallit Received: 7/15/04, 
Revised: 8/11/05, Accepted: 9/26/05, Published: 10/11/05}
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\centerline{\bf Abstract}
\noindent
In this paper, the authors establish some
identities involving inverses of binomial coefficients and
generalize an identity.

\pagestyle{myheadings}
 \markright{\smalltt INTEGERS: \smallrm ELECTRONIC
 JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5(1) (2005), \#A22\hfill}

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\noindent
{1. \bf Introduction}

\noindent
For convenience, we first give some notation. The binomial
coefficients are defined by
$$
{n\choose m}=\cases{\displaystyle\frac{n!}{m!(n-m)!}, \ \ n\geq
m,\cr
                  0,\quad\quad\quad\quad\quad    n<m,
                 }
$$
where $n$ and $m$ are nonnegative integers.

Binomial coefficients are classical combinatorial numbers, which
play an important role in many subjects such as probability,
statistics, and number theory. There are many identities related to
binomial coefficients. However, computations involving the inverses
of binomial coefficients are often difficult. For previous literature 
dealing
with identities related to the inverses of binomial coefficients,
see [1--7]. In this paper, we offer some new identities involving inverses 
of binomial coefficients. To do so, we shall make use of the following 
formula (see [2])
\mbox{}\\[4pt]
\parskip=2pt
$$
{n\choose k}^{-1}=(n+1)\int^1_0t^k(1-t)^{n-k}dt. \ \ \    \eqno(1)
$$
\mbox{}\\[4pt]
\parskip=6pt
 In particular, we generalize the well-known identity (due to Euler)
\mbox{}\\[4pt]
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$$
\sum_{n=1}^{\infty}\frac{1}{n^2{2n\choose n}}=\frac{\pi^2}{18}. \
\ \ \  \eqno(2)
$$
\newpage

\parskip=10pt
\noindent
{\bf 2. Main Results}

\noindent
In this section, we give the main results of this paper.

\noindent
{\bf Theorem} Let $m$ be a positive integer. Then
$$
\sum_{n=1}^{\infty}\frac{1}{n^2{2mn\choose
mn}}=-\frac{m}{2}\int^1_0\frac{\ln[1-t^m(1-t)^m]dt}{t(1-t)}, \ \
\eqno(3)
$$
$$
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2{2mn\choose
mn}}=-\frac{m}{2}\int^1_0\frac{\ln[1+t^m(1-t)^m]dt}{t(1-t)}, \ \
\eqno(4)
$$
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{1}{n^2(n+1){2mn\choose
mn}}&&=-\frac{m}{2}\int^1_0\frac{\ln[1-t^m(1-t)^m]dt}{t(1-t)}
\nonumber\\
&& \ \ \
+\frac{m}{2}\int^1_0\frac{t^m(1-t)^m+\ln[1-t^m(1-t)^m]}{t^{m+1}
(1-t)^{m+1}}dt,\nonumber 
\end{eqnarray}
\vskip -50pt
$$ \ \ \   \eqno(5) $$ \vskip -10pt
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2(n+1){2mn\choose
mn}}&&=-\frac{m}{2}\int^1_0\frac{\ln[1+t^m(1-t)^m]dt}{t(1-t)}\nonumber\\
&&\ \ \
+\frac{m}{2}\int^1_0\frac{t^m(1-t)^m-\ln[1+t^m(1-t)^m]}{t^{m+1}(1-t)^{m+1}}dt.\nonumber
\end{eqnarray}
\vskip -50pt
$$ \ \ \   \eqno(6) $$ 

\noindent
{\it Proof.} We first prove (3).
By definition, we have
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{1}{n^2{2mn\choose
mn}}&&=\frac{m}{2}\sum_{n=1}^{\infty}\frac{1}{n(2mn-1){2mn-2\choose
mn-1}}\nonumber\\
&&=\frac{m}{2}\sum_{n=0}^{\infty}\frac{1}{(n+1)(2mn+2m-1){2mn+2m-2\choose
mn+m-1}}.\nonumber
\end{eqnarray}
It follows from (1) that
$$
\sum_{n=1}^{\infty}\frac{1}{n^2{2mn\choose
mn}}=\frac{m}{2}\sum_{n=0}^{\infty}\frac{1}{n+1}\int^1_0t^{mn+m-1}(1-t)^{mn+m-1}dt.
$$
Noticing that
$$
\sum_{n=0}^{\infty}\frac{t^{mn+m}(1-t)^{mn+m}}{n+1}=-\ln[1-t^m(1-t)^m]
\ \    \eqno(7)
$$
converges uniformly for $t\in[0, 1]$, we have (3).

Now we show that (5) holds.
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{1}{n^2(n+1){2mn\choose
mn}}&&=\frac{m}{2}\sum_{n=1}^{\infty}\frac{1}{n(n+1)(2mn-1){2mn-2\choose
mn-1}} \nonumber\\
&&=\frac{m}{2}\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\int^1_0t^{mn-1}(1-t)^{mn-1}dt\nonumber\\
&&=\frac{m}{2}\bigg(\sum_{n=1}^{\infty}\frac{\int^1_0t^{mn-1}(1-t)^{mn-1}dt}{n}-\sum_{n=1}
^{\infty}\frac{\int^1_0t^{mn-1}(1-t)^{mn-1}dt}{n+1}\bigg).\nonumber
\end{eqnarray}
By using (7), we have
\begin{eqnarray}
\sum_{n=1}^{\infty}\frac{1}{n^2(n+1){2mn\choose mn}}&&=-\frac{m}{2}\int^1_0\frac{\ln[1-t^m(1-t)^m]dt}{t(1-t)}\nonumber\\
&& \ \ \
+\frac{m}{2}\int^1_0\frac{t^m(1-t)^m+\ln[1-t^m(1-t)^m]}{t^{m+1}(1-t)^{m+1}}dt.\nonumber
\end{eqnarray}

The proofs of equalities (4) and (6) follow the same pattern and
are omitted. \hfill $\Box$

To see that (3) is a generalization of (2), let
$I(a)=-\displaystyle\frac{1}{2}\int^1_0\frac{\ln[1-at(1-t)]dt}{t(1-t)}
\ \ (0\leq a\leq 1).$ When $a>0$,
$$
I^{\prime}(a)=\frac{1}{2}\int^1_0\frac{dt}{at^2-at+1}=\frac{2}{a}\sqrt{\frac{a}{4-a}}\arctan\sqrt{\frac{a}{4-a}}.
$$
Then
$
I(a)=2\left(\arctan\sqrt{\frac{a}{4-a}}\,\,\right)^2+c, \ \ {\rm where}
\ \ c \ \ {\rm is\ \ a \ \ constant}.
$
Since $\displaystyle\lim_{a\rightarrow 0}I(a)=0$, we get $
I(a)=2\left(\arctan\sqrt{\displaystyle\frac{a}{4-a}}\,\right)^2.$
Hence,
$$
\sum_{n=1}^{\infty}\frac{1}{n^2{2n\choose
n}}=I(1)=\frac{\pi^2}{18}.
$$

By computing the integrals in (4)-(6), we can obtain other
identities involving inverses of binomial coefficients. For example,
if $m=1$ in (4), we have
$$
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2{2n\choose
n}}=-\frac{1}{2}\int^1_0\frac{\ln[1+t(1-t)]dt}{t(1-t)}.
$$

Put
$J(a)=-\displaystyle\frac{1}{2}\int^1_0\frac{\ln[1+at(1-t)]dt}{t(1-t)}
\ \ (0\leq a\leq 1).$ When $a>0$,
$$
J^{\prime}(a)=\frac{1}{2}\int^1_0\frac{dt}{at^2-at-1}=\frac{1}{\sqrt{a^2+4a}}\ln\bigg|\frac{\sqrt{a+4}-\sqrt
a} {\sqrt{a+4}+\sqrt a}\bigg|.
$$
Using some calculus, we find that
$$
J(a)=-\frac{1}{2}\bigg[\ln\frac{\sqrt{a+4}-\sqrt a}{\sqrt{a+4}+\sqrt
a}\bigg]^2+c_1.
$$
Since $\displaystyle\lim_{a\rightarrow 0}J(a)=0$, we have that
$c_1=0$. Thus
$$
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2{2n\choose
n}}=J(1)=-2\bigg(\ln\frac{\sqrt 5-1}{2}\bigg)^2. \ \ \ \eqno(8)
$$

Let
$$
K(a)=\frac{1}{2}\int^1_0\frac{at(1-t)-\ln[1+at(1-t)]}{t^2(1-t)^2}dt
\ \ \ (0\leq a\leq 1).
$$
Then, when $0<a\leq 1$,
$$
K^{\prime}(a)=-\frac{a}{2}\int^1_0\frac{dt}{at^2-at-1}.
$$
By calculus, we obtain
$$
K^{\prime}(a)=-\frac{\sqrt
a}{\sqrt{a+4}}\bigg[\ln\bigg(\sqrt{\frac{a+4}{a}}-1\bigg)-\ln\bigg(\sqrt{\frac{a+4}{a}}+1\bigg)\bigg],
$$
\begin{eqnarray}
K(a)&=&-\bigg\{\frac{\sqrt{a+4}+\sqrt a}{\sqrt{a+4}-\sqrt
a}\ln\bigg(\frac{\sqrt{a+4}-\sqrt a}{\sqrt{a+4}+\sqrt
a}\bigg)+\frac{\sqrt{a+4}+\sqrt a}{\sqrt{a+4}-\sqrt
a}+\bigg[\ln\bigg(\frac{\sqrt{a+4}-\sqrt a}{\sqrt{a+4}+\sqrt
a}\bigg)\bigg]^2\nonumber\\
&&-\frac{\sqrt{a+4}-\sqrt a}{\sqrt{a+4}+\sqrt
a}\ln\bigg(\frac{\sqrt{a+4}-\sqrt a}{\sqrt{a+4}+\sqrt
a}\bigg)+\frac{\sqrt{a+4}-\sqrt a}{\sqrt{a+4}+\sqrt
a}\bigg\}+c_2.\nonumber
\end{eqnarray}
Because $\displaystyle\lim_{a\rightarrow 0}K(a)=0$, we find that
$c_2=2$. By means of (8), we have
$$
\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2(n+1){2n\choose
n}}=-6\bigg(\ln\frac{\sqrt 5-1}{2}\bigg)^2-\sqrt
5\ln\bigg(\frac{3-\sqrt 5}{2}\bigg)-1.
$$

In the same way, from (5) we obtain
$$
\sum_{n=1}^{\infty}\frac{1}{n^2(n+1){2n\choose n}}=\frac{\sqrt
3\pi}{3}-\frac{\pi^2}{18}-1.
$$

\noindent
{\bf Acknowledgment}

\noindent
The authors wish to thank the anonymous referee for his/her
valuable suggestions for this paper.

\newpage
\parskip=5pt
\noindent
{\bf References}
\footnotesize
\begin{enumerate}
 \item Nicolae Pavelescu. ``Problem C:1280.'' {\it Gaz. Mat.} {\bf
  97.6} (1992): 230.
 \item Juan Pla. ``The Sum of Inverses of Binomial Coefficients
  Revisited.'' {\it The Fibonacci Quarterly} {\bf 35.4} (1997):
  342--345.
 \item Andrew M. Rockett. ``Sums of the Inverses of Binomial
  Coefficients.'' {\it The Fibonacci Quarterly} {\bf 19.5} (1981):
  433--437.
 \item B. Sury. ``Sum of the reciprocals of the Binomial
  Coefficients." {\it European J. Combin.} {\bf 14.4} (1993):
  351--353.
 \item Tiberiu Trif. ``Combinatorial Sums and Series Involving
  Inverses of Binomial Coefficients.'' {\it The Fibonacci
  Quarterly} {\bf 38.1} (2000): 79--84.
 \item WMC Problems Group. ``Problem 10494.'' {\it Amer. Math.
  Monthly} {\bf 103.1} (1996): 74.
 \item B. Sury, Tianming Wang, and Feng-Zhen Zhao. ``Identities Involving Reciprocals of Binomial
  Coefficients." {\it Journal of Integer Sequences} {\bf 7(2)} (2004): Article 04.2.8.
\end{enumerate}
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