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\begin{center}
\vskip 1cm{\LARGE\bf Identities Involving Reciprocals of Binomial Coefficients}
\vskip 1cm
\large
B. Sury\\
Stat-Math Unit, Indian Statistical Institute\\
8th Mile Mysore Road\\
Bangalore 560 059\\
India\\
\href{mailto:sury@isibang.ac.in}{\tt sury@isibang.ac.in} \\
\ \\
Tianming Wang \&amp; 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 deal with several
combinatorial sums and some infinite series which involve the reciprocals
of binomial coefficients. Many binomial identities as well as some
polynomial identities are proved.
\end{abstract}

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

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\newtheorem{Remark}[Definition]{Remark}


\noindent
\section{Introduction}
\vskip 5mm

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

In many subjects, such as combinatorial analysis, graph theory,
and number theory, binomial coefficients often appear naturally
and play an important role.
However, it is well known that it is difficult to compute the
values of combinatorial sums involving inverses of
binomial coefficients. For some investigations in this respect,
see [1--8].
It is the purpose of this paper to deal with several finite
combinatorial sums and some infinite series involving the reciprocals
of binomial coefficients.
In \cite{Su}, the first author used the identity
$$\beta(p,q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)},$$
to observe that
$$\frac{1}{{n \choose r}}
= \frac{r! (n-r)!}{n!} = \frac{\Gamma(r+1) \Gamma(n-r+1)}{\Gamma(n+1)}
= (n+1) \int_0^1 t^r (1-t)^{n-r} dt.$$
Starting with this observation, it was proved in \cite{Su} that 
the following two identities hold:
\begin{eqnarray*}
\sum\limits^n_{r=0} \frac{1}{{n \choose r}} = \frac{n+1}{2^n}
\sum\limits_ {r=0}^n \frac{2^r}{r+1}= \frac{n+1}{2^n}
\sum\limits_{j~\mbox{odd}} {n+1 \choose j} \frac{1}{j}.
\end{eqnarray*}
In this paper, we exploit this method much further and prove,
among other things, a polynomial identity and several interesting
combinatorial identities where the binomial coefficients occur in
the denominator. \vskip 8mm

\noindent
\section{A polynomial identity and applications}
\vskip 5mm

\begin{Theorem}
In the ring $\Q[T]$ of rational polynomials, 
the identity
\begin{eqnarray}
\sum\limits^n_{r=m} \frac{T^r (1-T)^{n-r}}{{n \choose r}}
&amp;=&amp; (n+1) \sum\limits^n_{r=m} \frac{T^{n+1} (1-T)^{n-r}}{r+1}
\nonumber \\
&amp;&amp; +(n+1) \sum\limits_{r=0}^{n-m} \frac{T^{n-r} (1-T)^{n-m+1}}
{(m+r+1) {m+r \choose r}}
\end{eqnarray}
holds for $m \leq n$.
An equivalent form is that for $\lambda \neq -1$,
\begin{eqnarray}
\sum\limits^n_{r=m} \frac{\lambda^r}{{ n \choose r}}
&amp;=&amp;(n+1) \sum\limits^{n-m}_{r=0} \frac{\lambda^{m+r}}{(\lambda+1)^{r+1}}
\sum\limits_{i=0}^{n-m-r}  {n-m-r \choose i} \frac{(-1)^i}{m+1+i}
\nonumber \\
&amp;&amp;+(n+1) \frac{\lambda^{n+1}}{(\lambda+1)^{n+2}}
\sum\limits^n_{r=m} \frac{(\lambda+1)^ {r+1}}{r+1}.
\end{eqnarray}
\end{Theorem}

\begin{proof}
Let us consider, for a fixed real number $\lambda$, $I_{m,n}
(\lambda)=\sum\limits^n_{r=m} \frac{\lambda^r}{ {n \choose r}}$.
Then,
\begin{eqnarray}
I_{m,n}(\lambda)&amp;=&amp; \frac{1}{n!} \sum^n_{r=m} \lambda^r \Gamma
(r+1) \Gamma (n-r+1)\nonumber\\
&amp;=&amp;(n+1)\displaystyle \sum^n_{r=m} \lambda^r \beta(r+1,
n-r+1)\nonumber \\
&amp;=&amp; (n+1) \sum^n_{r=m} \lambda^r \int^1_0 t^r (1-t)^{n-r}
dt\nonumber\\
&amp;=&amp;(n+1)\int^1_0 \sum^n_{r=m} (t\lambda)^r (1-t)^{n-r} dt
\nonumber \\
&amp;=&amp; (n+1) \int^1_0
\frac{(t\lambda)^{n+1}-(t\lambda)^m(1-t)^{n-m+1}}{t\lambda -(1-t)}
dt.\nonumber
\end{eqnarray}
Putting $s=t\lambda-(1-t)$, one gets $I_{m,n} (\lambda)=(n+1)
(I_1+I_2)$ where
$$
\mbox{}\hspace{-2cm}I_1= \frac{\lambda^{n+1}}{(\lambda+1)^{n+2}}
\int^\lambda_{-1} \frac{(s+1)^{n+1} -(s+1)^m}{s} ds$$ and
$$I_2=\frac{\lambda^m}{(\lambda+1)^{n+2}}
\int^\lambda_{-1}
\frac{(s+1)^m}{s}(\lambda^{n-m+1}-(\lambda-s)^{n-m+1}) ds.
$$
Now, by writing
$$\displaystyle\frac{(s+1)^{n+1}-(s+1)^m}{s}=\sum^n_{r=m}(s+1)^r$$
and interchanging the order of the summation and the integration,
one obtains $$I_1=\displaystyle\frac{\lambda^
{n+1}}{(\lambda+1)^{n+2}} \sum^n_{r=m} \frac{(\lambda+1)^{r+1}}{r+1}.$$
Similarly, $$I_2=\displaystyle\sum^{n-m}_{r=0}
\frac{\lambda^{m+r}}{(\lambda+1)^{r+1}} \sum^{n-m-r}_{i=0} (-1)^i
{n-m-r \choose i} \frac{1}{m+1+i}.$$

\noindent
Evidently, the above manipulations are valid when $\lambda$ is any
real number different from $-1$.
Therefore, we have, for $\lambda \neq -1$,
\begin{eqnarray*}
\sum\limits^n_{r=m} \frac{\lambda^r}{ {n \choose r}}
&amp;=&amp;(n+1) \sum\limits^{n-m}_{r=0} \frac{\lambda^{m+r}}{(\lambda+1)^{r+1}}
\sum\limits_{i=0}^{n-m-r}  {n-m-r \choose i} \frac{(-1)^i}{m+1+i}
\\
&amp;&amp;+ (n+1) \frac{\lambda^{n+1}}{(\lambda+1)^{n+2}}
\sum\limits^n_{r=m} \frac{(\lambda+1)^ {r+1}}{r+1}.
\end{eqnarray*}
This proves (2).
The particular case $m=0$ gives
\begin{eqnarray}
\sum\limits^n_{r=0} \frac{\lambda^r}{ {n \choose r}} = (n+1)
\sum\limits^n_{r=0} \frac{\lambda^{n+1} + \lambda^{n-r}}{(r+1)
(1+\lambda)^{n-r+1}}.
\end{eqnarray}

\noindent
From (2), one can easily get a recursion formula for
$I_{m,n}(\lambda)$. When $m=0$, the
recursive formula is
$$
(1+\lambda^{-1}) I_{0,n}=\lambda^n +\lambda^{-1}
+\bigg(1+\frac{1}{n}\bigg) I_{0,n-1}.
$$
For $0 &lt; \lambda &lt; 1$, this formula proves also that $I_{0,n}
(\lambda) \rightarrow 1$ as $n \rightarrow \infty$. Now, using (2)
with $\lambda$ replaced by $\frac{\theta}{1-\theta}$ for $\theta
\neq 1$ (which can be done because $\frac{\theta}{1-\theta}$ takes
all values except $-1$), we have
\begin{eqnarray}
\sum\limits^n_{r=m} \frac{\theta^r (1-\theta)^{n-r}}{ {n \choose r}}
&amp;=&amp; (n+1) \sum\limits_{r=0}^{n-m} \theta^{n-r} (1-\theta)^{n-m+1}
 \sum\limits_{i=0}^m (-1)^i {m \choose i} \frac{1}{r+i+1}
\nonumber \\
&amp;&amp;+(n+1) \sum\limits^n_{r=m} \frac{\theta^{n+1}
(1-\theta)^{n-r}}{r+1}.
\end{eqnarray}

\noindent
To show that this is a polynomial identity, let
us denote by $P(T)$ and $Q(T)$ the polynomials over $\Q$ which correspond
to the two sides of (4).

We have $P(\theta)=Q(\theta)$ for all $1 \neq \theta \in \Q$. This
means that $P$ and $Q$ have to coincide in $\Q[T]$ and this is indeed a
polynomial identity.

This is not yet the first identity (1) in the theorem. To obtain
that, we make a few observations of independent interest. First, if we
compare
the coefficients of $\theta^m$ on both sides of the identity (4)
above, we get
\begin{eqnarray}
\frac{1}{{n \choose m}} =(n+1) \sum\limits_{r=0}^m (-1)^{m-r}
 {m \choose r} \frac{1}{n-r+1}.
\end{eqnarray}

\noindent
In terms of $k=n-m$, this is also equivalent to the identity
\begin{eqnarray}
\sum\limits^k_{r=0} (-1)^r {k \choose r} \frac{1}{m+r+1} =\frac{1}
{(k+m+1)  {k+m \choose m}}.
\end{eqnarray}
\noindent If we use this in the above polynomial identity (4), we
get the identity (1) in Theorem 2.1.
\end{proof}

\begin{Corollary} 
For arbitrary natural numbers $m,n$
$$ \sum\limits^n_{r=0} (-1)^r {n \choose r} \frac{1}{m+r+1} =\frac{1}
{(n+m+1)  {n+m \choose m}}.$$
\end{Corollary}
\noindent {\it Some consequences of Theorem 2.1 are :} \vskip 2mm

\begin{Corollary}
\begin{eqnarray}
&amp;&amp;\sum\limits^n_{r=m} \frac{(-1)^{n-r}}{ {n \choose r}} = (n+1)
\sum\limits_{r=0}^{n-m} (-1)^r \frac{{n-m+1 \choose r}}{
{m+r \choose r}(m+r+1)}.\\
&amp;&amp;\sum\limits^n_{r=m} \frac{(-1)^r}{ {n \choose r}} = \bigg((-1)^n
+ \frac{(-1)^m}{ {n+1 \choose m}} \bigg) \frac{n+1}{n+2}.
\end{eqnarray}
Consequently, $\displaystyle\sum_r \frac{1}{ {2n \choose 2r}}
\rightarrow 2$ and $\displaystyle\sum_r \frac{1}{ {2n \choose
2r+1}} \rightarrow 0$ as $n \rightarrow \infty$.
\begin{eqnarray}
\sum\limits^n_{r=m} \frac{1}{ {n \choose r}^2}
&amp;=&amp; \frac{(n+1)^2}{(m+1)(n-m+2)} \sum\limits_{i=0}^{n-m} \frac{1}
{ {m+i+1 \choose i} {2n-m+2-i \choose n-i}}
\nonumber \\
&amp;&amp;+ \frac{(n+1)^2}{n+2} \sum\limits_{i=m}^n \frac{1}{(i+1) {2n+2-i
\choose n-i}}.
\end{eqnarray}
\begin{eqnarray}
\sum\limits_{r \geq 0} \frac{(-1)^r}{ {n \choose r}}  {n-r
\choose s-r} =(-1)^n (n+1) \sum\limits_{i \geq 0} \frac{(-1)^i}{i+1}
 {n+1 \choose s+i-n}.
\end{eqnarray}
\begin{eqnarray}
\sum\limits_{i=0}^n \frac{1}{i+1}  {n-i \choose j} =
\sum\limits_{i \geq 0} \frac{(-1)^i}{i+1}  {n+1 \choose i+j+1}.
\end{eqnarray}
In particular, for a prime $p \geq 3$, the numerator of
$1+\frac{1}{2}+ \cdots +
\frac{1}{p-1}$ is a multiple of $p$.\\
\begin{eqnarray}
\sum\limits_{r \geq 0} \frac{1}{2r+1}{ {n+1 \choose 2r+1}} &amp;=&amp;
\sum\limits^n_{i=0} \frac{2^i}{i+1}. \\
\sum\limits_{r \geq 1} \frac{1}{2r} {n+1 \choose 2r} &amp;=&amp;
\sum\limits^n_ {i=0} \frac{2^i-1}{i+1}.
\end{eqnarray}
\end{Corollary}
\begin{proof}
We compare the coefficients of $\theta^n$ in (4) and get (7).
If we further use the expression (5), we get the second identity
(8). The special case $m=0$ of (8) gives
\begin{eqnarray}
\sum\limits_{r=0}^n \frac{(-1)^r}{ {n \choose r}} =(1+(-1)^n)
\frac{n+1}{n+2}.
\end{eqnarray}
This shows that 
$$\sum_r \frac{1}{ {2n \choose
2r}} \rightarrow 2$$
and
$$\sum_r
\frac{1}{ {2n \choose 2r+1}}
\rightarrow 0$$
as $n \rightarrow \infty.$
The next identity (9) follows if we integrate (1) as a function of
$T$ from $0$ to $1$, and use the beta-gamma relation.

Equating the coefficients of $T^s$ for $s \leq n$ in (1),
one arrives at (10).

Further, for $m=0$, since the right hand side of (1)
has terms where $T$ occurs with powers higher
than $n$, one could compare the powers of $T^{n+j+1}$ to get (11).

Applying (11) with $n=p-1$ and $j=0$ yields the assertion on numerators.
When $j=0$, the identity (11) becomes:
\begin{eqnarray}
\sum\limits^n_{i=0} \frac{1}{i+1} = \sum\limits_{r \geq 0} \frac{1}{2r+1}
 {n+1 \choose 2r+1} -\sum\limits_{r \geq 1}
\frac{1}{2r} {n+1 \choose 2r}.
\end{eqnarray}
Combining this with the following identities from \cite{Su} 
\begin{eqnarray}
\sum\limits^n_{r=0} \frac{1}{ {n \choose r}} = \frac{n+1}{2^n} \sum\limits_
{r=0}^n \frac{2^r}{r+1}= \frac{n+1}{2^n} \sum\limits_{j~\mbox{odd}}
 {n+1 \choose j} \frac{1}{j},
\end{eqnarray}
we get the last two identities (12) and (13).
\end{proof}

\section{Combinatorial binomial identities}
\vskip 3mm

\noindent
In this section, we use the basic method of Theorem 2.1 to prove various
combinatorial identities where binomial coefficients occur in the
denominator. We start with the following identity which was discovered by
D. H. Lehmer \cite{Leh}; his proof is different.
\vskip 2mm

\begin{Theorem}
If $|x| &lt; 1$, then
\begin{eqnarray*}
\sum\limits_{m \geq 1} \frac{(2x)^{2m}}{m  {2m \choose m}} &amp;=&amp;
\frac{2x} {\sqrt {1-x^2}} \sin^{-1} x .
\end{eqnarray*}
\end{Theorem}
\begin{proof}
For $|x| &lt; 1,$
\begin{eqnarray}
\sum_{m \geq 1} \frac{(2x)^{2m}}{m {2m \choose m}} &amp;=&amp;\sum_{m \geq
1} (2x)^{2m} \beta (m+1,m)\nonumber\\
&amp;=&amp; \sum\limits_{m \geq 1} (2x)^{2m} \int_0^1 t^m (1-t)^{m-1} dt =
\int_0^1 \frac{4x^2t}{1-4x^2t (1-t)}dt.\nonumber
\end{eqnarray}
after interchanging the sum and the integral and evaluating the
sum. This is further equal to 
$$\int_{-x}^x
\frac{s}{s^2+(1-x^2)}ds + x \int^x_{-x} \frac{1}{s^2+(1-x^2)}ds.$$
on changing variables as $s=x(2t-1)$. Thus, we get
$$
\sum\limits_{m \geq 1} \frac{(2x)^{2m}}{m {2m \choose m}} =
\frac{2x} {\sqrt {1-x^2}} \tan^{-1} \left( \frac{x}{\sqrt {1-x^2}}
\right)=\frac{2x} {\sqrt {1-x^2}} \sin^{-1} x.
$$
\end{proof}

\begin{Corollary}
We have
$$\sum_{m \geq 1} \frac{1}{m {2m \choose m}} = \frac{\pi \sqrt{3}}{9}.$$
$$\sum_{m \geq 1} \frac{1}{{2m \choose m}} = \frac{1}{3} +
\frac{2 \pi}{9 \sqrt{3}}.$$
$$\sum_{m \geq 1} \frac{1}{m^2 {2m \choose m}} = \frac{\pi^2}{18}.$$
\end{Corollary}
\begin{proof}
The first equality is the special case $x=1/2$ of Lehmer's
identity. The second is obtained by differentiating Lehmer's
identity and then putting $x=1/2$. To get the third identity, one
integrates and then puts $x=1/2$.
\end{proof}

\begin{Theorem}
If $n$ and $p$ are positive integers with $p&gt;1$, then
\begin{eqnarray}
\sum\limits_{k=0}^{pn}(-1)^k{pn\choose k}{2pn\choose
2k}^{-1}=\frac{(2pn+1)(1+(-1)^{pn})}{2(pn+1)}.
\end{eqnarray}

\begin{eqnarray}
&amp;&amp; \sum\limits_{k=0}^{pn}(-1)^kk{pn\choose k}{2pn\choose 2k}^{-1}
\nonumber \\
&amp;=&amp;-\frac{pn(2pn+1)}{8}\bigg(\frac{1-(-1)^{pn}}{pn+2}-
\frac{2(1+(-1)^{pn})}{pn+1}
+\frac{1-(-1)^{pn}} {pn}\bigg) .
\end{eqnarray}

\begin{eqnarray}
&amp;&amp; \sum_{k=0}^{pn}(-1)^kk^2{pn\choose k}{2pn\choose
2k}^{-1}\nonumber\\
&amp;=&amp;-\frac{(2pn+1)pn}{8}\bigg(\frac{1-(-1)^{pn}}{pn+2}+\frac{2(1+(-1)^{pn})}{pn+1}
+\frac{1 -(-1)^{pn}}{pn}\bigg) \nonumber \\
&amp;&amp; +\frac{(2pn+1)pn(pn-1)}{32}\bigg(\frac{1+(-1)^{pn}}{pn-1}
-\frac{4(1-(-1)^{pn})}{pn} \nonumber \\
&amp;&amp;+ \frac{6(1+(-1)^{pn})}{pn+1}-\frac{4(1-(-1)^{pn})}{pn+2}
+\frac{1+(-1)^{pn}}{pn+3}\bigg).
\end{eqnarray}
\end{Theorem}

\begin{proof}
We give the proofs of (17) and (18). The proof of (19)
is similar and is omitted here.

We have
\begin{eqnarray}
&amp;&amp; \sum_{k=0}^{pn}(-1)^k{pn\choose k}{2pn\choose 2k}^{-1}\nonumber
\\
&amp;=&amp;\sum_{k=0}^{pn}(-1)^k{pn\choose
k}(2pn+1)\int^1_0t^{2k}(1-t)^{2pn-2k}dt\nonumber\\
&amp;=&amp;(2pn+1)\int^1_0\bigg((1-t)^{2pn}\sum_{k=0}^{pn}{pn\choose
k}{\bigg (}-\frac{t^2}{(1-t)^2}{\bigg )}^k\bigg)dt\nonumber\\
&amp;=&amp;(2pn+1)\int^1_0(1-t)^{2pn}{\bigg (}1-\frac{t^2}{(1-t)^2}{\bigg
)}^{pn}dt\nonumber\\
&amp;=&amp;(2pn+1)\int^1_0(1-2t)^{pn}dt=
\frac{(2pn+1)(1+(-1)^{pn})}{2(pn+1)}.\nonumber
\end{eqnarray}

\noindent
Similarly, once again we obtain
\begin{eqnarray}
&amp;&amp; \sum_{k=0}^{pn}(-1)^kk{pn\choose k}{2pn\choose
2k}^{-1} \nonumber \\
&amp;=&amp;(2pn+1)\int^1_0\bigg((1-t)^{2pn}\sum_{k=0}^{pn}k{pn\choose
k}{\bigg (}-\frac{t^2}{(1-t)^2}{\bigg )}^k\bigg)dt\nonumber\\
&amp;=&amp;-pn(2pn+1)\int^1_0\frac{t^2}{(1-t)^2}{\bigg
(}1-\frac{t^2}{(1-t)^2}{\bigg )}^{pn-1}(1-t)^{2pn}dt\nonumber\\
&amp;=&amp;-pn(2pn+1)\int^1_0t^2(1-2t)^{pn-1}dt.\nonumber
\end{eqnarray}
Thus, equality (18) holds.
\end{proof}

\noindent
We note that (17) becomes Theorem 1.2 of \cite{Tri} when $p=2$.

\begin{Theorem}
If $m$, $n$, $p$ and $q$ are nonnegative integers with $p\geq q$,
then
\begin{eqnarray}
&amp;&amp; \ \ \  \sum_{k=0}^n(-1)^k{m+n+p\choose m+k+q}^{-1}\nonumber\\
&amp;&amp;=\frac{m+n+p+1}{m+n+p+2}\bigg({m+n+p+1\choose
m+q}^{-1}+(-1)^n{m+n+p+1\choose m+n+q+1}^{-1}\bigg).\ \ \ \ \ \ \
\end{eqnarray}
\end{Theorem}
\begin{proof} We have
\begin{eqnarray}
&amp;&amp;\sum_{k=0}^n(-1)^k{m+n+p\choose m+k+q}^{-1}\nonumber\\
&amp;=&amp;\sum_{k=0}^n(-1)^k(m+n+p+1)\int^1_0t^{m+k+q}(1-t)^{n+p-k-q}dt\nonumber\\
&amp;=&amp;(m+n+p+1)\int^1_0\bigg(t^{m+q}(1-t)^{n+p-q}\sum_{k=0}^n\bigg(\frac{-t}{1-t}\bigg)^k\bigg)dt
\nonumber\\
&amp;=&amp;(m+n+p+1)\bigg(\int^1_0t^{m+q}(1-t)^{n+p-q+1}dt\nonumber\\
&amp;&amp;+(-1)^n\int^1_0t^{m+n+q+1}(1-t)^{p-q}dt\bigg)\nonumber\\
&amp;=&amp;\frac{m+n+p+1}{m+n+p+2}\bigg({m+n+p+1\choose m+q}^{-1}
+(-1)^n{m+n+p+1\choose m+n+q+1}^{-1}\bigg).\nonumber
\end{eqnarray}
We note that Theorem 1.5 of \cite{Tri} is a special case of
(20).
\end{proof}

\begin{Theorem}
If $n$ and $m$ are positive integers, then
\begin{eqnarray}
\sum\limits_{k=0}^n(-1)^kk{m+n\choose
m+k}^{-1}&amp;=&amp;\frac{m+n+1}{m+n+3}\bigg(\frac{(-1)^n(n+1)(m+n+3)}{m+n+2}-
\nonumber\\
&amp;&amp;{m+n+2\choose m+1}^{-1}
 -(-1)^n\bigg).
\end{eqnarray}

\begin{eqnarray}
\sum_{k=0}^n(-1)^kk^2{m+n\choose
m+k}^{-1}&amp;=&amp;(m+n+1)\bigg(\frac{(-1)^n(n+1)^2}{m+n+2}-
\frac{(-1)^n(2n+3)}{m+n+3}\nonumber\\
&amp;&amp;+\frac{2}{m+n+4}\bigg({m+n+3\choose m+2}^{-1}+(-1)^n\bigg)\nonumber\\
&amp;&amp;-\frac{1}{m+n+3}{m+n+2\choose m+1}^{-1}\bigg).
\end{eqnarray}
\end{Theorem}
\begin{proof} It is clear that
\begin{eqnarray}
\sum_{k=0}^n(-1)^kk{m+n\choose
m+k}^{-1}&amp;=&amp;\sum_{k=0}^n(-1)^kk(m+n+1)\int^1_0t^{m+k}(1-t)^{n-k}dt\nonumber\\
&amp;=&amp;(m+n+1)\int^1_0\bigg(t^m(1-t)^n\sum_{k=0}^nk\bigg(\frac{-t}{1-t}\bigg)^k\bigg)dt\nonumber\\
&amp;=&amp;(m+n+1)\bigg(\int^1_0(-1)^n(n+1)t^{m+n+1}dt\nonumber\\
&amp;&amp;-\int^1_0t^{m+1}(1-t)^{n+1}dt-\int^1_0(-1)^nt^{m
+n+2}dt\bigg).\nonumber
\end{eqnarray}
Then equality (21) holds. \noindent Similarly, we can verify
(22).
\end{proof}
\begin{Theorem}
For real $|t| &lt; 1$ the infinite series $\sum\limits_{r \geq m}
\frac{t^{n+r}}{ {n+r \choose r}}$ converges to

\begin{eqnarray}
n \sum\limits_{i=1}^{n-1}  {n-1 \choose i} \frac{(t-1)^{n-1-i}}
{i  {m+i \choose i}} -n \sum\limits^m_{i=1} {m \choose i}
\frac{(t-1)^{n-1+i}}{i {n-1+i \choose i}}\nonumber \\
 + n (t-1)^{n-1} \sum\limits_{i=m+1}^{n-1} \frac{1}{i}
+ n(t-1)^{n-1} \log \bigg(\frac{1}{1-t}\bigg).
\end{eqnarray}
\end{Theorem}
\begin{proof}
The proof is completely similar to that of Theorem 2.1. The
only point that has to be taken care of here is that we compute
the sequence of partial sums $\displaystyle\sum_{r=m}^M
\frac{t^{n+r}}{{n+r \choose r}}$ by the above method and observe
that it converges as $M \rightarrow \infty$, to
$$n \sum\limits_{i=1}^{n-1} {n-1 \choose i} \frac{(t-1)^{n-1-i}}{i
{m+i \choose i}} -n \sum\limits_{i=1}^n  {m \choose i}
\frac{(t-1)^{n-1+i}}{i {n-1+i \choose i}} $$
$$
+n (t-1)^{n-1} \sum\limits_{i=m+1}^{n-1} \frac{1}{i}
 +n(t-1)^{n-1} \log (\frac{1}{1-t}).$$
\end{proof}

\begin{Corollary}
\begin{eqnarray}
\sum\limits^\infty_{r=m} \frac{1}{ {n+r \choose r}} =\frac{n}
{(n-1) {m+n-1 \choose n-1}}.
\end{eqnarray}
\begin{eqnarray}
\sum\limits^\infty_{r=0} \frac{1}{(r+1)(r+2) \cdots (r+n)} &amp;=&amp;
\frac{1} {(n-1) (n-1)!}.
\end{eqnarray}
\begin{eqnarray}
\sum\limits^\infty_{r=0} \frac{(-1)^r}{ {n+r \choose r}} &amp;=&amp;
2^{n-1} n \bigg( \log 2 - \sum\limits_{i=1}^{n-1}
\frac{1}{i}\bigg)
-n \sum\limits_{i=1}^{n-1} (-1)^i {n-1 \choose i} \frac{2^{n-1-i}}{i}. \\
\sum\limits_{r=0}^\infty \frac{1}{2^r {n+r \choose r}} &amp;=&amp; 2n
(-1)^{n-1} \bigg( \log 2 + \sum\limits_{i=1}^{n-1} \frac{1 +
(-2)^i {n-1 \choose i}}{i}\bigg).
\end{eqnarray}
\end{Corollary}
\begin{proof}
It is easy to see that the right hand side in Theorem 3.11 has a
finite limit at $t= \pm{1}$ and, hence the theorem is valid for $t
= \pm 1$ also. The specializations $t=1,-1, \frac{1}{2}$ give,
respectively, (24), (26) and (27); (25) is just the case $m=0$ of
(24).
\end{proof}

\begin{Proposition}
\begin{eqnarray}
\sum_{n \geq 2} \frac{(-1)^n}{(2n-2)(2n-1)2n}
= \frac{\pi-3}{4}.\\
\sum_{n \geq 0} \frac{1}{(3n+1)(3n+2)(3n+3)}
= \frac{\pi \sqrt{3} - 3 \log 3}{12}.\\
\sum_{n \geq 0} \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)} = \frac{6 \log
2 - \pi}{24}.
\end{eqnarray}
\end{Proposition}

\begin{proof}
Now,
\begin{eqnarray}
&amp;&amp;\sum_{k \geq 2} (-1)^k \frac{(2k-3)!}{(2k)!}=\sum_{k \geq 2}
(-1)^k \frac{\Gamma(2k-2)}{\Gamma(2k+1)}= \frac{1}{2} \sum_{k \geq
2} (-1)^k \frac{\Gamma(2k-2)\Gamma(3)}{\Gamma(2k+1)}\nonumber\\
&amp;=&amp; \frac{1}{2} \sum_{k \geq 2} (-1)^k \beta(2k-2,3)= \frac{1}{2}
\sum_{k \geq 2} (-1)^k \int_0^1 t^{2k-3} (1-t)^2 dt\nonumber\\
&amp;=&amp; \frac{1}{2} \int_0^1 \frac{(1-t)^2}{t^3} \sum_{k \geq 2}
(-t^2)^k dt = \frac{1}{2} \int_0^1 \frac{t(1-t)^2}{1+t^2}
dt.\nonumber
\end{eqnarray}
This is an easy exercise to evaluate and turns out to be
$\frac{\pi - 3}{4}.$

The second sum is
\begin{eqnarray}
\sum_{n \geq 0} \frac{(3n)!}{(3n+3)!}&amp; = &amp;\frac{1}{2} \sum_{n \geq
0} \frac{\Gamma(3n+1) \Gamma(3)}{\Gamma(3n+4)}=\frac{1}{2} \sum_{n
\geq 0} \beta(3n+1,3)\nonumber\\
&amp;=&amp; \frac{1}{2} \int_0^1 \sum_{n \geq 0} t^{3n} (1-t)^2 dt =
\frac{1}{2} \int_0^1 \frac{1-t}{1+t+t^2} dt.\nonumber
\end{eqnarray}
One can easily compute this to be
$$
\sum_{n \geq 0} \frac{1}{(3n+1)(3n+2)(3n+3)}
= \frac{\pi \sqrt{3} - 3 \log 3}{12}.
$$
The last one is similar.
\end{proof}

\begin{Proposition}
$$
\mbox{}\hspace{-4cm} \sum \frac{1}{ {6n \choose 6r}} + \sum \frac{1}{ {6n \choose
6r+3}} - \sum \frac{1}{{6n \choose 6r+1}} - \sum \frac{1} {{6n
\choose 6r+4}}$$
$$ = 3 + \frac{1}{3n} +(6n+1) \sum\limits_{s &lt; n} \bigg( \frac{1}{3s} +
\frac{1}{6s+1} - \frac{1}{6s+2} - \frac{2}{6s+3} - \frac{1}{6s+4}
+ \frac{1}{6s+5}\bigg).
$$
\end{Proposition}
\begin{proof}
It is clear that the polynomial identity (1) of Theorem 2.1
holds for complex numbers $z$. Let $m=0$ and take $z$ to be a root
of $z(1-z)=1$; this means that $z$ is a primitive 6-th root of
unity. Let us write
\begin{eqnarray*}
L_i &amp;=&amp; \sum\limits_{r \equiv i~ {\rm mod}~ 3} \frac{1}{ {n \choose r}}
~~~\mbox{for}~~i=0,1,2 \\
R_i &amp;=&amp; \sum\limits_{\stackrel{1 \leq r \leq n+1}{r \equiv i~ {\rm mod}~6}}
\frac{1}{r}~~~ \mbox{for}~~r=0,1,2,3,4,5
\end{eqnarray*}
We have then
\begin{eqnarray}
L_0-L_1+z (L_1-L_2)=(n+1) z^n (2R_0+R_1-R_2-2R_3-R_4+R_5).
\end{eqnarray}
The identity of the proposition is the special case where $n
\equiv 0$ {\rm mod} $6$.
\end{proof}

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 \bibitem{Pav} Nicolae Pavelescu. Problem C:1280. {\it Gaz. Mat.} {\bf
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 \bibitem{Pla} Juan Pla. The sum of inverses of binomial coefficients
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 \bibitem{Roc} Andrew M. Rockett. Sums of the inverses of binomial
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 \bibitem{Su} B. Sury.  Sum of the reciprocals of the binomial
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\end{thebibliography}


\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: }
binomial coefficients, combinatorial identities,
polynomial identities.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received September 29 2003;
revised version received March 26 2004; July 4 2004.
Published in {\it Journal of Integer Sequences}, July 6 2004.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.math.uwaterloo.ca/JIS/}.
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\end{document}

                                                                                

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