\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amscd}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics,amsmath,amssymb}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{latexsym}
\usepackage{epsf}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{8.9in}

\newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

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

\def\QATOP#1#2{{#1 \atop #2}}%
\def\QDATOP#1#2{{\displaystyle {#1 \atop #2}}}%
%\allowdisplaybreak


\begin{center}
\vskip 1cm{\LARGE\bf General Properties Involving Reciprocals of
\\
\vskip .05in Binomial Coefficients} \vskip 1cm \large
Anthony Sofo \\
School of Computer Science and Mathematics\\
Victoria University \\
P. O. Box 14428 \\
Melbourne, VIC 8001 \\
Australia \\
\href{mailto:anthony.sofo@vu.edu.au}{\tt anthony.sofo@vu.edu.au} \\
\end{center}

\vskip .2 in
\begin{abstract}
Using the properties of the Beta function, we investigate the
representation of infinite series involving the reciprocals of
binomial coefficients. We confirm and generalize some of the
recent results of Sury, Wang and Zhao.
\end{abstract}


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

\theoremstyle{plain}
\newtheorem{mainthm}{Theorem}
\newtheorem{bigthm}{Theorem}
\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}[thm]{Proposition}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{defn}[thm]{Definition}
\newtheorem{defs}[thm]{Definitions}
\newtheorem{example}[thm]{Example}


\theoremstyle{remark}
\newtheorem*{remark}{Remark}

%\setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in}
%\setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in}
%\setlength{\textheight}{8.9in}

%\newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}}


%\begin{document}

\section{Introduction}

The binomial coefficients are defined by%
\begin{equation*}
\left(
\begin{array}{c}
n \\
m%
\end{array}%
\right) =\left\{
\begin{array}{ll}
\dfrac{n!}{m!\left( n-m\right) !}; & n\geq m, \\
&  \\
0; & n<m%
\end{array}%
\right.
\end{equation*}
for $n$ and $m$ non-negative integers.

Binomial coefficients play an important role in many areas of
mathematics, including number theory, statistics and probability.
Reciprocal binomial coefficients are also prolific in the
mathematical literature and many results on reciprocals of
binomial coefficient identities may be seen in the
papers of Mansour \cite{Man}, Pla \cite{Pla}, Rockett \cite{Roc}, Sury \cite%
{Sur}, Sury, Wang and Zhao \cite{SWZ}, Trif \cite{Tri}, and Zhao
and Wang \cite{ZW}.

Sury \cite{Sur} used the Beta function%
\begin{equation*}
B\left( p,q\right) =\frac{\Gamma \left( p\right) \Gamma \left( q\right) }{%
\Gamma \left( p+q\right) }
\end{equation*}%
to observe that%
\begin{align*}
\frac{1}{\left( \QATOP{n}{m}\right) }& =\frac{m!\left( n-m\right) !}{n!} \\
& =\frac{\Gamma \left( m+1\right) \Gamma \left( n-m+1\right)
}{\Gamma \left(
n+1\right) } \\
& =\left( n+1\right) \int_{0}^{1}t^{m}\left( 1-t\right) ^{n-m}dt
\end{align*}%
Utilising the integral identity for the inverse binomial
coefficients, Sury
and Trif further showed that%
\begin{equation*}
\sum_{m=0}^{n}\frac{1}{\left( \QATOP{n}{m}\right) }=\frac{n+1}{2^{n}}%
\sum_{m=0}^{n}\frac{2^{m}}{m+1}=\frac{n+1}{2^{n}}\sum_{j\ \text{odd}}\frac{1%
}{j}\left( \QATOP{n+1}{j}\right) .
\end{equation*}%
Sury, Wang and Zhao \cite{SWZ} proved the following theorem.

\begin{theorem}
\label{t1}In the ring of $Q\left[ T\right] $ of rational
polynomials, the
identity%
\begin{multline}
\qquad \sum_{r=m}^{n}\frac{T^{r}\left( 1-T\right) ^{n-r}}{\left( \QATOP{n}{r}%
\right) }  \label{1.1} \\
=\left( n+1\right) \sum_{r=m}^{n}\frac{T^{n+1}\left( 1-T\right) ^{n-r}}{r+1}%
+\left( n+1\right) \sum_{r=0}^{n-m}\frac{T^{n-r}\left( 1-T\right) ^{n-m+1}}{%
\left( m+r+1\right) \left( \QATOP{m+r}{r}\right) }\qquad
\end{multline}%
holds for $m\leq n.$ An equivalent form is that for $\lambda \neq -1$%
\begin{multline}
\sum_{r=m}^{n}\frac{\lambda ^{r}}{\left( \QATOP{n}{r}\right)
}=\left( n+1\right) \sum_{r=0}^{n-m}\frac{\lambda ^{m+r}}{\left(
\lambda +1\right) ^{r+1}}\sum_{i=0}^{n-m-r}\left(
\QDATOP{n-m-r}{i}\right) \frac{\left(
-1\right) ^{i}}{m+1+i}  \label{1.2} \\
+\left( n+1\right) \frac{\lambda ^{n+1}}{\left( \lambda +1\right) ^{n+2}}%
\sum_{r=m}^{n}\frac{\left( \lambda +1\right) ^{r+1}}{r+1}.
\end{multline}
\end{theorem}

By the use of Theorem \ref{t1} and noting that for $\left\vert
x\right\vert
<1,$%
\begin{equation*}
\sum_{r=1}^{\infty }\frac{\left( 2x\right) ^{2r}}{r\left( \QATOP{2r}{r}%
\right) }=\frac{2x\arcsin \left( x\right) }{\sqrt{1-x^{2}}}=\int_{0}^{1}%
\frac{4x^{2}t}{1-4x^{2}t\left( 1-t\right) }dt.
\end{equation*}%
Sury, Wang and Zhao \cite{SWZ} showed, among other results, that%
\begin{gather}
\sum_{n=0}^{\infty }\frac{1}{\left( n+1\right) \left( n+2\right)
\cdots \left( n+j\right) }=\frac{1}{\left( j-1\right) \left(
j-1\right) !},
\label{1.3} \\
\sum_{n=0}^{\infty }\frac{1}{\left( 3n+1\right) \left( 3n+2\right)
\left(
3n+3\right) }=\frac{\pi \sqrt{3}-3\ln 3}{12},  \label{1.4} \\
\sum_{n=0}^{\infty }\frac{1}{\left( 4n+1\right) \left( 4n+2\right)
\left( 4n+3\right) \left( 4n+4\right) }=\frac{6\ln 2-\pi }{24}
\label{1.5}
\end{gather}%
and%
\begin{multline}
\qquad \sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{\left( \QATOP{n+j}{n}%
\right) }=j2^{j-1}\left( \ln 2-\sum_{r=1}^{j-1}\frac{1}{r}\right)
-j\sum_{r=1}^{j-1}\left( -1\right) ^{r}\left( \QATOP{j-1}{r}\right) \frac{%
2^{j-1-r}}{r},  \label{1.6} \\
\text{for \ }j=1,2,\dots\qquad
\end{multline}%
Identities (\ref{1.3}) to (\ref{1.6}) are reciprocal binomial
identities of
the form%
\begin{equation}
S\left( a,j\right) =\frac{1}{j!}\sum_{n=0}^{\infty }\frac{1}{\left( \QATOP{%
an+j}{an}\right) }  \label{1.7}
\end{equation}%
and
\begin{equation}
T\left( a,j\right) =\frac{1}{j!}\sum_{n=0}^{\infty }\frac{\left(
-1\right) ^{n}}{\left( \QATOP{an+j}{an}\right) }  \label{1.8}
\end{equation}%
for $j=1,2,3\dots ,$ $a\in \mathbb{R}^{+}\backslash \left\{
0\right\} .$

The identity (\ref{1.3})%
\begin{equation*}
S\left( 1,j\right) =\frac{1}{\left( j-1\right) \left( j-1\right)
!}
\end{equation*}%
and%
\begin{align*}
S\left( 2,j\right) & =\frac{2^{j-2}}{\left( j-1\right) !}\left[
\ln 2+\sum_{r=1}^{j-2}\left( -1\right) ^{r}\left(
\QATOP{j-2}{r}\right) \left(
\frac{2^{r}-1}{r\cdot 2^{r}}\right) \right] \\
& =j!\ _{3}F_{2}\left[ \left.
\begin{array}{c}
\frac{1}{2},1,1, \\[5pt]
\frac{1+j}{2},\frac{2+j}{2}%
\end{array}%
\right\vert 1\right]
\end{align*}%
and others were also previously given by Sofo \cite{Sof}.

In this paper we shall extend the range of identities for $S\left(
a,j\right) ,$ $T\left( a,j\right) ,$ give particular closed form
representations to $S\left( \frac{1}{b},j\right) $ and $T\left( \frac{1}{b}%
,j\right) $ for $b=2,3,\dots .$

Finally, we shall give a generalization to both $S\left( a,j\right) $ and $%
T\left( a,j\right) .$

\section{Identity Representations}

Transform techniques are used extensively in the analysis of
series and in their representation in closed form.

In his work, Wheelon \cite{Whe}, and later Sofo \cite{Sof}
essentially
showed that%
\begin{align*}
S\left( a,j\right) & =\frac{1}{j!}\sum_{n=0}^{\infty }\frac{1}{\left( \QATOP{%
an+j}{an}\right) } \\
& =\sum_{n=0}^{\infty }\frac{1}{\prod_{k=1}^{j}\left( an+k\right) } \\
& =\frac{1}{\left( j-1\right) !}\int_{x=0}^{1}\frac{\left( 1-x\right) ^{j-1}%
}{1-x^{a}}dx \\
& =\frac{1}{j!}\ _{j+1}F_{j}\left[ \left.
\begin{array}{c}
1,\frac{1}{a},\frac{2}{a},\frac{3}{a},\dots ,\frac{j}{a} \\[5pt]
1+\frac{1}{a},1+\frac{2}{a},1+\frac{3}{a},\dots ,1+\frac{j}{a}%
\end{array}%
\right\vert 1\right] .
\end{align*}

We can use the ideas of Sury, Wang and Zhao \cite{SWZ} by
implementing the Beta function to state the following theorem.

\begin{theorem}
\label{t2.1}Let $a\in \mathbb{R}^{+}\backslash \left\{ 0\right\} $ and $%
j=2,3,4,\dots $. Then%
\begin{align}
S\left( a,j\right) & =\frac{1}{j!}\sum_{n=0}^{\infty }\frac{1}{\left( \QATOP{%
an+j}{an}\right) }  \label{2.1} \\
& =\sum_{n=0}^{\infty }\frac{1}{\prod_{k=1}^{j}\left( an+k\right)
}
\label{2.2} \\
& =\frac{1}{\left( j-1\right) !}\int_{x=0}^{1}\frac{\left( 1-x\right) ^{j-1}%
}{1-x^{a}}dx  \label{2.3} \\
& =\frac{1}{j!}\ _{j+1}F_{j}\left[ \left.
\begin{array}{c}
1,\frac{1}{a},\frac{2}{a},\frac{3}{a},\dots ,\frac{j}{a} \\[5pt]
1+\frac{1}{a},1+\frac{2}{a},1+\frac{3}{a},\dots ,1+\frac{j}{a}%
\end{array}%
\right\vert 1\right]  \label{2.4}
\end{align}%
and similarly%
\begin{align}
T\left( a,j\right) & =\frac{1}{j!}\sum_{n=0}^{\infty }\frac{\left(
-1\right)
^{n}}{\left( \QATOP{an+j}{an}\right) }  \label{2.5} \\
& =\sum_{n=0}^{\infty }\frac{\left( -1\right)
^{n}}{\prod_{k=1}^{j}\left(
an+k\right) }  \label{2.6} \\
& =\frac{1}{\left( j-1\right) !}\int_{x=0}^{1}\frac{\left( 1-x\right) ^{j-1}%
}{1+x^{a}}dx  \label{2.7} \\
& =\frac{1}{j!}\ _{j+1}F_{j}\left[ \left.
\begin{array}{c}
1,\frac{1}{a},\frac{2}{a},\frac{3}{a},\dots ,\frac{j}{a} \\[5pt]
1+\frac{1}{a},1+\frac{2}{a},1+\frac{3}{a},\dots ,1+\frac{j}{a}%
\end{array}%
\right\vert -1\right] .  \label{2.8}
\end{align}
\end{theorem}

\begin{proof}
Consider the alternating case%
\begin{align*}
T\left( a,j\right) & :=\frac{1}{j!}\sum_{n=0}^{\infty
}\frac{\left(
-1\right) ^{n}}{\left( \QATOP{an+j}{an}\right) } \\
& =\frac{1}{\left( j-1\right) !}\sum_{n=0}^{\infty }\left( -1\right) ^{n}%
\frac{\Gamma \left( j\right) \Gamma \left( an+1\right) }{\Gamma
\left(
an+j+1\right) } \\
& =\frac{1}{\left( j-1\right) !}\sum_{n=0}^{\infty }\left(
-1\right)
^{n}B\left( an+1,j\right) \\
& =\frac{1}{\left( j-1\right) !}\sum_{n=0}^{\infty
}\int_{x=0}^{1}\left( -1\right) ^{n}x^{an}\left( 1-x\right)
^{j-1}dx.
\end{align*}%
Interchanging the sum and integral, we have%
\begin{align*}
T\left( a,j\right) & =\frac{1}{\left( j-1\right) !}\int_{x=0}^{1}%
\sum_{n=0}^{\infty }\left( -1\right) ^{n}x^{an}\left( 1-x\right) ^{j-1}dx \\
& =\frac{1}{\left( j-1\right) !}\int_{x=0}^{1}\frac{\left( 1-x\right) ^{j-1}%
}{1+x^{a}}dx
\end{align*}%
which is the result (\ref{2.7}).%
\begin{equation*}
B\left( \alpha ,\beta \right) =\int_{u=0}^{1}u^{\alpha -1}\left(
1-u\right) ^{\beta -1}du,\quad \text{for }\alpha >0\text{ and
}\beta >0
\end{equation*}%
is the classical Beta function.

From the ratio of successive terms of (\ref{2.5})%
\begin{equation*}
\frac{V_{n+1}}{V_{n}}=-\frac{\prod_{j=1}^{k}\left( an+k\right) }{%
\prod_{j=1}^{k}\left( an+a+k\right) }
\end{equation*}%
we arrive at the result (\ref{2.8}).

The proof of $S\left( a,j\right) $ follows a similar argument as
that used for $T\left( a,j\right) $ and will not be done here.
\end{proof}

In the next theorem we consider a particular case for the value of
$a.$

\begin{theorem}
\label{t2.2}In the case that $a=\frac{1}{b},$ for $b$ an even
positive
integer, then $T\left( \frac{1}{b},j\right) $ forms the rational numbers%
\begin{align}
T\left( \frac{1}{b},j\right) & =\sum_{n=0}^{\infty }\frac{\left(
-1\right)
^{n}}{\prod_{k=1}^{j}\left( \frac{n}{b}+k\right) }  \label{2.9} \\
& =\frac{1}{\left( j-1\right) }\sum_{\mu =0}^{b-1}\frac{\left(
-1\right) ^{\mu }}{\prod_{\nu =1}^{j-1}\left( \nu +\frac{\mu
}{b}\right) }  \notag
\end{align}%
and similarly for $b=2,3,\dots $,
\begin{align}
S\left( \frac{1}{b},j\right) & =\sum_{n=0}^{\infty }\frac{1}{%
\prod_{k=1}^{j}\left( \frac{n}{b}+k\right) }  \label{2.10} \\
& =\frac{1}{\left( j-1\right) }\sum_{\mu
=0}^{b-1}\frac{1}{\prod_{\nu =1}^{j-1}\left( \nu +\frac{\mu
}{b}\right) }.  \notag
\end{align}
\end{theorem}

\begin{proof}
Consider%
\begin{align*}
T\left( \frac{1}{b},j\right) & =\sum_{n=0}^{\infty }\frac{\left(
-1\right)
^{n}}{\prod_{k=1}^{j}\left( \frac{n}{b}+k\right) } \\
& =\frac{1}{\left( j-1\right) !}\int_{x=0}^{1}\frac{\left( 1-x\right) ^{j-1}%
}{1+x^{\frac{1}{b}}}dx \\
& =\frac{1}{\left( j-1\right) !}\int_{x=0}^{1}\left( 1-x\right)
^{j-2}\left(
\frac{1-x}{1+x^{\frac{1}{b}}}\right) dx \\
& =\frac{1}{\left( j-1\right) !}\int_{x=0}^{1}\left( 1-x\right)
^{j-2}\sum_{\mu =0}^{b-1}\left( -1\right) ^{\mu }x^{\frac{\mu }{b}}dx \\
& =\frac{1}{\left( j-1\right)
!}\int_{x=0}^{1}\sum_{r=0}^{j-2}\left( -1\right) ^{r}\left(
\QATOP{j-2}{r}\right) x^{r}\sum_{\mu =0}^{b-1}\left(
-1\right) ^{\mu }x^{\frac{\mu }{b}}dx \\
& =\frac{1}{\left( j-1\right) !}\sum_{r=0}^{j-2}\left( -1\right)
^{r}\left(
\QATOP{j-2}{r}\right) \sum_{\mu =0}^{b-1}\frac{\left( -1\right) ^{\mu }}{r+1+%
\frac{\mu }{b}} \\
& =\frac{1}{\left( j-1\right) !}\sum_{\mu =0}^{b-1}\frac{\left(
-1\right)
^{\mu }}{\left( 1+\frac{\mu }{b}\right) \left( \QATOP{j-1+\frac{\mu }{b}}{j-2%
}\right) } \\
& =\frac{1}{\left( j-1\right) }\sum_{\mu =0}^{b-1}\frac{\left(
-1\right) ^{\mu }}{\prod_{\nu =1}^{j-1}\left( \nu +\frac{\mu
}{b}\right) },
\end{align*}%
which is the result (\ref{2.9}).

From (\ref{2.8}) it is also interesting to note that for $b$ an
even
positive integer%
\begin{equation*}
_{j+1}F_{j}\left[ \left.
\begin{array}{c}
1,b,2\cdot b,3\cdot b,\dots ,j\cdot b \\[5pt]
1+b,1+2\cdot b,1+3\cdot b,\dots ,1+j\cdot b%
\end{array}%
\right\vert -1\right] =j\sum_{\mu =0}^{b-1}\frac{\left( -1\right) ^{\mu }}{%
\prod_{\nu =1}^{j-1}\left( \nu +\frac{\mu }{b}\right) }
\end{equation*}%
The result (\ref{2.10}) can be proved in the same way and will not
be detailed here.

Again from (\ref{2.4}) it is interesting to note that for
$b=2,3,\dots $
\begin{equation*}
_{j+1}F_{j}\left[ \left.
\begin{array}{c}
1,b,2\cdot b,3\cdot b,\dots ,j\cdot b \\[5pt]
1+b,1+2\cdot b,1+3\cdot b,\dots ,1+j\cdot b%
\end{array}%
\right\vert 1\right] =j\sum_{\mu =0}^{b-1}\frac{1}{\prod_{\nu
=1}^{j-1}\left( \nu +\frac{\mu }{b}\right) }.
\end{equation*}
\end{proof}

\section{Examples}

In the case when $a$ is a positive integer, by known properties of
the
hypergeometric function we may state that%
\begin{multline*}\qquad
_{j+1}F_{j}\left[ \left.
\begin{array}{c}
1,\frac{1}{a},\frac{2}{a},\frac{3}{a},\dots ,\frac{j}{a} \\[5pt]
1+\frac{1}{a},1+\frac{2}{a},1+\frac{3}{a},\dots ,1+\frac{j}{a}%
\end{array}%
\right\vert -1\right] \\
=\ _{a+1}F_{a}\left[ \left.
\begin{array}{c}
1,\frac{1}{a},\frac{2}{a},\frac{3}{a},\dots ,\frac{a}{a} \\[5pt]
\frac{1+j}{a},\frac{2+j}{a},\frac{3+j}{a},\dots ,\frac{a+j}{a}%
\end{array}%
\right\vert -1\right] .\qquad
\end{multline*}%
Consequently, we may write%
\begin{align*}
T\left( a,j\right) & =\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{%
\prod_{k=1}^{j}\left( an+k\right) } \\
& =\frac{1}{j!}\ \ _{a+1}F_{a}\left[ \left.
\begin{array}{c}
1,\frac{1}{a},\frac{2}{a},\frac{3}{a},\dots ,\frac{a}{a} \\[5pt]
\frac{1+j}{a},\frac{2+j}{a},\frac{3+j}{a},\dots ,\frac{a+j}{a}%
\end{array}%
\right\vert -1\right] .
\end{align*}

\begin{enumerate}
\item[(i)] For $a=1$ we have,%
\begin{align*}
T\left( 1,j\right) & =\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{%
\prod_{k=1}^{j}\left( n+k\right) } \\
& =\frac{1}{j!}\ \ _{2}F_{1}\left[ \left.
\begin{array}{c}
1,1 \\[5pt]
1+j%
\end{array}%
\right\vert -1\right] \\
& =\frac{2^{j-1}\ln 2}{\left( j-1\right) !}+\frac{2^{j-1}}{\left(
j-1\right) !}\sum_{r=1}^{j-1}\left( -1\right) ^{r}\left(
\QATOP{j-1}{r}\right) \left(
\frac{2^{r}-1}{r\cdot 2^{r}}\right) \\
& =\left\{
\begin{array}{l}
2\ln 2-1;\hfill \text{for \ }j=2 \\
\\
\dfrac{2^{j-2}}{\left( j-2\right) !}\left\{ _{3}F_{2}\left[ \left.
\begin{array}{c}
1,1,2-j \\[5pt]
2,2%
\end{array}%
\right\vert -1\right] -\ \ _{3}F_{2}\left[ \left.
\begin{array}{c}
1,1,2-j \\[5pt]
2,2%
\end{array}%
\right\vert -\frac{1}{2}\right] \right\} ; \\[10pt]
\hfill \text{for \ }j>2.%
\end{array}%
\right.
\end{align*}%
This confirms the result (1.6), in the paper of Sury, Wang and Zhao \cite%
{SWZ}.

\item[(ii)] For $a=2$ and $j=4m+5,$ $m=0,1,2,\dots ,$ we may
extract the
following result%
\begin{align*}
T\left( 2,4m+5\right) & :=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{%
\prod_{k=1}^{4m+5}\left( 2n+k\right) } \\
& =\frac{1}{\left( 4m+5\right) !}\ _{3}F_{2}\left[ \left.
\begin{array}{c}
\frac{1}{2},1,1 \\[5pt]
2m+3,\frac{4m+7}{2}%
\end{array}%
\right\vert -1\right] \\
& =\frac{2^{2m+2}}{\left( 4m+4\right) !}\left\{ \left( -1\right) ^{m+1}\frac{%
\pi }{4}+\sum_{p=0}^{m}\frac{\left( -1\right) ^{p}}{2m-2p+1}\right. \\
& \qquad \left. +\sum_{r=1}^{2m+2}\left( -\frac{1}{2}\right)
^{r}\left(
\QATOP{2m+2}{r}\right) \sum_{s=0}^{r-1}\left( \QATOP{r-1}{s}\right) \frac{1}{%
2m+2s+3-r}\right\} \\
& =\frac{-\left( -1\right) ^{m}4^{m}\pi }{\left( 4m+4\right) !}+\frac{%
2^{2m+2}}{\left( 4m+4\right) !}\left\{ \sum_{p=0}^{m}\frac{\left(
-1\right)
^{p}}{2m-2p+1}\right. \\
& \qquad \left. +\sum_{r=1}^{2m+2}\frac{\left( -\frac{1}{2}\right) ^{r}}{%
2m+3-r}\left( \QATOP{2m+2}{r}\right) \ _{2}F_{1}\left[ \left.
\begin{array}{c}
1-r,m+\frac{3}{2}-\frac{r}{2} \\[5pt]
m+\frac{5}{2}-\frac{r}{2}%
\end{array}%
\right\vert -1\right] \right\} .
\end{align*}%
A rearrangement of the above result highlights an identity for $\pi ;$%
\begin{multline*}
\left( -1\right) ^{m}\frac{\pi }{4}=\sum_{p=0}^{m}\frac{\left( -1\right) ^{p}%
}{2m-2p+1}+\sum_{r=1}^{2m+2}\left( -\frac{1}{2}\right) ^{r}\left( \QATOP{2m+2%
}{r}\right) \sum_{s=0}^{r-1}\frac{\left( \QATOP{r-1}{s}\right)
}{2m+2s+3-r}
\\
-\frac{\left( 4m+4\right) !}{2^{2m+2}}\sum_{n=0}^{\infty
}\frac{\left( -1\right) ^{n}}{\prod_{k=1}^{4m+5}\left( 2n+k\right)
};\quad m=0,1,2,\dots .
\end{multline*}%
In particular, for $m=0$%
\begin{equation*}
\frac{\pi }{4}=\frac{5}{6}-6\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{%
\left( 2n+1\right) \left( 2n+2\right) \left( 2n+3\right) \left(
2n+4\right) \left( 2n+5\right) },
\end{equation*}%
for $m=1$%
\begin{equation*}
\frac{\pi }{4}=\frac{109}{140}+2520\sum_{n=0}^{\infty
}\frac{\left( -1\right) ^{n}}{\prod_{k=1}^{9}\left( 2n+k\right) }.
\end{equation*}
\end{enumerate}

\noindent \textbf{Note: }For $n=0,$ the first term bounds $\pi $ as follows:%
\begin{equation*}
\frac{3958}{1260}<\pi <\frac{3959}{1260}<\frac{22}{7}.
\end{equation*}

\begin{remark}
The series (\ref{2.2}), $S\left( a,j\right) $ and (\ref{2.6}),
$T\left( a,j\right) $ can be expressed in terms of the Lerch
transcendent.

In particular, from (\ref{2.6})%
\begin{align*}
T\left( a,j\right) & :=\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{%
\prod_{k=1}^{j}\left( an+k\right) } \\
& =\sum_{n=0}^{\infty }\left( -1\right) ^{n}\sum_{k=1}^{j}\frac{A_{j,k}}{%
\left( an+k\right) },
\end{align*}%
where%
\begin{align*}
A_{j,k}& =\lim_{n\rightarrow \left( -\frac{k}{a}\right) }\left\{ \frac{an+k}{%
\prod_{k=1}^{j}\left( an+k\right) }\right\} \\
& =\frac{\left( -1\right) ^{k+1}}{\left( k-1\right) !\left(
j-k\right) !}.
\end{align*}%
Hence%
\begin{align}
T\left( a,j\right) & =\sum_{n=0}^{\infty }\left( -1\right) ^{n}\sum_{k=1}^{j}%
\frac{\left( -1\right) ^{k+1}}{\left( k-1\right) !\left(
j-k\right) !\left(
an+k\right) }  \label{3.1} \\
& =\sum_{k=1}^{j}\frac{\left( -1\right) ^{k+1}}{\left( k-1\right)
!\left( j-k\right) !}\sum_{n=0}^{\infty }\frac{\left( -1\right)
^{n}}{\left(
an+k\right) }  \notag \\
& =\frac{1}{2a}\sum_{k=1}^{j}\frac{\left( -1\right) ^{k+1}}{\left(
k-1\right) !\left( j-k\right) !}\left\{ \psi \left( \frac{1}{2}+\frac{k}{2a}%
\right) -\psi \left( \frac{k}{2a}\right) \right\} ,  \notag
\end{align}%
where $\psi \left( z\right) $ is the Psi, or digamma function.
\end{remark}

The Lerch transcendent, $\phi \left( z,s,\alpha \right) $ is defined as%
\begin{equation*}
\phi \left( z,s,\alpha \right) =\sum_{n=0}^{\infty
}\frac{z^{n}}{\left( n+\alpha \right) ^{s}},
\end{equation*}%
where the $\left( n+\alpha \right) =0$ term is excluded from the
sum.

The polygamma functions $\psi ^{\left( k\right) }\left( z\right)
,$ $k\in
\mathbb{N}$ are defined by%
\begin{equation*}
\psi ^{\left( k\right) }\left( z\right)
:=\frac{d^{k+1}}{dz^{k+1}}\log \Gamma \left( z\right)
=\frac{d^{k}}{dz^{k}}\left( \frac{\Gamma ^{\prime
}\left( z\right) }{\Gamma \left( z\right) }\right) ,\quad k\in \mathbb{N}%
_{0}:=\mathbb{N}\cup \left\{ 0\right\} ,
\end{equation*}%
where $\psi ^{\left( 0\right) }\left( z\right) =\psi \left(
z\right) ,$
denotes the Psi, or digamma function, defined by%
\begin{equation*}
\psi \left( z\right) =\frac{d}{dz}\log \Gamma \left( z\right)
=\frac{\Gamma
^{\prime }\left( z\right) }{\Gamma \left( z\right) }\quad \text{or\quad }%
\log \Gamma \left( z\right) =\int_{1}^{z}\psi \left( t\right) dt.
\end{equation*}%
From (\ref{3.1})%
\begin{align*}
T\left( a,j\right) & =\sum_{k=1}^{j}\frac{\left( -1\right)
^{k+1}}{\left(
k-1\right) !\left( j-k\right) !}\cdot \frac{1}{a}\sum_{n=0}^{\infty }\frac{%
\left( -1\right) ^{n}}{\left( n+\frac{k}{a}\right) } \\
& =\sum_{k=1}^{j}\frac{\left( -1\right) ^{k+1}}{\left( k-1\right)
!\left( j-k\right) !a}\phi \left( -1,1,\frac{k}{a}\right) .
\end{align*}%
Similar technical details can be written about the series
(\ref{2.2}), however, they will not be detailed here.

\section{Generalization}

Both series (\ref{2.1}) and (\ref{2.5}) for $S\left( a,j\right) $ and $%
T\left( a,j\right) $ can be generalized in the following manner.

\begin{theorem}
\label{t4.1}For $m\geq 1$ and $a>0$ and $j$ a positive integer, then%
\begin{align}
S\left( a,j,m\right) & =\frac{1}{j!}\sum_{n=0}^{\infty }\frac{\left( \QATOP{%
n+m-1}{n}\right) }{\left( \QATOP{an+j}{an}\right) }=\frac{1}{j!}%
\sum_{n=0}^{\infty }\frac{\left( \QATOP{n+m-1}{m-1}\right) }{\left( \QATOP{%
an+j}{j}\right) }  \label{4.1} \\
& =\frac{1}{\left( j-1\right) !}\int_{0}^{1}\frac{\left( 1-x\right) ^{j-1}}{%
\left( 1-x^{a}\right) ^{m}}dx  \label{4.2} \\
& =\frac{1}{j!}\ _{j+1}F_{j}\left[ \left.
\begin{array}{c}
m,\frac{1}{a},\frac{2}{a},\frac{3}{a},\dots ,\frac{j}{a} \\[5pt]
1+\frac{1}{a},1+\frac{2}{a},1+\frac{3}{a},\dots ,1+\frac{j}{a}%
\end{array}%
\right\vert 1\right]  \label{4.3} \\
& =\sum_{n=0}^{\infty }\left( m\right) _{n}\left/
n!\prod_{k=1}^{j}\left( an+k\right) \right.  \label{4.31}
\end{align}%
and%
\begin{align}
T\left( a,j,m\right) & =\frac{1}{j!}\sum_{n=0}^{\infty
}\frac{\left(
-1\right) ^{n}\left( \QATOP{n+m-1}{n}\right) }{\left( \QATOP{an+j}{an}%
\right) }  \label{4.4} \\
& =\frac{1}{\left( j-1\right) !}\int_{0}^{1}\frac{\left( 1-x\right) ^{j-1}}{%
\left( 1+x^{a}\right) ^{m}}dx  \label{4.5} \\
& =\frac{1}{j!}\ _{j+1}F_{j}\left[ \left.
\begin{array}{c}
m,\frac{1}{a},\frac{2}{a},\frac{3}{a},\dots ,\frac{j}{a} \\[5pt]
1+\frac{1}{a},1+\frac{2}{a},1+\frac{3}{a},\dots ,1+\frac{j}{a}%
\end{array}%
\right\vert -1\right]  \label{4.6} \\
& =\sum_{n=0}^{\infty }\left( -1\right) ^{n}\left( m\right)
_{n}\left/ n!\prod_{k=1}^{j}\left( an+k\right) \right. .
\label{4.7}
\end{align}
\end{theorem}

\begin{proof}
Consider%
\begin{align*}
T\left( a,j,m\right) & =\frac{1}{j!}\sum_{n=0}^{\infty }\left( -1\right) ^{n}%
\frac{\left( \QATOP{n+m-1}{n}\right) }{\left( \QATOP{an+j}{an}\right) }=%
\frac{1}{j!}\sum_{n=0}^{\infty }\left( -1\right) ^{n}\frac{\left( \QATOP{%
n+m-1}{m-1}\right) }{\left( \QATOP{an+j}{j}\right) } \\
& =\frac{1}{\left( j-1\right) !}\sum_{n=0}^{\infty }\left(
-1\right) ^{n}\left( \QATOP{n+m-1}{n}\right) \frac{\Gamma \left(
an+1\right) \Gamma
\left( j\right) }{\Gamma \left( an+1+j\right) } \\
& =\frac{1}{\left( j-1\right) !}\sum_{n=0}^{\infty }\left(
-1\right)
^{n}\left( \QATOP{n+m-1}{n}\right) B\left( an+1,j\right) \\
& =\frac{1}{\left( j-1\right) !}\sum_{n=0}^{\infty }\left(
-1\right) ^{n}\left( \QATOP{n+m-1}{n}\right)
\int_{0}^{1}x^{an}\left( 1-x\right) ^{j-1}dx,
\end{align*}%
interchanging sum and integral, we have%
\begin{align*}
T\left( a,j,m\right) & =\frac{1}{\left( j-1\right)
!}\int_{0}^{1}\left(
1-x\right) ^{j-1}\sum_{n=0}^{\infty }\left( -1\right) ^{n}\left( \QATOP{n+m-1%
}{n}\right) x^{an}dx \\
& =\frac{1}{\left( j-1\right) !}\int_{0}^{1}\frac{\left( 1-x\right) ^{j-1}}{%
\left( 1+x^{a}\right) ^{m}}dx,
\end{align*}%
which is (\ref{4.5}).

Next, by considering the ratio of terms in (\ref{4.4}), the
hypergeometric identity (\ref{4.6}) follows. Notice that in the
case when $a$ is a positive integer, because of the properties of
the hypergeometric function, we may
also write%
\begin{equation*}
T\left( a,j,m\right) =\frac{1}{j!}\ _{a+1}F_{a}\left[ \left.
\begin{array}{c}
m,\frac{1}{a},\frac{2}{a},\frac{3}{a},\dots ,\frac{a}{a} \\[5pt]
\frac{1+j}{a},\frac{2+j}{a},\frac{3+j}{a},\dots ,\frac{a+j}{a}%
\end{array}%
\right\vert -1\right] .
\end{equation*}%
To deduce (\ref{4.7}) we may write%
\begin{align*}
T\left( a,j,m\right) & =\frac{1}{j!}\sum_{n=0}^{\infty }\left( -1\right) ^{n}%
\frac{\Gamma \left( m+n\right) }{\Gamma \left( m\right) \Gamma
\left( n+1\right) }\cdot \frac{j!\Gamma \left( an+1\right)
}{\Gamma \left(
an+1+j\right) } \\
& =\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}\left( m\right) _{n}}{%
n!\left( an+1\right) _{j}},
\end{align*}%
where%
\begin{equation*}
\left( p\right) _{\alpha }=p\left( p+1\right) \cdots \left(
p+\alpha -1\right) =\frac{\Gamma \left( p+\alpha \right) }{\Gamma
\left( p\right) }
\end{equation*}%
is Pochhammer's symbol.

Now we can state%
\begin{equation*}
T\left( a,j,m\right) =\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}}{n!}%
\cdot \frac{\left( m\right) _{n}}{\prod_{k=1}^{j}\left(
an+k\right) }.
\end{equation*}%
The identities (\ref{4.2}) and (\ref{4.3}) of $S\left( a,j,m\right) $ in (%
\ref{4.1}) follow in a similar fashion as above and will not be
detailed here.
\end{proof}

Some examples now follow, with the minimum of detail.

\begin{example}
\begin{equation*}
S\left( 1,j,m\right) =\frac{1}{j!}\sum_{n=0}^{\infty }\frac{\left( \QATOP{%
n+m-1}{n}\right) }{\left( \QATOP{n+j}{n}\right) }=\frac{1}{\left(
j-m\right) \left( j-1\right) !},
\end{equation*}%
hence%
\begin{align*}
\sum_{n=0}^{\infty }\frac{\left( \QATOP{n+m-1}{n}\right) }{\left( \QATOP{n+j%
}{n}\right) }& =\frac{j}{j-m}\quad \text{for \ }j\neq m \\
& =\ _{2}F_{1}\left[ \left.
\begin{array}{c}
1,m \\[5pt]
1+j%
\end{array}%
\right\vert 1\right] ,
\end{align*}%
this generalizes the result (\ref{1.3}).
\end{example}

\begin{example}
\begin{equation*}
S\left( 2,9,\frac{9}{2}\right) =\frac{1}{9!}\sum_{n=0}^{\infty
}\frac{\left( \QATOP{n+\frac{7}{2}}{n}\right) }{\left(
\QATOP{2n+9}{2n}\right) },
\end{equation*}%
hence%
\begin{align*}
\sum_{n=0}^{\infty }\frac{\left( \QATOP{n+\frac{7}{2}}{n}\right)
}{\left(
\QATOP{2n+9}{2n}\right) }& =\frac{9\pi }{2}-\frac{456}{35} \\
& =\sum_{n=0}^{\infty }\frac{\left( \frac{9}{2}\right) _{n}}{n!\left( \QATOP{%
2n+9}{2n}\right) }=\ _{3}F_{2}\left[ \left.
\begin{array}{c}
\frac{1}{2},1,\frac{9}{2} \\[5pt]
2,\frac{11}{2}%
\end{array}%
\right\vert 1\right] .
\end{align*}%
Also,%
\begin{equation*}
T\left( 2,9,\frac{9}{2}\right) =\frac{1}{9!}\sum_{n=0}^{\infty
}\left(
-1\right) ^{n}\frac{\left( \QATOP{n+\frac{7}{2}}{n}\right) }{\left( \QATOP{%
2n+9}{2n}\right) },
\end{equation*}%
hence%
\begin{align*}
\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}\left( \QATOP{n+\frac{7}{2}}{n%
}\right) }{\left( \QATOP{2n+9}{2n}\right) }& =9\ln \left( \sqrt{2}+1\right) +%
\frac{2559\sqrt{2}}{35}-\frac{552}{5} \\
& =\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}\left(
\frac{9}{2}\right)
_{n}}{n!\left( \QATOP{2n+9}{2n}\right) } \\
& =\ _{3}F_{2}\left[ \left.
\begin{array}{c}
\frac{1}{2},1,\frac{9}{2} \\[5pt]
2,\frac{11}{2}%
\end{array}%
\right\vert -1\right] .
\end{align*}
\end{example}

\begin{example}
\begin{equation*}
S\left( 3,5,4\right) =\frac{1}{5!}\sum_{n=0}^{\infty }\frac{\left( \QATOP{n+3%
}{n}\right) }{\left( \QATOP{3n+5}{3n}\right) }
\end{equation*}%
hence%
\begin{align*}
\sum_{n=0}^{\infty }\frac{\left( \QATOP{n+3}{n}\right) }{\left( \QATOP{3n+5}{%
3n}\right) }& =\frac{100\sqrt{3}\pi }{243}-\frac{10}{9} \\
& =\ _{4}F_{3}\left[ \left.
\begin{array}{c}
\frac{1}{3},\frac{2}{3},1,4 \\[5pt]
2,\frac{7}{3},\frac{8}{3}%
\end{array}%
\right\vert 1\right] .
\end{align*}%
Also%
\begin{equation*}
T\left( 3,5,4\right) =\frac{1}{5!}\sum_{n=0}^{\infty }\frac{\left(
-1\right) ^{n}\left( \QATOP{n+3}{n}\right) }{\left(
\QATOP{3n+5}{3n}\right) }
\end{equation*}%
hence%
\begin{align*}
\sum_{n=0}^{\infty }\frac{\left( -1\right) ^{n}\left(
\QATOP{n+3}{n}\right)
}{\left( \QATOP{3n+5}{3n}\right) }& =\frac{10}{9}+\frac{40\ln 2}{27}-\frac{%
160\sqrt{3}\pi }{729} \\
& =\ _{4}F_{3}\left[ \left.
\begin{array}{c}
\frac{1}{3},\frac{2}{3},1,4 \\[5pt]
2,\frac{7}{3},\frac{8}{3}%
\end{array}%
\right\vert -1\right] .
\end{align*}
\end{example}

\noindent \textbf{Note: }The series (\ref{4.1}) and (\ref{4.4})
can be generalized further, these results will be reported in
another forum.



\providecommand{\href}[2]{#2}

\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
% \MRhref is called by the amsart/book/proc definition of \MR.
\providecommand{\MRhref}[2]{\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}}


\begin{thebibliography}{9}
\bibitem{Man} T. Mansour, Combinatorial identities and inverse binomial
coefficients, \textit{Adv. Appl. Math.}, \textbf{28} (2002),
196--202.

\bibitem{Pla} J. Pla, The sum of inverses of binomial coefficients
revisited, \textit{Fibonacci Quart}., \textbf{35} (1997),
342--345.

\bibitem{Roc} A. M. Rockett, Sums of the inverses of binomial coefficients,
\textit{Fibonacci Quart}., \textbf{19} (1981), 433--437.

\bibitem{Sof} A. Sofo, \textit{Computational Techniques for the Summation of
Series}, Kluwer Academic/Plenum Publishers, 2003.

\bibitem{Sur} B. Sury, Sum of the reciprocals of the binomial coefficients,
\textit{European J. Combin}., \textbf{14} (1993), 351--535.

\bibitem{SWZ} B. Sury, T. Wang and F. Z. Zhao, \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL7/Sury/sury99.html}{Identities involving
reciprocals of binomial coefficients}, \href{http://www.cs.uwaterloo.ca/journals/JIS/}{\textit{Journal of Integer
Sequences}}, \textbf{7} (2004), Article 04.2.8.

\bibitem{Tri} T. Trif, Combinatorial sums and series involving inverses of
binomial coefficients, \textit{Fibonacci Quart}., \textbf{38}
(2000), 79--84.

\bibitem{Whe} A.D. Wheelon, On the summation of infinite series in closed
form, \textit{Journal of Applied Physics}, \textbf{25} (1954),
113--118.

\bibitem{ZW} F. Zhao and T. Wang, Some results for sums of the inverse of
binomial coefficients, \href{http://www.integers-ejcnt.org/}{\textit{Integers: Electronic Journal of
Combinatorial Number Theory}}, \textbf{5}(1) (2005), \#22.
\end{thebibliography}

\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords:} Binomial coefficients, combinatorial
identities, integral representations.

\bigskip
\hrule
\bigskip

\vspace*{+.1in} \noindent Received February 20 2006; revised version
received  July 27 2006. Published in {\it Journal of Integer
Sequences}, September 7 2006.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}