\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://oeis.org/#1}{\underline{#1}}}

\begin{document}

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

\begin{center}
\vskip 1cm{\LARGE\bf On Sums Involving Binomial Coefficients
}
\vskip 1cm
\large
S. Amghibech\\ 
Rue Wilfrid-L\'egar\'e \\
Qu\'ebec, QC  \\
Canada\\
\href{mailto:amghibech@hotmail.com}{\tt amghibech@hotmail.com} \\
\end{center}

\vskip .2 in
\begin{abstract}
We give closed forms for the series
$\sum_{m=1}^{\infty}\frac{(2x)^{2m+2k}}{m^{2}(m+k)  {2m \choose m}}$ and
$\sum_{m=1}^{\infty}\frac{(2x)^{2m}(-1)^{m+k}}{m^{2}(m+k)  {2m \choose
m}}$ for integers $k \geq 0$.  
\end{abstract}

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


\newcommand{\norm}[1]{\left\Vert#1\right\Vert}
\newcommand{\abs}[1]{\left\vert#1\right\vert}
\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\Real}{\mathbb R}
\newcommand{\eps}{\varepsilon}
\newcommand{\To}{\longrightarrow}
\newcommand{\BX}{\mathbf{B}(X)}
\newcommand{\A}{\mathcal{A}}


\section{Introduction}

    D. H. Lehmer \cite{Leh} studied various series with binomial
coefficients in the denominator, for example,
\begin{eqnarray}
\sum\limits_{m \geq 1} \frac{(2x)^{2m}}{m  {2m \choose m}} &=&
\frac{2x} {\sqrt {1-x^2}} \arcsin x ,
\end{eqnarray}
valid for $|x| < 1$.    In this note we consider some related results.

\section{Main Results}

The main results of the paper can be stated as follows:
\begin{theorem}\label{theorem1}

\begin{itemize}

\item[(a)] For $t\in (-1,1)$ we have
 \begin{eqnarray*}
 \sum_{m = 1}^{\infty} \frac{(2t)^{2m+2k}}{m(2m+2k)  {2m \choose m}}  & = &
   \arcsin(t)^{2}{2k \choose k} +\sum_{j=1}^{k}{2k\choose
   k-j}\frac{(-1)^{j+1}}{j^{2}} \\
     & + & \sum_{j=1}^{k}(-1)^{j}{2k\choose
     k-j}\left(\frac{2\arcsin t\sin(2j\arcsin t)}{j}+\frac{\cos(2j\arcsin t)}{j^{2}}
     \right) .
\end{eqnarray*}

\item[(b)]
If we replace $t$ by $\sqrt{-1}t$, we get another form
 \begin{eqnarray*}
 \sum_{m = 1}^{\infty} \frac{(-1)^{m+k}(2t)^{2m+2k}}{m(2m+2k)  {2m \choose m}}  & = &
   -\log(t+\sqrt{1+t^{2}})^{2}{2k \choose k} +\sum_{j=1}^{k}{2k\choose
   k-j}\frac{(-1)^{j+1}}{j^{2}} \\
   +\sum_{j=1}^{k}(-1)^{j}{2k\choose
     k-j}  &  &
     \left(\frac{(-t+\sqrt{t^{2}+1})^{2j}-(t+\sqrt{t^{2}+1})^{2j}}{j} \right)\log(t+\sqrt{t^{2}+1})\\
     +\sum_{j=1}^{k}(-1)^{j}{2k\choose
     k-j} &  &  \left( \frac{(-t+\sqrt{t^{2}+1})^{2j}+(t+\sqrt{t^{2}+1})^{2j})}{2j^{2}}
     \right) .
\end{eqnarray*}
\end{itemize}
\end{theorem}

\begin{theorem}\label{theorem2}

\begin{itemize}

\item[(a)] For $t\in (-1,1)$ we have
  \begin{eqnarray*}
 \sum_{m = 1}^{\infty} \frac{(2t)^{2m+2k}}{m^{2}(2m+2k)  {2m \choose m}}  & = &
  \frac{ -(\arcsin t)^{2}}{k}{2k \choose k}
  +\frac{(2t)^{2k}}{k}(\arcsin t)^{2}-\sum_{j=1}^{k}{2k\choose
   k-j}\frac{(-1)^{j+1}}{kj^{2}} \\
     & - & \sum_{j=1}^{k}(-1)^{j}{2k\choose  k-j}\left(\frac{2\arcsin t\sin(2j\arcsin t)}{kj}+\frac{\cos(2j\arcsin t)}{kj^{2}}
     \right) .
\end{eqnarray*}

\item[(b)] The identity in (a) is valid if we replace $t$ by $it$ and
 $\arcsin(it)$ by $i\log(1+\sqrt{1+t^{2}})$:
   \begin{eqnarray*}
 \sum_{m = 1}^{\infty} \frac{(2t)^{2m+2k}(-1)^{m+k}}{m^{2}(2m+2k)  {2m \choose m}}  & = &
  \left( \frac{ 1}{k}{2k \choose k}
  -\frac{(2t)^{2k}}{k}\right)(\log(1+\sqrt{1+t^{2}}))^{2}-\sum_{j=1}^{k}{2k\choose
   k-j}\frac{(-1)^{j+1}}{kj^{2}} \\
    -\sum_{j=1}^{k}(-1)^{j}{2k\choose
     k-j}  &  &
     \left(\frac{(-t+\sqrt{t^{2}+1})^{2j}-(t+\sqrt{t^{2}+1})^{2j}}{jk} \right)\log(t+\sqrt{t^{2}+1})\\
     -\sum_{j=1}^{k}(-1)^{j}{2k\choose
     k-j} &  &  \left(
     \frac{(-t+\sqrt{t^{2}+1})^{2j}+(t+\sqrt{t^{2}+1})^{2j})}{2kj^{2}}
     \right) .
\end{eqnarray*}
\end{itemize}
 \end{theorem}


We use the basic method  to prove various combinatorial identities
where binomial coefficients occur in the denominator. As a
particular consequence of our  results we get the sums
$S_{1}(k),S_{2}(k),T_{1}(k),T_{2}(k)$ computed in \cite{Yan} in a
compact form.

\section{Proofs of the Theorems}

\noindent {\it Proof of Theorem \ref{theorem1}.}

We start with the following identity, which was
discovered by D. H. Lehmer \cite{Leh}.
If $|x| < 1$, then
\begin{eqnarray}\label{1}
\sum\limits_{m \geq 1} \frac{(2x)^{2m}}{m  {2m \choose m}} &=&
\frac{2x} {\sqrt {1-x^2}} \arcsin x .
\end{eqnarray}
If we replace $x$ by $\sqrt{-1}x$, we get another form of the
identity (\ref{1}):
\begin{eqnarray*}
\sum\limits_{m \geq 1} \frac{(-1)^{m}(2x)^{2m}}{m  {2m \choose m}} &=&
\frac{-2x} {\sqrt {1+x^2}} \log(x+\sqrt{1+x^{2}})
\end{eqnarray*}

If we multiply both members of the equation (\ref{1}) by $x^{2k-1}$ and then
integrate, we obtain
\begin{eqnarray*}
\sum\limits_{m \geq 1} \frac{2^{2m}t^{2m+2k}}{m(2m+2k)  {2m \choose m}} &=&
\int_{0}^{t}\frac{2x^{2k}} {\sqrt {1-x^2}} \arcsin x dx .
\end{eqnarray*}
The left-hand side of this equation can be written
$$
\int_{0}^{\arcsin t}2x(\sin(x))^{2k}dx=\frac{(-1)^{k}}{2^{2k-1}}
\sum_{j=0}^{2k}{2k \choose j}(-1)^{j}\int_{0}^{\arcsin
t}x\exp(2ix(k-j))dx
$$
from which we get
\begin{eqnarray*}
   \int_{0}^{\arcsin t}2x(\sin(x))^{2k}dx  & = &
   \frac{(\arcsin t)^{2}}{2^{2k}}{2k \choose k}  \\
     & + &\frac{1}{2^{2k-1}} \sum_{j=1}^{k}(-1)^{j}{2k\choose
     k-j}\left(\frac{\tau\sin(2j\tau)}{j}+\frac{\cos(2j\tau)}{2j^{2}}-
     \frac{1}{2j^{2}}\right)
\end{eqnarray*}
where $\tau=\arcsin(t)$.
Finally we get
\begin{eqnarray*}
 \sum\limits_{m \geq 1} \frac{(2t)^{2m+2k}}{m(2m+2k)  {2m \choose m}}  & = &
   (\arcsin t)^{2}{2k \choose k} +\sum_{j=1}^{k}{2k\choose
   k-j}\frac{(-1)^{j+1}}{j^{2}} \\
     & + & \sum_{j=1}^{k}(-1)^{j}{2k\choose
     k-j}\left(\frac{2(\arcsin t)\sin(2j\arcsin t)}{j}+\frac{\cos(2j\arcsin t)}{j^{2}}
     \right),
\end{eqnarray*}
which gives Theorem~\ref{theorem1}.

\bigskip

\noindent {\it Proof of Theorem~\ref{theorem1}.}

\medskip

If we integrate (\ref{1}) from $0$ to $x$ we obtain
$$
\sum\limits_{m \geq 1} \frac{(2x)^{2m}}{m^{2}  {2m \choose m}}
=2(\arcsin x)^{2}
$$
If we multiply both members of this equation by $x^{2k-1}$ and then
integrate, we obtain
\begin{eqnarray*}
\sum\limits_{m \geq 1} \frac{2^{2m}t^{2m+2k}}{m^{2}(2m+2k)  {2m \choose m}} &=&
\int_{0}^{t}2x^{2k-1}  (\arcsin x)^{2}dx .
\end{eqnarray*}
The left-hand side of can be written
$$
\int_{0}^{t}2x^{2k-1}  (\arcsin x)^{2}dx=\frac{t^{2k}}{k}(\arcsin t)^{2}
-\frac{1}{k}\int_{0}^{t}\frac{2x^{2k}}{\sqrt{1-x^{2}}}\arcsin x dx.
$$
By using the proof of Theorem~\ref{theorem1} we get Theorem~\ref{theorem2}.


\begin{thebibliography}{9}

 \bibitem{Leh} D. H. Lehmer. Interesting series involving the central binomial
coefficient. {\it Amer. Math. Monthly} {\bf 92} (1985), 449--457.

\bibitem{Yan} Jin-Hua Yang and Feng-Zhen Zhao,  
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Zhao/zhao20.html}{Sums
involving binomial coefficients}, {\it Journal of Integer Sequences}
{\bf 9} (2006), Article 06.4.2.
\end{thebibliography}

\bigskip
\hrule
\bigskip

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

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

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 26 2006;
revised version received  January 11 2007.
Published in {\it Journal of Integer Sequences}, January 15 2007.

\bigskip
\hrule
\bigskip

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


\end{document}