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\begin{center}
\vskip 1cm{\LARGE\bf Asymptotic Expansions of Certain Sums \\
\vskip .1in
Involving Powers of Binomial Coefficients }
\vskip 1cm \large Chun-Xue
Zhao\footnote{This work was supported by the Science Research
Foundation of Dalian University of Technology
(2008).} and
Feng-Zhen Zhao$^1$ \\
School of Mathematical Sciences \\
Dalian University of Technology \\
Dalian 116024 \\
China \\
\href{mailto:zhaochunxue66@163.com}{\tt zhaochunxue66@163.com} \\
\href{mailto:fengzhenzhao@yahoo.com.cn.}{\tt fengzhenzhao@yahoo.com.cn}\\
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\vskip .2 in

\begin{abstract}
In this paper, we investigate the computation of sums involving
binomial coefficients. We give asymptotic expansions of certain sums
concerning powers of binomial coefficients by using the
multi-Laplace asymptotic integral theorem.
\end{abstract}

\newtheorem{theorem}{Theorem}
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%************************* section 1 *******************************************


\section{Introduction}
Sums related to binomial coefficients are involved in many subjects
such as combinatorial analysis, graph theory, number theory,
statistics and probability. For example, there is an application of
sums involving inverses of binomial coefficients (see \cite{ref4}).
For the computation of sums concerning inverses of binomial
coefficients, we may see for instance \cite{ref10, ref12, ref13,
ref14, ref15, ref16}. Hence, it is important to discuss the
computation of sums concerning binomial coefficients. In this paper,
we study the computation of certain sums involving powers of
binomial coefficients.

For convenience, we first give some notation and definitions
involved in this paper. The binomial coefficient $\binom{n}{m}$ is
defined by
$${n\choose m}=\begin{cases}\displaystyle\frac{n!}{m!(n-m)!}, &amp; n\geq
m;\cr\ 0, &amp; n&lt;m.\end{cases}
$$
where n and m are nonnegative integers.

Throughout this paper, $H_n(n\geq1)$ stands for the harmonic number,
and the definition of $H_n(n\geq 0)$ is
\begin{align*}
&amp;H_{0}=0,\quad  H_{n}=\sum_{k=1}^{n}\frac{1}{k},\quad n=1,2,\ldots.
\end{align*}
The generating function of $H_n$ is
\begin{align*}
\sum_{n=1}^{\infty}H_nt^n=-\frac{\ln(1-t)}{1-t}.
\end{align*}
The harmonic numbers have been generalized by several authors (see
\cite{ref1, ref2, ref5, ref6, ref7, ref8, ref9}):
\begin{align*}
&amp;H_{n}^{[0]}=\frac{1}{n},\quad
H_{n}^{[r]}=\sum_{k=1}^{n}H_{k}^{[r-1]}\quad(n,r\geq
1),\\
&amp;H_{n,0}=1,\ H_{n,r}=\sum_{1\leq n_{1}&lt;\cdots&lt;n_{r}\leq
n}\frac{1}{n_{1}n_{2}\cdots n_{r}}\quad (n,r\geq
1),\\
&amp;H(n,r)=\sum_{1\leq
n_{0}+n_{1}+\cdots+n_{r}\leq n}\frac{1}{n_{0}n_{1}\cdots n_{r}}\quad (n\geq
1,r\geq 0).
\end{align*}
It is clear that $H(n,0)=H_n=H_{n,0}=H_n^{[1]}.$ The generating functions of the above generalized harmonic numbers are as follows:
\begin{eqnarray}
\sum_{n=1}^{\infty}
 H_n^{[r]}  t^n&amp;=&amp;\frac{-\ln{(1-t)}}{(1-t)^r},\label{eq:E1}\\
\sum_{n=r+1}^{\infty}
 H_{n,r}  t^n&amp;=&amp;\frac{(-1)^{r} (\ln{(1-t)})^r}{r!(1-t)},
 \label{eq:E2}\\
\sum_{n=r+1}^{\infty}H(n,r)
t^n&amp;=&amp;\frac{(-1)^{r+1}(\ln(1-t))^{r+1}}{1-t}. \label{eq:E3}
\end{eqnarray}
In this paper, we denote the Stirling numbers of the first kind and
the second kind by $s(n,k)$, $S(n,k)$, respectively. The generating
functions of the two kinds of Stirling numbers are as follows:
\begin{align*}
\sum_{n=k}^{\infty}s(n,k)\frac{t^n}{n!}=\frac{\ln^k{(1+t)}}{k!}, \ \ \ \ \ \  \sum_{n=k}^{\infty}S(n,k)\frac{t^n}{n!}=\frac{(e^t-1)^k}{k!}.
\end{align*}
It is well known that it is difficult to give the accurate values of
sums involving inverses of binomial coefficients. However, the
asymptotic expansions of certain sums related to inverses of
binomial coefficients can be obtained. For the computation of sums
related to inverses of binomial coefficients, integral
representations is an effective method. We know that ${n\choose
m}^{-1}$ is related to an integral (see \cite{ref13}):
\begin{eqnarray}
{n\choose m}^{-1}=(n+1)\int_0^1x^m(1-x)^{n-m}dx. \label{eq:E4}
\end{eqnarray}
We will derive the main results by using \eqref{eq:E4} and the next
lemma (see\cite{ref17}):
\begin{lemma}
\label{lem:1} {Let $D$ be a continuous bounded domain, and let
$\phi(x)=\phi(x_1,x_2,\ldots,x_n)$ and $f(x)=f(x_1,x_2,\ldots,x_n)$
be the real functions defined in $D$
and satisfy the following conditions:\\
(i) The second-order partial derivatives $f_{ik}\ \left(\frac{\partial^{2}f}{\partial{x_i}\partial{x_k}}\ (i,k=1,2,\ldots,n)\right)$ exist and $f(x)&gt;0$ $(x\in D);$\\
(ii) The product $\phi(x)[f(x)]^\lambda\ (\lambda&gt;0)$ is absolutely integrable in $D$;\\
(iii) The function $f(x)$ reaches the effective maximum at the
interior point $\varepsilon
=(\varepsilon_1,\varepsilon_2,\ldots,\varepsilon_n)$ in $D$, i.e.,
$f_k(\varepsilon)=0,$
$H_k[-f(\varepsilon)]&gt;0\ (k=1,2,\ldots,n),$ where, $H_k[-f(x)]=det(-f_{ij}(x))_{1\leq i,j\leq k};$\\
(iv) The function $\phi(x)$ is continuous at point $\varepsilon$ and $\phi(\varepsilon)\neq0.$\\
Then when $ \lambda \to \infty $,
\begin{eqnarray*}
\int\!\!\!\idotsint\limits_D{\phi(x)[f(x)]^\lambda}\, dx_1dx_2\cdots
dx_n \sim
\frac{\phi(\varepsilon)[f(\varepsilon)]^{\lambda+\frac{n}{2}}}{\sqrt
{H_n[-f(\varepsilon)]}}{\left(\frac{2\pi}{\lambda}\right)}^\frac{n}{2}.
\end{eqnarray*}}
\end{lemma}
 In this paper, we discuss the
computation of certain sums related to powers of binomial
coefficients. In Section \ref{sec2}, we study the asymptotic
expansions of sums such as
$$\sum_{n=0}^{\infty}\frac{1}{(2n+2k+1)^2\binom{2n+2k}{n+k}^2},
$$
and in Section \ref{sec3}, we investigate the asymptotic expansions
of sums involving binomial coefficients and generalized harmonic
numbers.

\section{Asymptotic Expansions of a Class of Sums Involving Binomial Coefficients}
\label{sec2}

In this section, we give the asymptotic expansions of sums involving
inverses of binomial coefficients.
\begin{theorem}
\label{thm:2} {Let t, p, j and k be integers with $p&gt;0$, $j\geq0$ and $k\geq0$, we have the asymptotic expansions as follows:\\
(i) when k is fixed,
\begin{eqnarray}
&amp;\sum_{n=0}^p \frac{\binom{p}{n}}{(2n+2k+1)^2\binom{2n+2k}{n+k}^2
}\sim \frac{17^{p+1}\pi}{4^{2k+2p+1}p}\quad (p \rightarrow
\infty),\label{eq:E5}
\end{eqnarray}
(ii) when p is fixed,
\begin{eqnarray}
&amp;\sum_{n=0}^p \frac{\binom{p}{n}}{(2n+2k+1)^2\binom{2n+2k}{n+k}^2
}\sim \frac{17^p\pi}{4^{2k+2p+1}k}\quad (k \rightarrow
\infty),\label{eq:E6}
\end{eqnarray}
(iii)\begin{eqnarray} &amp;\sum_{n=0}^{\infty} \frac{1}{(2n+2k+1)^2
\binom{2n+2k}{n+k}^2} \sim\frac{\pi}{4^{2k-1}15k}\quad (k \to
\infty),\label{eq:E7}
\end{eqnarray}
(iv) when k is fixed and $k\geq1$,
\begin{eqnarray}
&amp;\sum_{n=0}^{\infty} \frac{\binom{n+k-1}{n}}{(2n+2j+1)^2
\binom{2n+2j}{n+j}^2} \sim\frac{ \pi}{4^{2j-2k+1} 15^kj}\quad (j \to
\infty),\label{eq:E8}
\end{eqnarray}
(v)\begin{eqnarray} &amp;\sum_{n=0}^{\infty
}\frac{\binom{n+k}{n}}{(2n+1)^2 \binom{2n}{n}^2} \sim\frac{
4^{2k+1}\pi}{15^k(k+1)}\quad (k \to \infty),\label{eq:E9}
\end{eqnarray}
(vi)\begin{eqnarray} &amp;\sum_{n=0}^{\infty} \frac{n
\binom{n+k}{n}}{(2n+1)^2 \binom{2n}{n}^2} \sim \frac{
4^{2k+1}\pi(k+1)}{15^{k+1}(k+2)}\quad (k\to \infty).\label{eq:E10}
\end{eqnarray}}
\end{theorem}
\begin{proof}
In the proof, we will use the following formulas:
\begin{eqnarray}
\sum_{k=1}^n\binom{n}{k}x^k&amp;=&amp;(1+x)^n,\label{eq:E11}\\
\sum_{n=0}^\infty x^n&amp;=&amp;\frac{1}{1-x},\label{eq:E12}\\
\sum_{n=0}^\infty\binom{n+k-1}{n}x^n&amp;=&amp;\frac{1}{(1-x)^k},\label{eq:E13}\\
\sum_{n=0}^\infty\binom{n+k}{n}x^n&amp;=&amp;\frac{1}{(1-x)^{k+1}},\label{eq:E14}\\
\sum_{n=0}^\infty
n\binom{n+k}{n}x^n&amp;=&amp;\frac{(k+1)x}{(1-x)^{k+2}}.\label{eq:E15}
\end{eqnarray}
It follows from \eqref{eq:E4} and \eqref{eq:E11}--\eqref{eq:E15}
that
\begin{align*}
&amp;\sum_{n=0}^p\frac{\binom{p}{n}}{(2n+2k+1)^2\binom{2n+2k}{n+k}^2}\\
=&amp;\sum_{n=0}^p\binom{p}{n}\int_0^1x^{n+k}(1-x)^{n+k}\,dx\int_0^1y^{n+k}(1-y)^{n+k}\,dy\\
=&amp;\sum_{n=0}^p\binom{p}{n}\int_0^1\int_0^1[xy(1-x)(1-y)]^{n+k}\,dxdy\\
=&amp;\int_0^1\int_0^1{(xy(1-x)(1-y))}^k \sum_{n=0}^p\binom{p}{n}{(xy(1-x)(1-y))}^n\,dxdy\\
=&amp;\int_0^1\int_0^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^p\,dxdy,
\end{align*}
\begin{align*}
&amp;\sum_{n=0}^\infty \frac{1}{(2n+2k+1)^2
\binom{2n+2k}{n+k}^2}\\
=&amp;\sum_{n=0}^\infty\int_0^1\int_0^1[xy(1-x)(1-y)]^{n+k}\,dxdy\\
=&amp;\int_0^1\int_0^1\frac{{(xy(1-x)(1-y))}^k}{1-xy(1-x)(1-y)}\,dxdy,
\end{align*}
\begin{align*}
&amp;\sum_{n=0}^\infty\frac{\binom{n+k-1}{n}}{(2n+2j+1)^2\binom{2n+2j}{n+j}^2}\\
=&amp;\int_0^1\int_0^1\sum_{n=0}^\infty\binom{n+k-1}{n}{(xy(1-x)(1-y))}^{n+j}\,dxdy\\
=&amp;\int_0^1\int_0^1\frac{(xy(1-x)(1-y))^j}{(1-xy(1-x)(1-y))^k}\,dxdy,
\end{align*}
\begin{align*}
&amp;\sum_{n=0}^\infty\frac{\binom{n+k}{n}}{(2n+1)^2\binom{2n}{n}^2}\\
=&amp;\int_0^1\int_0^1\sum_{n=0}^\infty\binom{n+k}{n}{(xy(1-x)(1-y))}^n\,dxdy\\
=&amp;\int_0^1\int_0^1\frac{1}{(1-xy(1-x)(1-y))^{k+1}}\,dxdy,
\end{align*}
\begin{align*}
&amp;\sum_{n=0}^\infty\frac{n\binom{n+k}{n}}{(2n+1)^2\binom{2n}{n}^2}\\
=&amp;\int_0^1\int_0^1\sum_{n=0}^\infty n\binom{n+k}{n}{(xy(1-x)(1-y))}^n\,dxdy\\
=&amp;\int_0^1\int_0^1\frac{(k+1)x(1-x)y(1-y)}{(1-xy(1-x)(1-y))^{k+2}}\,dxdy.
\end{align*}
(i) Let $\phi(x,y)=(xy(1-x)(1-y))^k,\ f(x,y)=1+xy(1-x)(1-y).$
It is evident that $f(x,y)&gt;0$ for $(x,y)\in[0,1]\times[0,1],$
\begin{eqnarray*}
f_x&amp;=&amp;y-2xy-y^2+2xy^2,\\
f_y&amp;=&amp;x-x^2-2xy+2x^2y,\\
f_{xx}&amp;=&amp;2y^2-2y,\\
f_{yy}&amp;=&amp;2x^2-2x,\\
f_{xy}&amp;=&amp;1-2x-2y+4xy.
\end{eqnarray*}
When $f_x=f_y=0$, we get $x=\frac{1}{2},\ y=\frac{1}{2}.$ Then we
have
\begin{eqnarray*}
H_1\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&amp;=&amp;-f_{xx}\Big(\frac{1}{2},\frac{1}{2}\Big)\\
&amp;=&amp;\frac{1}{2},\\
H_2\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&amp;=&amp;\left|\begin{array}{cc}-f_{xx}\Big(\frac{1}{2},\frac{1}{2}\Big) &amp; -f_{xy}\Big(\frac{1}{2},\frac{1}{2}\Big) \\-f_{yx}\Big(\frac{1}{2},\frac{1}{2}\Big) &amp; -f_{yy}\Big(\frac{1}{2},\frac{1}{2}\Big)\end{array}\right|\\
&amp;=&amp;\left|\begin{array}{cc}\frac{1}{2} &amp; 0 \\0 &amp; \frac{1}{2}\end{array}\right|\\
&amp;=&amp;\frac{1}{4},\\
\phi\Big(\frac{1}{2},\frac{1}{2}\Big)&amp;=&amp;4^{-2k},\\
f\Big(\frac{1}{2},\frac{1}{2}\Big)&amp;=&amp;\frac{17}{16}.
\end{eqnarray*}
By using Lemma \ref{lem:1}, we have
\begin{align*}
&amp;\int_{0}^{1}\int_{0}^{1}(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^p\,dxdy\\
\sim&amp;\frac{4^{-2k}(17/16)^{p+1}}{1/2}\times\frac{2\pi}{p}\ \ \ \ \ \ \ \ (p\to\infty)\\
=&amp;\frac{17^{p+1}\pi}{4^{2k+2p+1}p}\ \ \ \ \ \ \ \ (p\to\infty).
\end{align*}
(ii) Let $\phi(x,y)=(1+xy(1-x)(1-y))^p,\ f(x,y)=xy(1-x)(1-y).$ It is
evident that
\begin{eqnarray*}
f_x&amp;=&amp;y-2xy-y^2+2xy^2,\\
f_y&amp;=&amp;x-x^2-2xy+2x^2y,\\
f_{xx}&amp;=&amp;2y^2-2y,\\
f_{yy}&amp;=&amp;2x^2-2x,\\
f_{xy}&amp;=&amp;1-2x-2y+4xy,\\
f(x,y)&amp;&gt;&amp;0,\quad \bigg((x, y)\in[\eta,1-\eta]\times[\eta,1-\eta],\quad
0&lt;\eta&lt;\frac{1}{16}\bigg).
\end{eqnarray*}
When $f_x=f_y=0$,  we get $x=\frac{1}{2},\ y=\frac{1}{2}.$ Then we
have
\begin{eqnarray*}
H_1\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&amp;=&amp;\frac{1}{2},\\
H_2\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&amp;=&amp;\frac{1}{4},\\
\phi\Big(\frac{1}{2},\frac{1}{2}\Big)&amp;=&amp;\bigg(\frac{17}{16}\bigg)^p,\\
f\Big(\frac{1}{2},\frac{1}{2}\Big)&amp;=&amp;\bigg(\frac{1}{2}\bigg)^4.
\end{eqnarray*}
By using Lemma \ref{lem:1}, we can derive
\begin{align*}
&amp;\int_{\eta}^{1-\eta}\int_{\eta}^{1-\eta}(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^p\,dxdy\\
\sim&amp;\frac{(17/16)^p 4^{-2k-2}}{1/2}\times\frac{2\pi}{k}\ \ \ \ \ \ \ \ (k\to\infty)\\
=&amp;\frac{17^p\pi}{4^{2k+2p+1}k}\ \ \ \ \ \ \ \ (k\to\infty).
\end{align*}
When $0&lt;\eta&lt;\frac{1}{16}$, it is obvious that
\begin{eqnarray*}
&amp;&amp;\quad\int_0^1\int_0^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\
&amp;&amp;=\int_0^\eta\int_0^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\
&amp;&amp;\quad+\int_{1-\eta}^1\int_0^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\
&amp;&amp;\quad+\int_\eta^{1-\eta}\int_0^\eta(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\
&amp;&amp;\quad+\int_\eta^{1-\eta}\int_{1-\eta}^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\
&amp;&amp;\quad+\int_\eta^{1-\eta}\int_\eta^{1-\eta}(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\
&amp;&amp;=O(\eta^k(1-\eta)^k)+\int_\eta^{1-\eta}\int_\eta^{1-\eta}(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy.
\end{eqnarray*}
When $0&lt;\eta&lt;\frac{1}{16},$
$\eta^k(1-\eta)^k=o\left(\frac{17^p\pi}{4^{2k+2p+1}k}\right) \quad
(k\to\infty)$. Then we can obtain
\begin{align*}
&amp;\sum_{n=0}^p \frac{\binom{p}{n}}{(2n+2k+1)^2\binom{2n+2k}{n+k}^2 }\sim \frac{17^p\pi}{4^{2k+2p+1}k}\ \ \ (k \rightarrow  \infty).
\end{align*}
(iii) Let $f(x,y)=xy(1-x)(1-y),\
\phi{(x,y)}=\frac{1}{1-xy(1-x)(1-y)}.$ By means of the method of
proving \eqref{eq:E6}, we derive
\begin{align*}
&amp;\sum_{n=0}^{\infty}
\frac{1}{(2n+2k+1)^2  \binom{2n+2k}{n+k}^2}
\sim\frac{\pi}{4^{2k-1}15k}\ \ \ (k \to
\infty).
\end{align*} \\
The proofs of \eqref{eq:E8}--\eqref{eq:E10} are similar to that of
\eqref{eq:E5} and \eqref{eq:E6}, and are omitted here.
\end{proof}

Now, we compare the accurate values with the asymptotic values. Let
$$A_{k,p}=\int_0^1\int_0^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy,\quad
B_{k,p}=\frac{17^{p+1}\pi}{4^{2k+2p+1}p}.$$

\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}\hline
p &amp; $A_{1,p}$ &amp; $B_{1,p}$ &amp; $A_{1,p}/B_{1,p}$\\
\hline
10 &amp; 0.0018 &amp; 0.0096 &amp; 0.1832\\
\hline
50 &amp; 0.0122 &amp; 0.0216 &amp; 0.5640\\
\hline
100 &amp; 3.0777 &amp; 3.5835 &amp; 0.8589\\
\hline
150 &amp; 44.8969 &amp; 49.5072 &amp; 0.9069\\
\hline
200 &amp; 715.9654 &amp; 769.4432 &amp; 0.9305\\
\hline
240 &amp; 6828.1 &amp; 7246.9 &amp; 0.9422\\
\hline
\end{tabular}
\end{center}
\caption{some values of $A_{1,p}$ and $B_{1,p}$}
\end{table}

From the above table, we note that $\frac{A_{1,p}}{B_{1,p}}$ is
close to $1$ with the increase of $p$.

By using the method of proving Theorem \ref{thm:2}, we have
\begin{theorem}
\label{thm:3} {Let t, p, j and k be integers with $p&gt;0$, $j\geq0$ and $k\geq0$, we have the asymptotic expansions as follows:\\
(i) when k is fixed,
\begin{align*}
&amp;\sum_{n=0}^p \frac{\binom{p}{n}}{(2n+2k+1)^3\binom{2n+2k}{n+k}^3 }\sim \frac{65^{p+\frac{3}{2}}\pi^\frac{3}{2}}{2^{6k+6p+3}p^\frac{3}{2}}\quad(p \rightarrow  \infty),
\end{align*}
(ii) when p is fixed,
\begin{align*}
&amp;\sum_{n=0}^p \frac{\binom{p}{n}}{(2n+2k+1)^3\binom{2n+2k}{n+k}^3 }\sim \frac{65^p\pi^\frac{3}{2}}{2^{6k+6p+3}k^\frac{3}{2}}\quad(k \rightarrow  \infty),
\end{align*}
(iii)\begin{align*}
&amp;\sum_{n=0}^{\infty}
\frac{1}{(2n+2k+1)^3  \binom{2n+2k}{n+k}^3}
\sim\frac{\pi^\frac{3}{2}}{63k^\frac{3}{2}2^{6k-3}}\quad(k \to
\infty),
\end{align*}
(iv) when k is fixed and $k\geq1$,
\begin{align*}
&amp;\sum_{n=0}^{\infty} \frac{\binom{n+k-1}{n}}{(2n+2j+1)^3
\binom{2n+2j}{n+j}^3} \sim\frac{ \pi^\frac{3}{2}}{2^{6j-6k+3}
63^kj^\frac{3}{2}}\quad(j \to \infty),
\end{align*}
(v)\begin{align*}
&amp;\sum_{n=0}^{\infty }\frac{\binom{n+k}{n}}{(2n+1)^3
\binom{2n}{n}^3} \sim\frac{\pi^\frac{3}{2} 2^{6k+3}}{63^{k-\frac{1}{2}}(k+1)^\frac{3}{2}}\quad
(k \to \infty),
\end{align*}
(vi)\begin{align*}
&amp;\sum_{n=0}^{\infty} \frac{n  \binom{n+k}{n}}{(2n+1)^3
\binom{2n}{n}^3} \sim \frac{\pi^\frac{3}{2}
2^{6k+3}(k+1)}{63^{k+\frac{1}{2}}(k+2)^\frac{3}{2}}\quad(k\to \infty).
\end{align*}}
\end{theorem}

\section{Asymptotic Expansions of Certain Sums Involving Binomial Coefficients and
Generalized Harmonic Numbers}
\label{sec3}

There are many identities between the harmonic number and the
binomial coefficient. For example (see \cite{ref11}),
$$
\sum_{n=1}^{\infty}\frac{H_n}{\binom{n+k}{k}}=\frac{k}{(k-1)^2}\quad
(k&gt;1).
$$
Now we give asymptotic expansions of certain sums involving the
generalized harmonic numbers and the binomial coefficients.
\begin{theorem}
\label{thm:4}
{Let $r$ be a positive integer, when $r\to \infty$, we have the following asymptotic expansions:\\
(i)\begin{eqnarray}
&amp;\sum_{n=1}^{\infty}H_n^{[r]}\frac{1}{(2n+1)^2\binom{2n}{n}^2}\sim
\frac{16^{r-1}4\pi}{15^{r-1}r}\ln\frac{16}{15},\label{eq:E16}
\end{eqnarray}
(ii)\begin{eqnarray}
&amp;\sum_{n=r+1}^{\infty}H_{n,r}\frac{(-1)^n}{(2n+1)^2\binom{2n}{n}^2}
\sim\frac{(-1)^r4\pi}{rr!}\left(\ln\frac{17}{16}\right)^{r+1},\label{eq:E17}
\end{eqnarray}
(iii)\begin{eqnarray}
&amp;\sum_{n=r+1}^{\infty}H(n,r)\frac{(-1)^n}{(2n+1)^2\binom{2n}{n}^2}
\sim\frac{(-1)^{r+1}4\pi}{r+1}\left(\ln\frac{17}{16}\right)^{r+2}.\label{eq:E18}
\end{eqnarray}}
\end{theorem}
\begin{proof}
It follows from \eqref{eq:E1}--\eqref{eq:E4} that
\begin{align*}
&amp;\sum_{n=1}^{\infty}H_n^{[r]}\frac{1}{(2n+1)^2\binom{2n}{n}^2}\\
=&amp;\sum_{n=1}^{\infty}H_n^{[r]}\int_0^1\int_0^1[xy(1-x)(1-y)]^{n}\,dxdy\\
=&amp;\int_0^1\int_0^1\frac{-\ln{(1-xy(1-x)(1-y))}}{(1-xy(1-x)(1-y))^r}\,dxdy,
\end{align*}
\begin{align*}
&amp;\sum_{n=r+1}^\infty H_{n,r}\frac{(-1)^n}{(2n+1)^2
\binom{2n}{n}^2}\\
=&amp;\sum_{n=r+1}^\infty H_{n,r}(-1)^n
\int_0^1\int_0^1[xy(1-x)(1-y)]^n\,dxdy\\
=&amp;\frac{(-1)^r}{r!}\int_0^1\int_0^1\frac{\ln^r{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy,
\end{align*}
\begin{align*}
&amp;\sum_{n=r+1}^{\infty}H(n,r)\frac{(-1)^n}{(2n+1)^2\binom{2n}{n}^2}\\
=&amp;\int_0^1\int_0^1\sum_{n=r+1}^{\infty}H(n,r)(-xy(1-x)(1-y))^n\,dxdy\\
=&amp;(-1)^{r+1}\int_0^1\int_0^1\frac{\ln^{r+1}{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy.
\end{align*}
Now we only give the detailed proof of \eqref{eq:E17}, the others
can be proved by the same method. Let
$$f(x,y)=\ln{(1+xy(1-x)(1-y))},\quad
\phi{(x,y)}=\frac{1}{1+xy(1-x)(1-y)}.
$$
When
$(x,y)\in[\eta,1-\eta]\times[\eta,1-\eta]$\ $(0&lt;\eta&lt;\frac{1}{2})$,\
$f(x,y)&gt;0,$ $f_x$,  $f_y$, $f_{xx}$, $f_{yy}$, $f_{xy}$ can be
computed easily. When $f_x=0,\ f_y=0,$ we get $x=\frac{1}{2},\
y=\frac{1}{2}$. Then we have
\begin{eqnarray*}
H_1\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&amp;=&amp;\frac{8}{17},\\
H_2\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&amp;=&amp;\left(\frac{8}{17}\right)^2,\\
\phi\Big(\frac{1}{2},\frac{1}{2}\Big)&amp;=&amp;\frac{16}{17},\\
f\Big(\frac{1}{2},\frac{1}{2}\Big)&amp;=&amp;\ln{\frac{17}{16}}.
\end{eqnarray*}
By using Lemma \ref{lem:1}, we obtain
\begin{align*}
&amp;\int_{\eta}^{1-\eta}\int_{\eta}^{1-\eta}\frac{\ln^r{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy\\
\sim&amp;\frac{(16/17)(\ln{\frac{17}{16}})^{r+1}}{8/17}\times\frac{2\pi}{r}\quad(r\to\infty)\\
=&amp;\frac{4\pi}{r}\left(\ln{\frac{17}{16}}\right)^{r+1}\quad(r\to\infty).
\end{align*}
We can easily derive that
\begin{align*}
&amp;\int_0^1\int_0^1\frac{\ln^r{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy-
\int_{\eta}^{1-\eta}\int_{\eta}^{1-\eta}\frac{\ln^r{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy\\
&lt;&amp;(\ln{(1+\eta^2(1-\eta)^2)})^r4\eta(1-\eta)\\
=&amp;(4\eta-4\eta^2)\ln^r{(1+\eta^2(1-\eta)^2)}.
\end{align*}
Let $\eta=\frac{1}{16},$ when $r\to \infty$,
$$
(4\eta-4\eta^2)\ln^r{(1+\eta^2(1-\eta)^2)}=o\left(\frac{4\pi}{r}
\left(\ln{\frac{17}{16}}\right)^{r+1}\right).
$$
Hence
\begin{align*}
&amp;\sum_{n=r+1}^{\infty}H_{n,r}\frac{(-1)^n}{(2n+1)^2\binom{2n}{n}^2}
\sim\frac{(-1)^r4\pi}{rr!}\left(\ln\frac{17}{16}\right)^{r+1}\quad(r\to\infty).
\end{align*}
\end{proof}

In the final of this section, we give some asymptotic expansions of sums
involving the Stirling numbers and the powers of binomial coefficients. We know that
the Stirling numbers and the binomial coefficients satisfy (see\cite{ref3}):\\
\begin{align*}
&amp;s(n,k)=\sum_{0\leq h \leq n-k}(-1)^h\binom{n-1+h}{n-k+h}\binom{2n-k}{n-k-h}S(n-k+h,h)\\
=&amp;\sum_{0\leq j \leq h\leq n-k}(-1)^{j+h}\binom{h}{j}\binom{n-1+h}{n-k+h}\binom{2n-k}{n-k-h}\frac{(h-j)^{n-k+h}}{h!}.
\end{align*}
For two kinds of  Stirling numbers and the powers of binomial
coefficients, we have
\begin{theorem}
\label{thm:5}
{when $k\to\infty,$ we have the following asymptotic expansions:\\
(i)\begin{align*}
&amp;\sum_{n=k}^{\infty}\frac{s(n,k)}{n!(2n+1)^2\binom{2n}{n}^2}
\sim\frac{17\pi}{4kk!}\left(\ln\frac{17}{16}\right)^{k+1},
\end{align*}
(ii)\begin{align*}
&amp;\sum_{n=k}^{\infty}\frac{S(n,k)}{n!(2n+1)^2\binom{2n}{n}^2}
\sim\frac{4\pi(e^{\frac{1}{16}}-1)^{k+1}}{kk!e^{\frac{1}{32}}}.
\end{align*}}
\end{theorem}
The proof of this theorem is similar to that of Theorem \ref{thm:4},
and is omitted here.

\section{Acknowledgments}

The authors would like to thank the anonymous referees for their
criticism and useful suggestions.

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\bigskip
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\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:  Primary
05A10; Secondary 11B65.

\noindent {\it Keywords}: binomial coefficients, Laplace's method,
harmonic numbers, Stirling number, asymptotic expansion.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 21 2009;
revised version received  March 19 2010.
Published in {\it Journal of Integer Sequences}, March 23 2010.

\bigskip
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\noindent
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