\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} \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}\\ \end{center} \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} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{thm}{Theorem} \newtheorem{cor}{Corollary} \newtheorem{lem}{Lemma}[section] \newcommand{\bvec}[1]{\mbox{\boldmath$#1$}} \renewcommand{\d}{\displaystyle} \renewcommand\thefootnote{\fnsymbol{footnote}} %************************* 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 $${n\choose m}=\begin{cases}\displaystyle\frac{n!}{m!(n-m)!}, & n\geq m;\cr\ 0, & n0$ $(x\in D);$\\ (ii) The product $\phi(x)[f(x)]^\lambda\ (\lambda>0)$ is absolutely integrable in $D$;\\ (iii) The function $f(x)$ reaches the effective maximum at the interior point $\varepsilon =(\varepsilon_1,\varepsilon_2,\ldots,\varepsilon_n)$ in $D$, i.e., $f_k(\varepsilon)=0,$ $H_k[-f(\varepsilon)]>0\ (k=1,2,\ldots,n),$ where, $H_k[-f(x)]=det(-f_{ij}(x))_{1\leq i,j\leq k};$\\ (iv) The function $\phi(x)$ is continuous at point $\varepsilon$ and $\phi(\varepsilon)\neq0.$\\ Then when $ \lambda \to \infty $, \begin{eqnarray*} \int\!\!\!\idotsint\limits_D{\phi(x)[f(x)]^\lambda}\, dx_1dx_2\cdots dx_n \sim \frac{\phi(\varepsilon)[f(\varepsilon)]^{\lambda+\frac{n}{2}}}{\sqrt {H_n[-f(\varepsilon)]}}{\left(\frac{2\pi}{\lambda}\right)}^\frac{n}{2}. \end{eqnarray*}} \end{lemma} In this paper, we discuss the computation of certain sums related to powers of binomial coefficients. In Section \ref{sec2}, we study the asymptotic expansions of sums such as $$\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>0$, $j\geq0$ and $k\geq0$, we have the asymptotic expansions as follows:\\ (i) when k is fixed, \begin{eqnarray} &\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} &\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} &\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} &\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} &\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} &\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&=&(1+x)^n,\label{eq:E11}\\ \sum_{n=0}^\infty x^n&=&\frac{1}{1-x},\label{eq:E12}\\ \sum_{n=0}^\infty\binom{n+k-1}{n}x^n&=&\frac{1}{(1-x)^k},\label{eq:E13}\\ \sum_{n=0}^\infty\binom{n+k}{n}x^n&=&\frac{1}{(1-x)^{k+1}},\label{eq:E14}\\ \sum_{n=0}^\infty n\binom{n+k}{n}x^n&=&\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*} &\sum_{n=0}^p\frac{\binom{p}{n}}{(2n+2k+1)^2\binom{2n+2k}{n+k}^2}\\ =&\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\\ =&\sum_{n=0}^p\binom{p}{n}\int_0^1\int_0^1[xy(1-x)(1-y)]^{n+k}\,dxdy\\ =&\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\\ =&\int_0^1\int_0^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^p\,dxdy, \end{align*} \begin{align*} &\sum_{n=0}^\infty \frac{1}{(2n+2k+1)^2 \binom{2n+2k}{n+k}^2}\\ =&\sum_{n=0}^\infty\int_0^1\int_0^1[xy(1-x)(1-y)]^{n+k}\,dxdy\\ =&\int_0^1\int_0^1\frac{{(xy(1-x)(1-y))}^k}{1-xy(1-x)(1-y)}\,dxdy, \end{align*} \begin{align*} &\sum_{n=0}^\infty\frac{\binom{n+k-1}{n}}{(2n+2j+1)^2\binom{2n+2j}{n+j}^2}\\ =&\int_0^1\int_0^1\sum_{n=0}^\infty\binom{n+k-1}{n}{(xy(1-x)(1-y))}^{n+j}\,dxdy\\ =&\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*} &\sum_{n=0}^\infty\frac{\binom{n+k}{n}}{(2n+1)^2\binom{2n}{n}^2}\\ =&\int_0^1\int_0^1\sum_{n=0}^\infty\binom{n+k}{n}{(xy(1-x)(1-y))}^n\,dxdy\\ =&\int_0^1\int_0^1\frac{1}{(1-xy(1-x)(1-y))^{k+1}}\,dxdy, \end{align*} \begin{align*} &\sum_{n=0}^\infty\frac{n\binom{n+k}{n}}{(2n+1)^2\binom{2n}{n}^2}\\ =&\int_0^1\int_0^1\sum_{n=0}^\infty n\binom{n+k}{n}{(xy(1-x)(1-y))}^n\,dxdy\\ =&\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)>0$ for $(x,y)\in[0,1]\times[0,1],$ \begin{eqnarray*} f_x&=&y-2xy-y^2+2xy^2,\\ f_y&=&x-x^2-2xy+2x^2y,\\ f_{xx}&=&2y^2-2y,\\ f_{yy}&=&2x^2-2x,\\ f_{xy}&=&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]&=&-f_{xx}\Big(\frac{1}{2},\frac{1}{2}\Big)\\ &=&\frac{1}{2},\\ H_2\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&=&\left|\begin{array}{cc}-f_{xx}\Big(\frac{1}{2},\frac{1}{2}\Big) & -f_{xy}\Big(\frac{1}{2},\frac{1}{2}\Big) \\-f_{yx}\Big(\frac{1}{2},\frac{1}{2}\Big) & -f_{yy}\Big(\frac{1}{2},\frac{1}{2}\Big)\end{array}\right|\\ &=&\left|\begin{array}{cc}\frac{1}{2} & 0 \\0 & \frac{1}{2}\end{array}\right|\\ &=&\frac{1}{4},\\ \phi\Big(\frac{1}{2},\frac{1}{2}\Big)&=&4^{-2k},\\ f\Big(\frac{1}{2},\frac{1}{2}\Big)&=&\frac{17}{16}. \end{eqnarray*} By using Lemma \ref{lem:1}, we have \begin{align*} &\int_{0}^{1}\int_{0}^{1}(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^p\,dxdy\\ \sim&\frac{4^{-2k}(17/16)^{p+1}}{1/2}\times\frac{2\pi}{p}\ \ \ \ \ \ \ \ (p\to\infty)\\ =&\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&=&y-2xy-y^2+2xy^2,\\ f_y&=&x-x^2-2xy+2x^2y,\\ f_{xx}&=&2y^2-2y,\\ f_{yy}&=&2x^2-2x,\\ f_{xy}&=&1-2x-2y+4xy,\\ f(x,y)&>&0,\quad \bigg((x, y)\in[\eta,1-\eta]\times[\eta,1-\eta],\quad 0<\eta<\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]&=&\frac{1}{2},\\ H_2\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&=&\frac{1}{4},\\ \phi\Big(\frac{1}{2},\frac{1}{2}\Big)&=&\bigg(\frac{17}{16}\bigg)^p,\\ f\Big(\frac{1}{2},\frac{1}{2}\Big)&=&\bigg(\frac{1}{2}\bigg)^4. \end{eqnarray*} By using Lemma \ref{lem:1}, we can derive \begin{align*} &\int_{\eta}^{1-\eta}\int_{\eta}^{1-\eta}(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^p\,dxdy\\ \sim&\frac{(17/16)^p 4^{-2k-2}}{1/2}\times\frac{2\pi}{k}\ \ \ \ \ \ \ \ (k\to\infty)\\ =&\frac{17^p\pi}{4^{2k+2p+1}k}\ \ \ \ \ \ \ \ (k\to\infty). \end{align*} When $0<\eta<\frac{1}{16}$, it is obvious that \begin{eqnarray*} &&\quad\int_0^1\int_0^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\ &&=\int_0^\eta\int_0^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\ &&\quad+\int_{1-\eta}^1\int_0^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\ &&\quad+\int_\eta^{1-\eta}\int_0^\eta(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\ &&\quad+\int_\eta^{1-\eta}\int_{1-\eta}^1(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\ &&\quad+\int_\eta^{1-\eta}\int_\eta^{1-\eta}(xy(1-x)(1-y))^k(1+xy(1-x)(1-y))^pdxdy\\ &&=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<\eta<\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*} &\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*} &\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 & $A_{1,p}$ & $B_{1,p}$ & $A_{1,p}/B_{1,p}$\\ \hline 10 & 0.0018 & 0.0096 & 0.1832\\ \hline 50 & 0.0122 & 0.0216 & 0.5640\\ \hline 100 & 3.0777 & 3.5835 & 0.8589\\ \hline 150 & 44.8969 & 49.5072 & 0.9069\\ \hline 200 & 715.9654 & 769.4432 & 0.9305\\ \hline 240 & 6828.1 & 7246.9 & 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>0$, $j\geq0$ and $k\geq0$, we have the asymptotic expansions as follows:\\ (i) when k is fixed, \begin{align*} &\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*} &\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*} &\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*} &\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*} &\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*} &\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>1). $$ Now we give asymptotic expansions of certain sums involving the generalized harmonic numbers and the binomial coefficients. \begin{theorem} \label{thm:4} {Let $r$ be a positive integer, when $r\to \infty$, we have the following asymptotic expansions:\\ (i)\begin{eqnarray} &\sum_{n=1}^{\infty}H_n^{[r]}\frac{1}{(2n+1)^2\binom{2n}{n}^2}\sim \frac{16^{r-1}4\pi}{15^{r-1}r}\ln\frac{16}{15},\label{eq:E16} \end{eqnarray} (ii)\begin{eqnarray} &\sum_{n=r+1}^{\infty}H_{n,r}\frac{(-1)^n}{(2n+1)^2\binom{2n}{n}^2} \sim\frac{(-1)^r4\pi}{rr!}\left(\ln\frac{17}{16}\right)^{r+1},\label{eq:E17} \end{eqnarray} (iii)\begin{eqnarray} &\sum_{n=r+1}^{\infty}H(n,r)\frac{(-1)^n}{(2n+1)^2\binom{2n}{n}^2} \sim\frac{(-1)^{r+1}4\pi}{r+1}\left(\ln\frac{17}{16}\right)^{r+2}.\label{eq:E18} \end{eqnarray}} \end{theorem} \begin{proof} It follows from \eqref{eq:E1}--\eqref{eq:E4} that \begin{align*} &\sum_{n=1}^{\infty}H_n^{[r]}\frac{1}{(2n+1)^2\binom{2n}{n}^2}\\ =&\sum_{n=1}^{\infty}H_n^{[r]}\int_0^1\int_0^1[xy(1-x)(1-y)]^{n}\,dxdy\\ =&\int_0^1\int_0^1\frac{-\ln{(1-xy(1-x)(1-y))}}{(1-xy(1-x)(1-y))^r}\,dxdy, \end{align*} \begin{align*} &\sum_{n=r+1}^\infty H_{n,r}\frac{(-1)^n}{(2n+1)^2 \binom{2n}{n}^2}\\ =&\sum_{n=r+1}^\infty H_{n,r}(-1)^n \int_0^1\int_0^1[xy(1-x)(1-y)]^n\,dxdy\\ =&\frac{(-1)^r}{r!}\int_0^1\int_0^1\frac{\ln^r{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy, \end{align*} \begin{align*} &\sum_{n=r+1}^{\infty}H(n,r)\frac{(-1)^n}{(2n+1)^2\binom{2n}{n}^2}\\ =&\int_0^1\int_0^1\sum_{n=r+1}^{\infty}H(n,r)(-xy(1-x)(1-y))^n\,dxdy\\ =&(-1)^{r+1}\int_0^1\int_0^1\frac{\ln^{r+1}{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy. \end{align*} Now we only give the detailed proof of \eqref{eq:E17}, the others can be proved by the same method. Let $$f(x,y)=\ln{(1+xy(1-x)(1-y))},\quad \phi{(x,y)}=\frac{1}{1+xy(1-x)(1-y)}. $$ When $(x,y)\in[\eta,1-\eta]\times[\eta,1-\eta]$\ $(0<\eta<\frac{1}{2})$,\ $f(x,y)>0,$ $f_x$, $f_y$, $f_{xx}$, $f_{yy}$, $f_{xy}$ can be computed easily. When $f_x=0,\ f_y=0,$ we get $x=\frac{1}{2},\ y=\frac{1}{2}$. Then we have \begin{eqnarray*} H_1\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&=&\frac{8}{17},\\ H_2\Big[-f\Big(\frac{1}{2},\frac{1}{2}\Big)\Big]&=&\left(\frac{8}{17}\right)^2,\\ \phi\Big(\frac{1}{2},\frac{1}{2}\Big)&=&\frac{16}{17},\\ f\Big(\frac{1}{2},\frac{1}{2}\Big)&=&\ln{\frac{17}{16}}. \end{eqnarray*} By using Lemma \ref{lem:1}, we obtain \begin{align*} &\int_{\eta}^{1-\eta}\int_{\eta}^{1-\eta}\frac{\ln^r{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy\\ \sim&\frac{(16/17)(\ln{\frac{17}{16}})^{r+1}}{8/17}\times\frac{2\pi}{r}\quad(r\to\infty)\\ =&\frac{4\pi}{r}\left(\ln{\frac{17}{16}}\right)^{r+1}\quad(r\to\infty). \end{align*} We can easily derive that \begin{align*} &\int_0^1\int_0^1\frac{\ln^r{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy- \int_{\eta}^{1-\eta}\int_{\eta}^{1-\eta}\frac{\ln^r{(1+xy(1-x)(1-y))}}{1+xy(1-x)(1-y)}\,dxdy\\ <&(\ln{(1+\eta^2(1-\eta)^2)})^r4\eta(1-\eta)\\ =&(4\eta-4\eta^2)\ln^r{(1+\eta^2(1-\eta)^2)}. \end{align*} Let $\eta=\frac{1}{16},$ when $r\to \infty$, $$ (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*} &\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*} &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)\\ =&\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*} &\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*} &\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. \begin{thebibliography}{99} \bibliographystyle{plain} \bibitem{ref1} V. 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