\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 Dixon's Formula and Identities Involving \\ \vskip .1in Harmonic Numbers} \vskip 1cm \large Xiaoxia Wang\footnote{This work is supported by Shanghai Leading Academic Discipline Project, Project No.\ S30104.} and Mei Li \\ Department of Mathematics\\ Shanghai University\\ Shanghai, China\\ \href{mailto:xiaoxiawang@shu.edu.cn}{\tt xiaoxiawang@shu.edu.cn} \\ \end{center} \vskip .2 in \begin{abstract} Inspired by the recent work of Chu and Fu, we derive some new identities with harmonic numbers from Dixon's hypergeometric summation formula by applying the derivation operator to the summation of binomial coefficients. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{thm}[theorem]{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{defin}[theorem]{Definition} \newenvironment{definition}{\begin{defin}\normalfont\quad}{\end{defin}} \newtheorem{examp}[theorem]{Example} 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%%%multline senza & %%%%%% \newcommand{\xmultz}[1]{\begin{multline*}#1%%%%%%%%%%% \end{multline*}}%%%multline senza & %%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF% %%%%%%%%%%%%%%%%%%%%% positions %%%%%%%%%%%%%%%%%%%%%% \newcommand{\centro}[1] {\begin{center}#1\end{center}} \newcommand{\sinistra}[1] {\begin{flushleft}#1\end{flushleft}} \newcommand{\destra}[1] {\begin{flushright}#1\end{flushright}} \newcommand{\centerbox}[2]{\centro{\begin{tabular}{|c|}\hline \parbox{#1}{\mbox{}\\[2.5mm]#2\mbox{}\\}\\\hline\end{tabular}}} \newcommand{\centrobmp}[3]{\vspace*{1mm} \centerbmp{#1}{#2}{#3}\vspace*{1mm}} \newcommand{\sav}[1]{\newsavebox{#1}% \sbox{#1}{\theequation}\\[-2mm]}%=\newline \newcommand{\usa}[1]{(\usebox{#1})}%{#1}={\some-nome} \newcommand{\summary}[2]{\begin{center}\parbox{#1} {{\sc\bf Summary}:\,{\small\it#2}}\end{center}} \newcommand{\riassunto}[2]{\begin{center}\parbox{#1} {{\sc\bf Riassunto}:\,{\small\it#2}}\end{center}} \newcommand{\grazia}[2]{\begin{center}\parbox{#1} {{\sc\bf Ringraziamento}:\,{\small\it#2}}\end{center}} \newcommand{\thank}[2]{\begin{center}\parbox{#1} {{\sc\bf Acknowledgement}:\,{\small\it#2}}\end{center}} \newcommand{\bbtm}[5]{\bibitem{kn:#1}{#2,}~{#3,}~\emph{#4}~{#5.}} \newcommand{\bbtmn}[4]{\bibitem{kn:#1}{#2,}~\emph{#3,}~{#4.}} \newcommand{\bbtmm}[4]{\bibitem{kn:#1}{#2,}~{#4.}} \newcommand{\cito}[1]{\cite{kn:#1}} \newcommand{\citu}[2]{\cite[#2]{kn:#1}} \newcommand{\mr}{\textbf{MR}\:} \newcommand{\zbl}{\textbf{Zbl}\:} \newcommand{\nota}[2]{\centerbox{#1}{\textbf{Achtung}\quad#2}} \newcommand{\graph}[3]{\begin{picture}(0,0) \put(#1,#2){#3}\end{picture}} \newcommand{\fwd}{\mbox{$\bigtriangleup\qdn\!\cdot\,\:$}} \newcommand{\bwd}{\bigtriangledown} \section{Introduction}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% For an indeterminate $x$ and a nonnegative integer $n$, the shifted factorial or {\it Pochhammer's symbol} is defined by \[(x)_0:=1\qdp\text{and}\qdp (x)_n:=\Gamma(x+n)/\Gamma(x)=x(x+1)\cdots(x+n-1), \qdp n=1,2,\cdots\] with the $\Gamma$-function given through Euler integral \[\Gamma(x)=\int_0^\infty u^{x-1}e^{-u}\text{d}u \xqdp \text{with}\xqdp \mathfrak{R}(x)>0.\] Following Bailey~\citu{bailey}{\S2.1} and Slater~\citu{slater}{\S2.1}, the generalized hypergeometric series is defined by \[{_pF_q}\ffnk{cccc}{z}{a_1,\+a_2,\+\cdots,\+a_p} {b_1,\+b_2,\+\cdots,\+b_q} =\sum_{n=0}^\infty\frac{(a_1)_n(a_2)_n\cdots(a_p)_n} {(b_1)_n(b_2)_n\cdots(b_q)_n} \frac{z^n}{n!}, \] where we suppose that none of the denominator parameters is a nonpositive integer, so that the series is well defined. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% The generalized harmonic numbers are defined by \bnm H_n^{\langle\ell\rangle}(x):=\sum_{k=1}^n\frac{1}{(k+x)^\ell} \quad\text{and}\quad H_n^{\langle\ell\rangle}:=H_n^{\langle\ell\rangle}(0)=\sum_{k=1}^n\frac{1}{k^\ell}, \quad\quad H_0^{\langle\ell\rangle}:=0, \enm with indeterminate $x$ and natural numbers $n$ and $\ell$. When $\ell=1$, they will be abbreviated as $H_n(x)$ and $H_n$ respectively. These numbers come naturally from the derivatives of binomial coefficients \bnm D_x{{n+x}\choose n}=H_n(x){{n+x}\choose n} \quad \text{and} \quad D_x{{n+x}\choose n}^{-1}=-H_n(x){{n+x}\choose n}^{-1}, \enm where the differential operator is defined as \bnm D_xf(x)=\frac{d}{dx}f(x), \enm with differentiable function $f(x)$. Obviously, the generalized harmonic numbers satisfy the following recurrence relation \bnm D_xH_n(x)=-H_n^{\langle2\rangle}(x) \quad \text{and} \quad D_xH_n^{\langle\ell\rangle}(x)=-\ell H_n^{\langle{\ell+1}\rangle}(x). \enm This fact can be traced back to Issac Newton~\cito{newt}, and has been explored recently in several papers~\cite{kn:chu13a, kn:chu93c,kn:chu05,kn:dri,kn:pau}. Chu and Fu \cito{chu11a} derived many identities involving harmonic numbers from Dougall--Dixon's summation formula. There exist ${n\choose k}^3$ or ${n\choose k}^4$ in the coefficients of identities. In recent work, Chen and Chu \cito{chu13a} established a general formula involving harmonic numbers and the Riemann zeta function. The purpose of this article is to present some new identities with harmonic numbers from Dixon's summation formula by applying the derivation operator to binomial coefficients. In this paper, there are ${2n\choose k}^3$, ${2n\choose k}$ and ${2n\choose k}^{-1}$ in the coefficients of the identities. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{The identities due to Dixon's~$_3F_2(1)$~ summation formula} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% In this section, we will obtain several identities with harmonic numbers from Dixon's summation formula. Dixon's summation theorem is presented as follows: \begin{thm}[Dixon \cito{slater}]\label{dixon} \bnm {_3F_2}\ffnk{cccc}{1}{a,\quad b,\quad c}{1+a-b,1+a-c} =\frac{\Gamma(1+\frac{1}{2}a)\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+\frac{1}{2}a-b-c)}{\Gamma(1+a) \Gamma(1+\frac{1}{2}a-b)\Gamma(1+\frac{1}{2}a-c)\Gamma(1+a-b-c)}. \enm \end{thm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% (\Rmnum{1}) \: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Letting $a=-2n, b=1+\lambda x$ and $c=1+\theta x$ in Theorem \ref{dixon}, we have \bnm {_3F_2}\ffnk{cccc}{1}{-2n,1+\lambda x,1+\theta x} {-\lambda x-2n,-\theta x-2n} =\frac{(2n)!(1+\lambda x)_n(1+\theta x)_n(1+\lambda x+\theta x)_{2n+1}} {n!(1+\lambda x)_{2n}(1+\theta x)_{2n}(1+\lambda x+\theta x)_{n+1}}. \enm Multiplying both sides by the binomial coefficients ${{{2n+\lambda x}\choose {2n}}{{2n+\theta x}\choose {2n}}}$, we reformulate the result as the following finite summation identity. \bmn \label{d-b-1} \sum_{k=0}^{2n}\frac{(-1)^k{{\lambda x+k}\choose k}{{\theta x+k}\choose k} {{2n-k+\lambda x}\choose {2n-k}}{{2n-k+\theta x}\choose {2n-k}}}{{{2n}\choose k}} =\frac{(2n+1){{n+\lambda x}\choose {n}}{{n+\theta x}\choose n} {{2n+1+\lambda x+\theta x}\choose {2n+1}}}{(n+1){{n+1+\lambda x+\theta x}\choose {n+1}}}. \emn Computing the derivation of identity \eqref{d-b-1} with respect to $x$ for one time, we get \begin{thm}\label{dixon2} \bnm \sum_{k=0}^{2n}\frac{(-1)^k{{\lambda x+k}\choose k}{{\theta x+k}\choose k} {{2n-k+\lambda x}\choose {2n-k}}{{2n-k+\theta x}\choose{2n-k}}\Omega(x)}{{{2n}\choose k}} =\frac{(2n+1){{n+\lambda x}\choose {n}}{{n+\theta x}\choose n} {{2n+1+\lambda x+\theta x}\choose {2n+1}}W(x)}{(n+1){{n+1+\lambda x+\theta x}\choose {n+1}}}, \enm where $ \Omega(x)$ and $W(x)$ are given respectively by \bnm &&\Omega (x)=\lambda H_{2n-k}(\lambda x)+\theta H_{2n-k}(\theta x) +\lambda H_k(\lambda x)+\theta H_k(\theta x);\\ &&W(x)=(\lambda+\theta)H_{2n+1}(\lambda x+\theta x) +\lambda H_{n}(\lambda x)+\theta H_n(\theta x)-(\lambda+\theta) H_{n+1}(\lambda x+\theta x). \enm \end{thm} \begin{proof} The derivatives of binomial coefficients are as follows \bnm D_x{{n+\lambda x}\choose n}=\lambda H_n(\lambda x){{n+\lambda x}\choose n} \quad \text{and} \quad D_x{{n+\lambda x}\choose n}^{-1}=-\lambda H_n(\lambda x){{n+ \lambda x}\choose n}^{-1}. \enm Applying the derivation operator on the left side of the identity \eqref{d-b-1} to $x$, changing the calculation of the orders of summation and derivation and simplifying the result, we have \bnm \+\+\sum_{k=0}^{2n}\frac{(-1)^k{{\lambda x+k}\choose k}{{\theta x+k}\choose k} {{2n-k+\lambda x}\choose {2n-k}}{{2n-k+\theta x}\choose{2n-k}}}{{{2n}\choose k}}\\ \+\+\times\Big\{\lambda H_k(\lambda x)+\theta H_k(\theta x)+\lambda H_{2n-k}(\lambda x)+\theta H_{2n-k}(\theta x)\Big\}. \enm By the same method, we evaluate the derivation of the right side of identity \eqref{d-b-1} to $x$ as follows \bnm \+\+\frac{(2n+1){{n+\lambda x}\choose {n}}{{n+\theta x}\choose n} {{2n+1+\lambda x+\theta x}\choose {2n+1}}}{(n+1){{n+1+\lambda x+\theta x}\choose {n+1}}}\\ \+\+\times\Big\{\lambda H_{n}(\lambda x)+\theta H_n(\theta x)+(\lambda+\theta)H_{2n+1}(\lambda x+\theta x) +(\lambda+\theta) H_{n+1}(\lambda x+\theta x)\Big\}. \enm Comparing both results, we get Theorem \ref{dixon2}. \end{proof} Setting $x=0$ in Theorem \ref{dixon2} and noting that \[\Omega(0)=(\lambda+\theta) (H_{2n-k}+H_k); \quad W(0)=(\lambda +\theta)(H_{2n+1}+H_{n}-H_{n+1}),\] we have the new identity with the harmonic numbers and ${2n\choose k}^{-1}$ as follows: \begin{corollary} \bnm \sum_{k=0}^{2n}\frac{(-1)^k }{{{2n}\choose k}}H_k =\frac{2n+1}{2(n+1)}\Big\{H_{2n+1}+H_n-H_{n+1}\Big\}. \enm \end{corollary} Furthermore, computing the derivation of the identity \eqref{d-b-1} with respect to $x$ for two times, we have another relation about the harmonic numbers. \begin{thm}\label{dixon3} \bnm \+\+\sum_{k=0}^{2n}\frac{(-1)^k{{\lambda x+k}\choose k}{{\theta x+k}\choose k} {{2n-k+\lambda x}\choose {2n-k}}{{2n-k+\theta x}\choose {2n-k}} \{\Omega^2(x)+\Omega'(x)\}}{{{2n}\choose k}} \\ \+\+=\frac{(2n+1){{n+\lambda x}\choose {n}}{{n+\theta x}\choose n}{{2n+1+\lambda x+\theta x}\choose {2n+1}} \{W^2(x)+W'(x)\}}{(n+1){{n+1+\lambda x+\theta x}\choose {n+1}}}, \enm where $\Omega'(x)$ and $W'(x)$ are given respectively by \bnm &&\Omega'(x)=-\lambda^2 H_{2n-k}^{\langle2\rangle}(\lambda x)-\theta^2 H_{2n-k}^{\langle2\rangle} (\theta x)- \lambda^2 H_k^{\langle2\rangle}(\lambda x)-\theta^2 H_k^{\langle2\rangle}(\theta x);\\ &&W'(x)=-(\lambda+\theta)^2H_{2n+1}^{\langle2\rangle}(\lambda x+\theta x) +(\lambda+\theta)^2 H_{n+1}^{\langle2\rangle}(\lambda x+\theta x) -\lambda^2H_{n}^{\langle2\rangle}(\lambda x)-\theta^2H_n^{\langle2\rangle}(\theta x). \enm \end{thm} Noting further that \bnm &&\Omega'(0)=-(\lambda^2+\theta^2)(H_{2n-k}^{\langle2\rangle}+H_k^{\langle2\rangle});\\ &&W'(0)=(\lambda+\theta)^2(H_{n+1}^{\langle2\rangle}-H_{2n+1}^{\langle2\rangle}) -(\lambda^2+\theta^2)H_n^{\langle2\rangle}, \enm we have the following new identity with harmonic numbers when $x=0$ in Theorem \ref{dixon3}. \begin{corollary} \label{d-c-1} \bnm \+\+\sum_{k=0}^{2n}\frac{(-1)^k}{{{2n}\choose k}} \Big\{(\lambda+\theta)^2(H_{2n-k}+H_k)^2 -(\lambda^2+\theta^2)(H_{2n-k}^{\langle2\rangle}+H_k^{\langle2\rangle})\Big\}\\ \+\+=\frac{2n+1}{n+1}\Big\{(\lambda +\theta)^2(H_{2n+1}+ H_n-H_{n+1})^2 +(\lambda+\theta)^2(H_{n+1}^{\langle2\rangle}-H_{2n+1}^{\langle2\rangle}) -(\lambda^2+\theta^2) H_n^{\langle2\rangle}\Big \}. \enm \end{corollary} Now, we present some examples with the harmonic numbers from Corollary \ref{d-c-1}. \begin{example}[$\lambda=0, \theta\neq 0$ or $\lambda \neq 0, \theta=0$ in Corollary \ref{d-c-1}] \bnm &&\sum_{k=0}^{2n}\frac{(-1)^k}{{{2n}\choose k}} \big\{(H_{2n-k}+H_k)^2-(H_{2n-k}^{\langle2\rangle}+H_k^{\langle2\rangle})\big\}\\ &&=\frac{2n+1}{n+1} \big\{(H_{2n+1}+ H_n-H_{n+1})^2+H_{n+1}^{\langle2\rangle}-H_{2n+1}^{\langle2\rangle} - H_n^{\langle2\rangle}\big\}. \enm \end{example} \begin{example} [$\lambda=-\theta\neq0$ in Corollary \ref{d-c-1}] \bnm \sum_{k=0}^{2n}\frac{(-1)^k}{{{2n}\choose k}}H_k^{\langle2\rangle} =\frac{2n+1}{2(n+1)}H_n^{\langle2\rangle}. \enm \end{example} %\begin{exam} [$\lambda=1$ and $\theta=1$ in $\mathbf{Corollary \:\:\ref{d-c-1}}$] %\bnm %\+\+\sum_{k=0}^{2n}\frac{(-1)^k}{{{2n}\choose k}}\Big\{(H_{2n-k}+H_k)^2-H_k^{\langle2\rangle}\Big\}\\ %\+\+=\frac{2n+1}{n+1}\Big\{(H_{2n+1}+H_n-H_{n+1})^2+H_{n+1}^{\langle2\rangle} %-H_{2n+1}^{\langle2\rangle}-\frac12H_n^{\langle2\rangle}\Big\}. %\enm %\end{exam} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% (\Rmnum{2}) \: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Letting $a=-2n, \:b=\lambda x-2n$ and $c=\theta x-2n$ in Dixon's Theorem \ref{dixon}, we have the following identity \bnm {_3F_2}\ffnk{cccc}{1}{-2n,\lambda x-2n,\theta x-2n}{1-\lambda x,\quad 1-\theta x} =\frac{(2n)!(1-\theta x-\lambda x)_{3n}}{n!(1-\lambda x)_{n}(1-\theta x)_n(1-\theta x-\lambda x)_{2n}}. \enm Dividing both sides of the above identity by the binomial coefficients ${{2n-\lambda x}\choose 2n}{{2n-\theta x}\choose 2n}$, we reformulate the result as the following finite summation identity. \bmn\label{d-2} \sum_{k=0}^{2n}\frac{(-1)^k{{2n}\choose k}^3} {{{k-\lambda x}\choose k}{{k-\theta x}\choose k}{{2n-k-\lambda x}\choose {2n-k}}{{2n-k-\theta x}\choose {2n-k}}} =\frac{(-1)^n{{2n}\choose n}{{3n}\choose n}{{3n-\lambda x-\theta x}\choose 3n}}{{{n-\lambda x}\choose n} {{2n-\lambda x}\choose {2n}}{{n-\theta x}\choose {n}}{{2n-\theta x}\choose {2n}}{2n-\theta x-\lambda x \choose 2n}}. \emn When $x=0$ in the above identity, we have the well known identity with binomial coefficients. \begin{corollary} \bnm \sum_{k=0}^{2n}(-1)^k{{2n}\choose k}^3 =(-1)^n{{2n}\choose n}{{3n}\choose n}. \enm \end{corollary} Evaluating the derivation of identity \eqref{d-2} with respect to $x$ for one time, we get the following result. \begin{thm}\label{dixon4} \bnm \sum_{k=0}^{2n}\frac{(-1)^k{{2n}\choose k}^3\Omega(x)} {{{k-\lambda x}\choose k}{{k-\theta x}\choose k}{{2n-k-\lambda x}\choose {2n-k}}{{2n-k-\theta x}\choose {2n-k}}} =\frac{(-1)^n{{2n}\choose n}{{3n}\choose n}{{3n-\lambda x-\theta x}\choose 3n}W(x)} {{{n-\lambda x}\choose n}{{2n-\lambda x}\choose {2n}}{{n-\theta x}\choose {n}} {{2n-\theta x}\choose {2n}}{2n-\theta x-\lambda x \choose 2n}}. \enm where $\Omega(x)$ and $W(x)$ are given respectively by \bnm \Omega(x)\+=\+\lambda H_{k}(-\lambda x)+\theta H_{k}(-\theta x)+\lambda H_{2n-k}(-\lambda x)+\theta H_{2n-k}(-\theta x);\\ W(x)\+=\+\lambda \big\{H_n(-\lambda x)+H_{2n}(-\lambda x)\big\}+\theta \big\{H_{n}(-\theta x)+H_{2n}(-\theta x)\big\}\\ \+\++(\theta+\lambda) \big\{H_{2n}(-\theta x-\lambda x)-H_{3n}(-\theta x-\lambda x)\big\}. \enm \end{thm} Letting $x=0$ in Theorem \ref{dixon4} and noting that \bnm &&\Omega(0)=(\lambda+ \theta )(H_{k}+H_{2n-k});\\ &&W(0)=(\lambda+\theta)(H_{n}+2H_{2n}-H_{3n}), \enm we have the following identity. \begin{corollary}\label{d-c-2} \bnm \sum_{k=0}^{2n}(-1)^k{{2n}\choose k}^3\{H_{k}+H_{2n-k}\} =(-1)^n{{2n}\choose n}{{3n}\choose n}\{H_{n}+2H_{2n}-H_{3n}\}, \enm \end{corollary} \noindent which is a special case of Example $1$ in \cito{chu11a}. Evaluating the derivation of identity \eqref{d-2} with respect to $x$ for two times, we obtain the identity as follows. \begin{thm}\label{dixon5} \bnm \sum_{k=0}^{2n}\frac{(-1)^k{{2n}\choose k}^3\big\{\Omega(x)^2+\Omega'(x)\big\}} {{{k-\lambda x}\choose k}{{k-\theta x}\choose k}{{2n-k-\lambda x}\choose {2n-k}}{{2n-k-\theta x}\choose {2n-k}}} =\frac{(-1)^n{{2n}\choose n}{{3n}\choose n}{{3n-\lambda x-\theta x}\choose 3n}\big\{W(x)^2+W'(x)\big\}} {{{n-\lambda x}\choose n}{{2n-\lambda x}\choose {2n}}{{n-\theta x}\choose {n}} {{2n-\theta x}\choose {2n}}{2n-\theta x-\lambda x \choose 2n}}. \enm where $\Omega'(x)$ and $W'(x)$ are given respectively by \bnm \Omega'(x)\+=\+\lambda^2H_{k}^{\langle2\rangle}(-\lambda x) +\lambda^2H_{2n-k}^{\langle2\rangle}(-\lambda x)+\theta^2H_{k}^{\langle2\rangle}(-\theta x) +\theta^2H_{2n-k}^{\langle2\rangle}(-\theta x);\\ W'(x)\+=\+\lambda^2\big\{H_n^{\langle2\rangle}(-\lambda x)+H_{2n}^{\langle2\rangle}(-\lambda x)\big\} +\theta^2\big\{H_{n}^{\langle2\rangle}(-\theta x)+H_{2n}^{\langle2\rangle}(-\theta x)\big\}\\ \+\++(\theta+\lambda)^2\big\{H_{2n}^{\langle2\rangle}(-\theta x-\lambda x) -H_{3n}^{\langle2\rangle}(-\theta x-\lambda x)\big\}. \enm \end{thm} Letting $x=0$ in Theorem \ref{dixon5} and noting that \bnm &&\Omega'(0)=(\lambda^2+\theta^2)(H_{k}^{\langle2\rangle}+H_{2n-k}^{\langle2\rangle});\\ &&W'(0)=(\lambda+\theta)^2(H_{2n}^{\langle2\rangle}-H_{3n}^{\langle2\rangle})+ (\lambda^2+\theta^2)(H_{n}^{\langle2\rangle}+H_{2n}^{\langle2\rangle}), \enm we derive the identity as follows. \begin{corollary}\label{d-c-3} \bnm \+\+\sum_{k=0}^{2n}(-1)^k{{2n}\choose k}^3 \Big\{(\lambda+ \theta )^2(H_{k}+H_{2n-k})^2 +(\lambda^2+\theta^2)(H_{k}^{\langle2\rangle}+H_{2n-k}^{\langle2\rangle})\Big\}\\ \+=\+(-1)^n{{2n}\choose n}{{3n}\choose n} \Big\{(\lambda+\theta)^2\big[(H_{n}+2H_{2n}-H_{3n})^2+H_{2n}^{\langle2\rangle}-H_{3n}^{\langle2\rangle}\big]+ (\lambda^2+\theta^2)(H_{n}^{\langle2\rangle}+H_{2n}^{\langle2\rangle})\Big\}. \enm \end{corollary} This identity is the same as Theorem 2 in \cito{chu11a}. Now we present some examples from Corollary \ref{d-c-3}. \begin{example}[$\lambda=0,\:\theta\neq 0$ or $\lambda\neq 0,\:\theta=0$ in Corollary \ref{d-c-3}] \bnm &&\sum_{k=0}^{2n}(-1)^k{{2n}\choose k}^3 \Big\{(H_{k}+H_{2n-k})^2+H_{k}^{\langle2\rangle}+H_{2n-k}^{\langle2\rangle}\Big\}\\ &&=(-1)^n{{2n}\choose n}{{3n}\choose n} \Big\{(H_{n}+2H_{2n}-H_{3n})^2+H_{n}^{\langle2\rangle}+2H_{2n}^{\langle2\rangle}-H_{3n}^{\langle2\rangle}\Big\}, \enm \end{example} \noindent which is equal to Example $3$ in \cito{chu11a}. \begin{example}[$\lambda=-\theta \neq 0$ in Corollary \ref{d-c-3}] \bnm \sum_{k=0}^{2n}(-1)^k{{2n}\choose k}^3 H_{k}^{\langle2\rangle} = \frac{(-1)^n}{2}{{2n}\choose n}{{3n}\choose n} \Big\{H_{n}^{\langle2\rangle}+H_{2n}^{\langle2\rangle}\Big\}. \enm \end{example} \begin{example}[$\lambda=1$ and $\theta=1$ in Corollary \ref{d-c-3}] \bnm &&\sum_{k=0}^{2n}(-1)^k{{2n}\choose k}^3 \Big\{(H_{k}+H_{2n-k})^2+H_{k}^{\langle2\rangle}\Big\}\\ &&=(-1)^n{{2n}\choose n}{{3n}\choose n}\Big\{(H_{n}+2H_{2n}-H_{3n})^2 +\frac12H_{n}^{\langle2\rangle}+\frac32H_{2n}^{\langle2\rangle}-H_{3n}^{\langle2\rangle}\Big\}. \enm \end{example} In fact, there are many identities involving harmonic numbers which can be obtained from Corollary \ref{d-c-3}. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% (\Rmnum{3}) \: %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Letting $a=-2n, b=-\lambda x-2n$ and $c=1-\theta x$ in Dixon's Theorem \:\:ref{dixon}, we have the following identity. \bnm {_3F_2}\ffnk{cccc}{1}{-2n,-\lambda x-2n,1-\theta x} {1+\lambda x,\quad \theta x-2n} =\frac{(2n)!(\lambda x+\theta x)_{n}(1-\theta x)_{n}} {n!(1+\lambda x)_n(1-\theta x)_{2n}}. \enm Dividing both sides of the above identity by the binomial coefficients ${{2n+\lambda x}\choose {2n}}/{{2n-\theta x}\choose {2n}}$, we reformulate the result as the following finite summation identity. \bmn\label{d-4} \sum_{k=0}^{2n}\frac{(-1)^k{{2n}\choose k}{{k-\theta x}\choose {k}}{{2n-k-\theta x}\choose {2n-k}}} {{{k+\lambda x}\choose {k}}{{2n-k+\lambda x}\choose {2n-k}}} =\frac{(\lambda x+\theta x){{n-\theta x}\choose {n}} {{n-1+\lambda x+\theta x}\choose {n-1}}}{n{{n+\lambda x}\choose {n}}{{2n+\lambda x}\choose {2n}}}. \emn Evaluating the derivation of identity \eqref{d-4} with respect to $x$ for two times, we have the following identity with the harmonic numbers. \begin{thm}\label{dixon9} \bnm &&\sum_{k=0}^{2n}\frac{(-1)^k{{2n}\choose k}{{k-\theta x}\choose {k}}{{2n-k-\theta x}\choose{2n-k}}} {{{k+\lambda x}\choose{k}}{{2n-k+\lambda x}\choose {2n-k}}}\Big\{\Omega^2(x)+\Omega'(x)\Big\}\\ &&\qquad =\frac{(\lambda x+\theta x){{n-\theta x}\choose {n}}{{n-1+\lambda x+\theta x}\choose {n-1}}} {n{{n+\lambda x}\choose {n}}{{2n+\lambda x}\choose {2n}}}\Big\{W(x)^2+\frac{2}{x}W(x)+W'(x)\Big\}, \enm where $\Omega(x)$, $\Omega'(x)$, $W(x)$ and $W'(x)$ are given respectively by \bnm &&\Omega(x)=-\theta H_k(-\theta x)-\theta H_{2n-k}(-\theta x)-\lambda H_k(\lambda x)-\lambda H_{2n-k}(\lambda x);\\ &&\Omega'(x)=-\theta^2H_{k}^{\langle2\rangle}(-\theta x)-\theta^2H_{2n-k}^{\langle2\rangle}(-\theta x) +\lambda^2H_{k}^{\langle2\rangle}(\lambda x)+\lambda^2H_{2n-k}^{\langle2\rangle}(\lambda x);\\ &&W(x)=(\lambda+\theta) H_{n-1}(\lambda x+\theta x)-\theta H_n(-\theta x)-\lambda H_n(\lambda x)-\lambda H_{2n}(\lambda x);\\ &&W'(x)=\lambda^2H_{2n}^{\langle2\rangle}(\lambda x) +\lambda^2H_{n}^{\langle2\rangle}(\lambda x)-\theta^2H_{n}^{\langle2\rangle}(-\theta x)-(\lambda+\theta)^2H_{n-1}^{\langle2\rangle}(\lambda x+\theta x). \enm \end{thm} Note that \bnm \+\+\Omega(0)=-(\lambda+\theta)(H_k+H_{2n-k}); \quad\quad W(0)=(\lambda+\theta)(H_{n-1}-H_{n})-\lambda H_{2n};\\ \+\+\Omega'(0)=(\lambda^2-\theta^2)(H_{k}^{\langle2\rangle}+H_{2n-k}^{\langle2\rangle});\quad W'(0)=\lambda^2H_{2n}^{\langle2\rangle} +(\lambda^2-\theta^2)H_{n}^{\langle2\rangle}-(\lambda+\theta)^2H_{n-1}^{\langle2\rangle}. \enm the case $x \to 0$ of Theorem \ref{dixon9} reads as the following new general formula. \begin{corollary}\label{d-c-4} \bnm \sum_{k=0}^{2n}(-1)^k{{2n}\choose k} \Big\{(\lambda+\theta)(H_{k}+ H_{2n-k})^2+2(\lambda-\theta)H_{k}^{\langle2\rangle}\Big\} =\frac{2}{n}\big\{(\lambda+\theta)(H_{n-1}-H_n)-\lambda H_{2n}\big\}. \enm \end{corollary} Now we present some examples with harmonic numbers from Corollary \ref{d-c-4}. \begin{example}[$\lambda=0$ and $\theta=1$ in Corollary \ref{d-c-4}] \bnm \sum_{k=0}^{2n}(-1)^k{{2n}\choose k}\Big\{(H_{k}+ H_{2n-k})^2-2H_{k}^{\langle2\rangle}\Big\} =-\frac{2}{n^2} . \enm \end{example} \begin{example}[$\lambda=1$ and $\theta=0$ in Corollary \ref{d-c-4}] \bnm \sum_{k=0}^{2n}(-1)^k{{2n}\choose k}\Big\{(H_{k}+ H_{2n-k})^2+2H_{k}^{\langle2\rangle}\Big\} =\frac{2}{n}\Big\{H_{n-1}-H_n-H_{2n}\Big\} . \enm \end{example} \begin{example}[$\lambda=1$ and $\theta=1$ in Corollary \ref{d-c-4}] \bnm \sum_{k=0}^{2n}(-1)^k{{2n}\choose k}(H_{k}+H_{2n-k})^2=\frac{1}{n}\{2H_{n-1}-2H_n-H_{2n}\}. \enm \end{example} \begin{example}[$\lambda=1$ and $\theta=-1$ in Corollary \ref{d-c-4}] \bnm \sum_{k=0}^{2n}(-1)^k{{2n}\choose k} H_{k}^{\langle2\rangle} =-\frac{H_{2n}}{2n}. \enm \end{example} Also there are many other identities can be obtained from Corollary \ref{d-c-4}. Here, we have just presented several of them as examples. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{thebibliography}{99} \bbtmn{bailey}{W. N. Bailey} {Generalized Hypergeometric Series} {Cambridge University Press, Cambridge, 1935} \bbtm{chu13a}{X. Chen and W. Chu} {Dixon's $_3F_2(1)$ and identities involving harmonic numbers and the Riemann zeta function} {Discrete Math.} {$\mathbf{310}$ (2010), 83--91} \bbtm{chu93c}{W. Chu} {A binomial coefficient identity associated with Beukers' conjecture on Ap\'{e}ry numbers} {Electr. J. Comb.} {$\mathbf{11}$ (2004), Paper R15} \bbtm{chu05}{W. Chu and L. D. Donno} {Hypergeometric series and harmonic number identities} {Adv. Appl. Math.} {$\mathbf{34}$ (2005), 123--137} \bbtm{chu11a}{W. Chu and Amy M. Fu} {Dougall-Dixon formula and harmonic number identities} {Ramanujan J.} {$\mathbf{18}$ (2009), 11--31} \bbtm{dri}{K. Driver, H. Prodinger, C. Schneider, and J. Weideman} {Pad\'{e} approximations to the logarithm. III. Alternative methods and additional results} {Ramanujan J.} {$\mathbf{12}$ (2006), 299--314} \bbtmn{newt}{I. Newton} {Mathematical Papers, Vol.\ III} {Cambridge University Press, London, 1969} \bbtm{pau}{P. Paule and C. Schneider} {Computer proofs of a new family of harmonic number identities} {Adv. Appl. Math.} {$\mathbf{31}$ (2003), 359--378} \bbtmn{slater}{L. J. Slater} {Generalized Hypergeometric Functions} {Cambridge University Press, Cambridge, 1966} \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B65; Secondary 33C20. \noindent \emph{Keywords: } binomial coefficients, harmonic numbers, derivation, hypergeometric series. \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received July 17 2010; revised versions received December 15 2010; December 20 2010. Published in {\it Journal of Integer Sequences}, January 4 2011. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}