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\begin{document}

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\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}


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\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
\[\Gamma(x)=\int_0^\infty u^{x-1}e^{-u}\text{d}u
\xqdp \text{with}\xqdp \mathfrak{R}(x)&gt;0.\]
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
&amp;&amp;\Omega (x)=\lambda H_{2n-k}(\lambda x)+\theta H_{2n-k}(\theta x)
+\lambda H_k(\lambda x)+\theta H_k(\theta x);\\
&amp;&amp;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
&amp;&amp;\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);\\
&amp;&amp;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
&amp;&amp;\Omega'(0)=-(\lambda^2+\theta^2)(H_{2n-k}^{\langle2\rangle}+H_k^{\langle2\rangle});\\
&amp;&amp;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
&amp;&amp;\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\}\\
&amp;&amp;=\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
&amp;&amp;\Omega(0)=(\lambda+ \theta )(H_{k}+H_{2n-k});\\
&amp;&amp;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
&amp;&amp;\Omega'(0)=(\lambda^2+\theta^2)(H_{k}^{\langle2\rangle}+H_{2n-k}^{\langle2\rangle});\\
&amp;&amp;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
&amp;&amp;\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\}\\
&amp;&amp;=(-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
&amp;&amp;\sum_{k=0}^{2n}(-1)^k{{2n}\choose k}^3
\Big\{(H_{k}+H_{2n-k})^2+H_{k}^{\langle2\rangle}\Big\}\\
&amp;&amp;=(-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
&amp;&amp;\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\}\\
&amp;&amp;\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
&amp;&amp;\Omega(x)=-\theta H_k(-\theta x)-\theta H_{2n-k}(-\theta x)-\lambda H_k(\lambda x)-\lambda H_{2n-k}(\lambda x);\\
&amp;&amp;\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);\\
&amp;&amp;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);\\
&amp;&amp;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.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\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/}.
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