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\title{{\bf A combinatorial identity of multiple\\  zeta
values with even arguments}\thanks{This work was supported by
Natural Science Foundation of China (Nos.11101245, 61271355,
11271208,
 61202362, 11301302), China Postdoctoral Science
Foundation (No.2013M530869), Natural Science Foundation of
Shandong (No. BS2013SF009).  }}

% input author, affilliation, address and support information as follows;
% the address should include the country, and does not have to include
% the street address

\author{Shifeng Ding, Lihua Feng, Weijun Liu\\
\small Department of Mathematics\\[-0.8ex]
\small Central South University\\[-0.8ex]
\small Changsha 410083, China\\
\small\tt dingssff@163.com}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{December 1, 2013}{?? ??, 2014}{,}\\
\small Mathematics Subject Classifications: 05A19, 11M06}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
% succinct statement of background followed by a listing of the
% principal new results that are to be found in the paper. The abstract
% should be informative, clear, and as complete as possible. Phrases
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\begin{abstract}
Let $\zeta(s_1,s_2,\cdots,s_k;\alpha)$ be the multiple Hurwitz
zeta function. Given two positive integers $k$ and $n$ with $k\leq
n$, let $E(2n, k;\alpha)$ be the sum of all multiple zeta values
with even arguments whose weight is $2n$ and whose depth is $k$.
In this note we  present some generating series for the numbers
$E(2n,k;\alpha)$.

% keywords are optional
\bigskip\noindent \textbf{Keywords:}\ Multiple zeta values;  Recursion algorithm; Generating
series
\end{abstract}

\section{Introduction}
The multiple zeta values
\begin{equation}\zeta(s_1,s_2,\cdots,s_m)=\sum_{n_1>n_2>\cdots>n_k>0}
\frac{1}{n_1^{s_1}n_2^{s_2}\cdots n_m^{s_m}},\nonumber
\end{equation}
are also called Euler-Zagier sums, where $s_1$, $s_2$, $\cdots$, $
s_m$ are positive integers with $s_1\geq2$. The multiple Hurwitz
zeta function $\zeta(s_1,s_2,\cdots,s_k;\alpha)$ is defined by the
multiple series
\[\zeta(s_1,s_2,\cdots,s_k;\alpha)=\sum_{n_1>n_2>\cdots>n_k\geq0}
\frac{1}{(n_1+\alpha)^{s_1}(n_2+\alpha)^{s_2}\cdots
(n_k+\alpha)^{s_k}},\] where
$\alpha\in\mathbb{C}\setminus\{0,-1,-2,\cdots\}$ and
$s_1,s_2,\cdots,s_k$ are positive integers with $s_1>1$. The
multiple zeta functions have attracted considerable interest in
recent years.

There are various relations among multiple zeta values. One of the
well-known Q-linear relations among multiple zeta values is the
sum formula\ ( see\
$\cite{Granville,Hoffman4,HoffmanAndMoen,Kaneko,Markett}$), which
states that
\[\sum_{\stackrel{j_1+\cdots+j_k=n}{\textnormal{\scriptsize {Each}}j_i\geq1,j_1>1}}
\zeta(j_1,j_2,\cdots,j_k)=\zeta(n).\]

In \cite{Masahiro Igarashi}, M. Igarashi proved a generalization
of the sum formula. Here we quote Igarashi's result in a slightly
different form.
\begin{theorem}{\textnormal{(\cite{Masahiro Igarashi})}}\ \ Let $\alpha$ be a complex number with
$\textnormal{Re}\ \alpha>0$ and let $n,k$ be positive integers
such that $n\geq k+1$. Then the following identity holds.
\begin{eqnarray}&&\sum_{\stackrel{j_1+j_2+\cdots+j_k=n}{\textnormal{Each}j_i\geq1,
j_1\geq2}}\zeta(j_1,j_2,\cdots,j_k;\alpha)\nonumber\\
&=&\frac{1}{(n-k-1)!}\sum_{N=0}^\infty\frac{1}{(N+\alpha)^k}\left.\left[\frac{(1-x)_{N}}
{(\alpha-x)_{N+1}}\right]^{(n-k-1)}\right|_{x=0},\nonumber
\end{eqnarray}
where $[\ \cdot \ ]^{(n-k-1)}$ denotes the $(n-k-1)$th derivative
of a function with respect to its variable $x$, and $(a)_N$
denotes the Pochhammer symbol.
\end{theorem}

Given two positive integers $n$ and $k$ with $n\geq k$, let $E(2n,
k;\alpha)$ be the sum of all the multiple zeta values of
even-integer arguments having weight $2n$ and depth $k$, i.e.,
\[E(2n,k;\alpha)=\sum_{\stackrel{j_1+\cdots+j_k=n}{j_1,j_2,\cdots,j_k\geq1}}
\zeta(2j_1,2j_2,\cdots,2j_k;\alpha).\] (The notation
$E(2n,k;\alpha)$ is similar to that in  \cite{Hoffman, Zhao}. In
\cite{Hoffman}, Hoffman defined a number$E(2n,k)$ which is
equivalent to the number $E(2n,k;1)$ in our notation, and in
\cite{Zhao}, the number $E(2n,k;1/2)/2^{2n}$ in our notation is
denoted by $T(2n,k)$.)

The result of Gangl, Kaneko and Zagier \cite{Gangl} that
\begin{eqnarray}E(2n,2;1)=\frac{3}{4}\zeta(2n), \ \ {\textnormal{for }}\
n\geq2.\nonumber
\end{eqnarray}
was extended by Shen and Cai \cite{Z.Y.Shen} to

\begin{equation}E(2n,3;1)=\frac{5}{8}\zeta(2n)-\frac{1}{4}\zeta(2)\zeta(2n-2),\
\ \  {\textnormal{for }}n\geq3,\nonumber
\end{equation}
\begin{equation}E(2n,4;1)=\frac{35}{64}\zeta(2n)-\frac{5}{16}\zeta(2)\zeta(2n-2),\
\ \ {\textnormal{for }}n\geq4.\nonumber
\end{equation}

Applying the theory of symmetric functions, Hoffman \cite{Hoffman}
and Zhao \cite{Zhao}, respectively, established the generating
functions for the numbers $E(2n,k;1)$ and $T(2n,k)$, respectively.
Hoffman \cite{Hoffman} proved that
\[1+\sum_{n\geq k\geq1}E(2n,k;1)t^ns^k=\frac{\sin(\pi\sqrt{1-s}\sqrt{t})}{\sqrt{1-s}\sin(\pi\sqrt{t})}.\]
And Zhao \cite{Zhao} proved that
\[1+\sum_{n\geq k\geq1}T(2n,k)t^ns^k=\frac{\cos(\pi\sqrt{(1-s)t}/2)}{\cos(\pi\sqrt{t}/2)}.\]
Based on these generating functions, some  formulas for the
numbers $E(2n,k;1)$ and $T(2n,k)$ for arbitrary $n\geq k$ have
been obtained. For example,
\begin{eqnarray}E(2n,k;1)=\frac{\zeta(2n)}{2^{2(k-1)}}{2k-1\choose k}-
\sum_{j=1}^{[\frac{k-1}{2}]}\frac{\zeta(2j)\zeta(2n-2j)}{2^{2k-3}(2j+1)B_{2j}}{2k-2j-1\choose
k}, \ \ \ \nonumber\end{eqnarray}
\[T(2n,k)=\frac{t(2n)}{2^{2(k-1)}k}{2k-2\choose k-1}-
\sum_{j=1}^{\lfloor\frac{k-1}{2}\rfloor}\frac{t(2j)t(2n-2j)}{2^{2k-3}(2^{2j}-1)kB_{2j}}{2k-2j-2\choose
k-1},\]
 where $B_{2j}$ is the $2j$th Bernoulli number.


In this note we use a method introduced by Granville
\cite{Granville} to establish an algorithm similar to that in
Theorem 1 for the calculation of the numbers $E(2n,k;\alpha)$. We
prove that the calculation of $E(2n,k;\alpha)$ involves the Euler
$\Gamma$-function and the direct formulas for $E(2n,k;1)$ and
$E(2n,k;1/2)$ can be deduced from the Bessel functions of the
first kind. The main results of this paper are the following
theorems.


\begin{theorem}\ Let $\alpha$ be a complex number with
$\alpha\in\mathbb{C}\setminus\{0,-1,-2,\cdots\}$. Let $N\geq0$
denote an integer and let $(a_0^{(N)},a_1^{(N)},a_2^{(N)},\cdots)$
 be an infinite series defined by the expansion
\begin{equation}\prod_{\stackrel{r=0}{r\not=N}}^\infty\left(1+\frac{x}{(r+\alpha)^2-(N+\alpha)^2}\right)=a_0^{(N)}+a_1^{(N)}x+\cdots+a_{k-1}^{(N)}x^{k-1}+\cdots.
\end{equation}
Then for positive integers $n$ and $k$ (suppose $n\geq k$), we
have
\begin{equation}E(2n,k;\alpha)=\sum_{N=0}^\infty\frac{a_{k-1}^{(N)}}{(N+\alpha)^{2n-2k+2}}.\end{equation}
\end{theorem}



\begin{theorem}\ Given an integer $N\geq0$ and a complex number
$\alpha\in\mathbb{C}\setminus\{0,-1,-2,\cdots\}$, for any
$x\in\mathbb{C}$  we have
\begin{eqnarray}&&\prod_{\stackrel{r=0}{r\not=N}}^\infty\left(1+\frac{x}{(r+\alpha)^2-(N+\alpha)^2}\right)\nonumber\\
&=&\frac{2(-1)^{N}(N+\alpha)\Gamma(2\alpha+N)}{N!}\frac{1}{x\Gamma(\alpha+\sqrt{(N+\alpha)^2-x})\Gamma(\alpha-\sqrt{(N+\alpha)^2-x})},
\nonumber
\end{eqnarray}
where $\Gamma$ denotes Euler's Gamma function.

\end{theorem}


\begin{corollary}\ Theorem 2 and Theorem 3 yield that: \begin{eqnarray}&&E(2n,k;\alpha)\nonumber\\
&=&\frac{2}{k!}\sum_{N=0}^\infty\frac{(-1)^N\Gamma(N+2\alpha)}{N!(N+\alpha)^{2n-2k+1}}
\left.\left[\frac{1}{\Gamma(\alpha+\sqrt{(N+\alpha)^2-x})
\Gamma(\alpha-\sqrt{(N+\alpha)^2-x})}\right]^{(k)}\right|_{x=0},\nonumber
\end{eqnarray}
where $[\ \cdot \ ]^{(k)}$ denotes the $k$th derivative of a
function with respect to its variable $x$ and\ $\Gamma$  denotes
the Gamma function.
\end{corollary}

\begin{corollary}\ For positive integers $n $ and $k$ (suppose $n\geq k$ ), we have
\begin{eqnarray}E(2n,k;1)&=&\frac{4(-1)^{k+1}\pi^{2k}}{k!}\sum_{N=0}^\infty
\frac{(-1)^N}{(N+1)^{2n-2k}}\left(\cos\sqrt{x}\right)^{(k+1)}|_{x=(N\pi+\pi)^2},\\
E(2n,k;\frac{1}{2})&=&\frac{2(-1)^{k}\pi^{2k-1}}{k!}\sum_{N=0}^\infty
\frac{(-1)^N}{(N+\frac{1}{2})^{2n-2k+1}}\left(\cos\sqrt{x}\right)^{(k)}|_{x=(N\pi+\pi/2)^2}.
\end{eqnarray}
\end{corollary}

\begin{corollary}\ The direct formulas for $E(2n,k;1)$ and
$E(2n,k;\frac{1}{2})$ are that:
\begin{eqnarray}&&E(2n,k;1)=\frac{1}{2^{2k-2}k!}
\sum_{j=0}^{\lfloor\frac{k-1}{2}\rfloor}\frac{(-1)^{j}(2k-1-2j)!(2\pi)^{2j}}{(2j+1)!(k-1-2j)!}\cdot\zeta(2n-2j),\\
&&E(2n,k;\frac{1}{2})=\frac{1}{2^{2k-2}k!}\sum_{j=0}^{\lfloor\frac{k-1}{2}\rfloor}\frac{(-1)^j(2k-2-2j)!(2\pi)^{2j}}{(2j)!(k-1-2j)!}
\zeta(2n-2j;\frac{1}{2}).
\end{eqnarray}
\end{corollary}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Some Lemmas}

This paper uses some properties of Euler's Gamma function
$\Gamma(s)$ and  the Bessel function $J_{p}(x)$ of the first kind.
Here we list some needed Lemmas.

There are several equivalent forms of the Gamma function
$\Gamma(s)$ developed by Euler, one by Weierstrass:

\[\frac{1}{\Gamma(s)}=se^{\gamma s}\prod_{m=1}^\infty\left(1+\frac{s}{m}\right)e^{-\frac{s}{m}},\ \ \forall s\in\mathbb{C},\]
where $\gamma$ denotes the Euler-Mascheroni constant defined by
$$\gamma=\lim_{m\rightarrow\infty}(1+\frac{1}{2}+\cdots+\frac{1}{m}-\log
m)=0.577215\cdots.$$ It is easy to deduce from Weierstrass'
definition of $\Gamma(s)$ the following lemma.

\begin{lemma}\ Let $\alpha_1,\alpha_2,\cdots,\alpha_k$ and
$\beta_1,\beta_2,\cdots,\beta_k$ be two groups of complex numbers
\textnormal{($k\geq1$)}. Suppose they satisfy the following
conditions{\textnormal{:\ (i)}}\
$\alpha_1+\alpha_2+\cdots+\alpha_k=\beta_1+\beta_2+\cdots+\beta_k$
and{\textnormal{ (ii)}} none of the $\beta_j$s is a negative
integer. Then we have
\[\prod_{m=1}^\infty\frac{(m+\alpha_1)(m+\alpha_2)\cdots(m+\alpha_k)}{(m+\beta_1)(m+\beta_2)\cdots(m+\beta_k)}
=\frac{\Gamma(1+\beta_1)\Gamma(1+\beta_2)\cdots\Gamma(1+\beta_k)}{\Gamma(1+\alpha_1)\Gamma(1+\alpha_2)\cdots\Gamma(1+\alpha_k)}.
\]
\end{lemma}

It is well known that $\Gamma(s)$ satisfies the functional
relation $\Gamma(s+1)=s\Gamma(s), \ \forall
s\in\mathbb{C}\setminus\{0,-1,-2,\cdots\}$, and if $N\geq0$ is an
integer then $\Gamma(s)$ has residue $\frac{(-1)^N}{N!}$ at
$s=-N$.

\begin{lemma}\textnormal{\cite{Remmert}} The following identities
hold for $s\in\mathbb{C}$:
\[\frac{1}{\Gamma(1+s)\Gamma(1-s)}=\frac{\sin\pi s}{\pi s},\ \ \ \
\ \ \
\frac{1}{\Gamma(\frac{1}{2}+s)\Gamma(\frac{1}{2}-s)}=\frac{\cos\pi
s}{\pi } .\ \]
\end{lemma}



We turn to introduce some properties of the  Bessel function of
the first kind with a half-integer index. The Bessel function
$J_p(x)$ is said to be of a half integer index if $p=
k\pm\frac{1}{2}$ with $k$ being an integer. It is well known that
the Bessel function of the first kind with a half integer index
can be represented by elementary functions.

\begin{lemma}\ \ $\frac{d^k}{dx^k}\cos\left(\sqrt{x}\right)=\sqrt{\frac{\pi}{2}}(-1)^k2^{-k}x^{\frac{1-2k}{4}}
J_{k-\frac{1}{2}}(\sqrt{x}), \ \ \textnormal{for} \ x>0\
\textnormal{and}\ \ k=0,1,2,\cdots.$
\end{lemma}

\begin{proof}\ The Bessel function $J_p(x)$ is defined by the
series
\begin{equation}J_p(x)=\left(\frac{x}{2}\right)^{p}\sum_{m=0}^\infty\frac{(-1)^m}{m!\Gamma(p+m+1)}\left(\frac{x}{2}\right)^{2m},\end{equation}
where the radius of convergence of the series is $+\infty$. It
follows from (7) that  (using the well-known identity
$\Gamma(n+1/2)=\sqrt{\pi}(2n)!/(n!2^{2n})$, $\forall n\geq0$)

\begin{equation}J_{k-\frac{1}{2}}(\sqrt{x})=\sqrt{\frac{2}{\pi}}2^kx^{\frac{2k-1}{4}}\sum_{m=0}^\infty\frac{(-1)^m(m+k)!}
{m!(2m+2k)!}x^m.\end{equation}

On the other hand, by Taylor's expansion of $\cos\sqrt{x}$, we
have
\begin{eqnarray}\frac{d^k}{dx^k}\cos\left(\sqrt{x}\right)=\sum_{m=k}^\infty\frac{(-1)^mm(m-1)\cdots(m-k+1)}
{(2m)!}x^{m-k}=\sum_{m=0}^\infty\frac{(-1)^{m+k}(m+k)!}
{(2m+2k)!m!}x^{m}.\end{eqnarray} Observing the right-hand sides of
(8) and (9), the result follows immediately.
\end{proof}


\begin{lemma}\ Let
$k\geq0$ be an integer and let  $x>0$. Then the Bessel function
$J_{k+1/2}(x)$ is represented by

\begin{eqnarray}J_{k+\frac{1}{2}}(x)&=&\sqrt{\frac{2}{\pi }}\frac{1}{\sqrt{x}}\left\{\sin x
\sum_{j=0}^{\lfloor\frac{k}{2}\rfloor}\frac{(-1)^j(2k-2j)!}{(2j)!(k-2j)!}\cdot\frac{1}{(2x)^{k-2j}}\right.\nonumber\\
&&\left.\ \ \ \ \ \ \ \ \ \ \ \ -\cos x
\sum_{j=0}^{\lfloor\frac{k-1}{2}\rfloor}\frac{(-1)^j(2k-1-2j)!}{(2j+1)!(k-1-2j)!}\cdot\frac{1}{(2x)^{k-1-2j}}\right\}.
\end{eqnarray}
\end{lemma}

\begin{proof}\ \ We omit the detailed steps. It is shown that\ (\cite{Y. Liu}) the
Bessel function $J_{k+1/2}(x)$  satisfies the relation
\begin{equation}J_{k+1/2}(x)=(-1)^k\sqrt{\frac{2}{\pi
x}}x^{k+1}\frac{d^k}{(xdx)^k}\left(\frac{\sin x}{x}\right),\ \
\textnormal{for}\ \ k=0,1,2,\cdots.\end{equation} It follows from
(11) that
\[J_{1/2}=\sqrt{\frac{2}{\pi
}}\frac{\sin x}{\sqrt{x}},\ \ \ \ \ J_{3/2}(x)=\sqrt{\frac{2}{\pi
}}\frac{1}{\sqrt{x}}\left(\frac{\sin x}{x}-\cos x\right),\cdots,\]
and so on. Then using induction on $k$ we can prove that (10) is
equivalent to (11).\end{proof}



\begin{lemma}Let $k\geq1$ be an integer. Then by Lemma 9 and Lemma 10, we have
\begin{eqnarray}\left(\cos{\sqrt{x}}\right)^{(k)}&=&\frac{(-1)^{k}}{2^{2k-1}}\frac{\sin\sqrt{x}}{\sqrt{x}}
\sum_{j=0}^{\lfloor\frac{k-1}{2}\rfloor}\frac{(-1)^j(2k-2-2j)!2^{2j}}{(2j)!(k-1-2j)!}\cdot\frac{1}{x^{k-1-j}}\nonumber\\
&& -\frac{(-1)^{k}}{2^{2k-2}}\cos \sqrt{x}
\sum_{j=0}^{\lfloor\frac{k-2}{2}\rfloor}\frac{(-1)^j(2k-3-2j)!2^{2j}}{(2j+1)!(k-2-2j)!}\cdot\frac{1}{x^{k-1-j}}.
\end{eqnarray}
\end{lemma}





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

 \section{ Proofs}
\textnormal{(}{\it Proof of Theorem 2 }\textnormal{)}

\begin{proof}\ The left side of (2) is


\begin{eqnarray}
&&\sum_{\stackrel{j_1+\cdots+j_k=n}{j_1,\cdots,j_k\geq1}}\sum_{n_1>n_2>\cdots>n_k\geq0}
\frac{1}{(n_1+\alpha)^{2j_1}(n_2+\alpha)^{2j_2}\cdots(n_k+\alpha)^{2j_k}}\nonumber\\
&&\ \ \ \
=\sum_{n_1>n_2>\cdots>n_k\geq0}\sum_{\stackrel{j_1+\cdots+j_k=n}{j_1,\cdots,j_k\geq1}}
\frac{1}{(n_1+\alpha)^{2j_1}(n_2+\alpha)^{2j_2}\cdots(n_k+\alpha)^{2j_k}}.\end{eqnarray}




The second sum in (13) is the coefficient of $x^{2n}$ in the
formal power series
\begin{eqnarray}&&\sum_{j=1}^\infty\left(\frac{x}{n_1+\alpha}\right)^{2j}
\sum_{j=1}^\infty\left(\frac{x}{n_2+\alpha}\right)^{2j}\cdots\sum_{j=1}^\infty\left(\frac{x}{n_k+\alpha}\right)^{2j}\nonumber\\
&&\ \
=\frac{x^{2k}}{[(n_1+\alpha)^2-x^2][(n_2+\alpha)^2-x^2]\cdots[(n_k+\alpha)^2-x^2]}\nonumber\\
&&\ \
=\sum_{j=1}^{k}\left(\frac{x^{2k}}{(n_j+\alpha)^2-x^2}\prod_{\stackrel{1\leq
m\leq k}{m\not=j}}\frac{1}{(n_m+\alpha)^2-(n_j+\alpha)^2}\right).
\end{eqnarray}
It follows that the coefficient of $x^{2n}$ above is
\[\sum_{j=1}^k\left(\frac{1}{(n_j+\alpha)^{2n-2k+2}}\prod_{\stackrel{1\leq
m\leq
k}{m\not=j}}\frac{1}{(n_m+\alpha)^2-(n_j+\alpha)^2}\right).\]
Hence the sum (13) is
\begin{equation}\sum_{n_1>n_2>\cdots>n_k\geq0}\sum_{j=1}^k\left(\frac{1}{(n_j+\alpha)^{2n-2k+2}}\prod_{\stackrel{1\leq
m\leq k}{m\not=j}}\frac{1}{(n_m+\alpha)^2-(n_j+\alpha)^2}\right).
\end{equation}

Now  we take each $n_j=N$ in turn. Then the sum (15) becomes that

\begin{equation}\sum_{N\geq0}\frac{1}{(N+\alpha)^{2n-2k+2}}\sum_{j=1}^k
P(N,j-1)Q(N,k-j),
\end{equation}
where the series $\{P(N,i)\}_{i=0}^\infty$ and
$\{Q(N,i)\}_{i=0}^\infty$ are defined in the following manner.
Define $P(N,0)=Q(N,0)\triangleq1$, and $Q(N,j)=0$, $\forall j> N$;
If $j>1$ then define $P(N,j-1)$ to be
\begin{equation}P(N,j-1)=\sum_{n_1>\cdots>n_{j-1}>N}\prod_{i=1}^{j-1}\frac{1}{[(n_i+\alpha)^2-(N+\alpha)^2]};\nonumber\end{equation}
While if  $1\leq j<k $ and $k-j\leq N$ , then define $Q(N,k-j)$ to
be
\begin{equation}Q(N,k-j)=\sum_{N>n_{j+1}>\cdots>n_k\geq0}\prod_{i=j+1}^{k}\frac{1}{[(n_i+\alpha)^2-(N+\alpha)^2]}.\nonumber\end{equation}
In other words, the series $\{P(N,i)\}_{i=0}^\infty$ and
$\{Q(N,i)\}_{i=0}^\infty$ are defined by  the following generating
functions:
\[1+P(N,1)x+P(N,2)x^2+\cdots=\prod_{r>N}\left(1+\frac{x}{(r+\alpha)^2-(N+\alpha)^2}\right),\]
and \[1+Q(N,1)x+Q(N,2)x^2+\cdots=\prod_{0\leq
r<N}\left(1+\frac{x}{(r+\alpha)^2-(N+\alpha)^2}\right).\]

Now the second sum \ $\sum_{j=1}^kP(N,j-1)Q(N,k-j)$ in (16) is the
coefficients of $x^{k-1}$ in the power series
\[[1+P(N,1)x+P(N,2)x^2+\cdots]\cdot[1+Q(N,1)x+Q(N,2)x^2+\cdots],\]
i.e.,
\[\sum_{j=1}^kP(N,j-1)Q(N,k-j)=a_{k-1}^{(N)},\]
where $a_{k-1}^{(N)}$ is given by (1).

Therefore, the sum (16) is
\[\sum_{N\geq0}\frac{a_{k-1}^{(N)}}{(N+\alpha)^{2n-2k+2}},\]
which gives the identity (2).
\end{proof}


\noindent\textnormal{(}{\it Proof of Theorem 3 }\textnormal{)}

\begin{proof}\ \ Consider the infinite product
\[\prod_{r=0}^\infty\left(1+\frac{x}{(r+\alpha)^2-y^2}\right),\]
where $y$ is a complex number with
$y\not\in\{\alpha,1+\alpha,2+\alpha,\cdots\}$.

By Lemma 7, we have
\begin{eqnarray}&&\prod_{r=0}^\infty\left(1+\frac{x}{(r+\alpha)^2-y^2}\right).\nonumber\\
&=&\frac{\alpha^2-y^2+x}{\alpha^2-y^2}\prod_{r=1}^\infty\frac{(r+\alpha+\sqrt{y^2-x})
(r+\alpha-\sqrt{y^2-x})}{(r+\alpha+y)(r+\alpha-y)}\nonumber\\
&=&\frac{\alpha^2-y^2+x}{\alpha^2-y^2}\frac{\Gamma(1+\alpha+y)\Gamma(1+\alpha-y)}{\Gamma(1+\alpha+\sqrt{y^2-x})\Gamma(1+\alpha-\sqrt{y^2-x})}\nonumber\\
&=&\frac{\Gamma(\alpha+y)\Gamma(\alpha-y)}{\Gamma(\alpha+\sqrt{y^2-x})\Gamma(\alpha-\sqrt{y^2-x})}.
\end{eqnarray}


 It follows that
\begin{eqnarray}\prod_{\stackrel{r=0}{r\not=N}}^\infty\left(1+\frac{x}{(r+\alpha)^2-y^2}\right)=
\frac{\Gamma(\alpha+y)\Gamma(\alpha-y)}{\Gamma(\alpha+\sqrt{y^2-x})\Gamma(\alpha-\sqrt{y^2-x})}\times\frac{(N+\alpha)^2-y^2}{(N+\alpha)^2-y^2+x}.
\end{eqnarray}
By the residue formula of the Gamma function, we have
\[\lim_{y\rightarrow
N+\alpha}\Gamma(\alpha-y)(N+\alpha-y)=\lim_{s\rightarrow-N}\Gamma(s)(s+N)=\frac{(-1)^N}{N!}.\]
Now on both sides of (18) we put $y\rightarrow N+\alpha$. Then we
get the required  identity in Theorem 3.
\end{proof}

\noindent\textnormal{(}{\it Proof of Corollary 5 and Corollary 6
}\textnormal{)}

\begin{proof}\ \ In Corollary 4, we take $\alpha=1$ and $\alpha=1/2$,
respectively. By Lemma 8, we have
\[\frac{1}{\Gamma(1+\sqrt{(N+1)^2-x})
\Gamma(1-\sqrt{(N+1)^2-x})}=\frac{\sin(\pi\sqrt{(N+1)^2-x})}{\pi\sqrt{(N+1)^2-x}},\]

\[\frac{1}{\Gamma(\frac{1}{2}+\sqrt{(N+\frac{1}{2})^2-x}\ )
\Gamma(\frac{1}{2}-\sqrt{(N+\frac{1}{2})^2-x}\
)}=\frac{1}{\pi}\cos\left(\pi\sqrt{(N+\frac{1}{2})^2-x}\right).\]
It is easy to verify that
\begin{eqnarray}&&\left.\left(\frac{\sin\left(\pi\sqrt{(N+1)^2-x}\right)}{\pi\sqrt{(N+1)^2-x}}\right)^{(k)}\right|_{x=0}
=(-1)^{k}\pi^{2k}\left.\left(\frac{\sin\sqrt{x}}{\sqrt{x}}\right)^{(k)}\right|_{x={(N\pi+\pi)^2}}\nonumber\\
&=&2(-1)^{k+1}\pi^{2k}\left.\left(\cos\sqrt{x}\right)^{(k+1)}\right|_{x={(N\pi+\pi)^2}},
\end{eqnarray}

\begin{eqnarray}&&\left.\left(\cos(\pi\sqrt{(N+\frac{1}{2})^2-x}\ )\right)^{(k)}\right|_{x=0}
=(-1)^{k}\pi^{2k}\left.(\cos\sqrt{x})^{(k)}\right|_{x={(N\pi+\pi/2)^2}}.
\end{eqnarray}

Substitute the right-hand side of (19) into  the right-hand side
of the identity in Corollary 4, then we get the identity (3).
Similarly, we can prove (4).

To prove Corollary 6, we use Lemma 11. It is deduced from (12)
that
\begin{eqnarray}&&\left.\left(\cos\sqrt{x}\right)^{(k+1)}\right|_{x={(N\pi+\pi)^2}}\nonumber\\
&&=\frac{(-1)^{(N+k+1)}}{(2\pi)^{2k}}\sum_{j=0}^{\lfloor\frac{k-1}{2}\rfloor}\frac{(-1)^j(2k-1-2j)!}{(2j+1)!(k-2j-1)!}\frac{(2\pi)^{2j}}{(N+1)^{2k-2j}},
\end{eqnarray}
\begin{eqnarray}&&\left.\left(\cos\sqrt{x}\right)^{(k)}\right|_{x={(N\pi+\pi/2)^2}}\nonumber\\
&&=\frac{(-1)^{(N+k)}}{(2\pi)^{2k-1}}\sum_{j=0}^{\lfloor\frac{k-1}{2}\rfloor}
\frac{(-1)^j(2k-2-2j)!}{(2j)!(k-2j-1)!}\frac{(2\pi)^{2j}}{(N+\frac{1}{2})^{2k-1-2j}}.
\end{eqnarray}
Substitute (21) and (22) into (3) and (4) respectively, then we
will get the desired identities (5) and (6).
\end{proof}

\section{Generalizations}

Similar to the definition of $E(2n,k;\alpha)$, we can define a
number $E(mn,k;\alpha)$ as follows. Let $m,n$ and $k$ be positive
integers with $m\geq2$ and $n\geq k$. Then define the number
$E(mn,k;\alpha)$ to be
\[E(mn,k;\alpha)=\sum_{\stackrel{j_1+\cdots+j_k=n}{j_1,j_2,\cdots,j_k\geq1}}
\zeta(mj_1,mj_2,\cdots,mj_k;\alpha).\]

If we check the proofs of Theorem 2 and Theorem 3 again, we may
find that the theorems can be generalized in the following way.
\begin{theorem} Let $\alpha$ be a complex number with
$\alpha\not\in\{0,-1,-2,\cdots\}$ and let $N\geq0$ denote an
arbitrary integer. Define a function $f_{N}(x)$ by
\[f_N(x)=\prod_{\stackrel{r=0}{r\not=N}}^\infty\left(1+\frac{x}{(r+\alpha)^m-(N+\alpha)^m}\right).
\]
Then for positive integers $n$ and $k$ (suppose $n\geq k$), we
have
\[E(mn,k;\alpha)=\frac{1}{(k-1)!}\sum_{N=0}^\infty\frac{[f_N(x)]^{(k-1)}|_{x=0}}{(N+\alpha)^{mn-mk+m}}.\]
Moreover, the function $f_N(x)$ can be represented by the Gamma
function, i.e.,
\[f_N(x)=\frac{(-1)^Nm(N+\alpha)^{m-1}}{N!}\cdot\frac{1}{x}\cdot
\frac{\prod_{i=1}^{m-1}\Gamma(\alpha-\theta_i(N+\alpha))}{\prod_{i=1}^m\Gamma(\alpha-\theta_i{\sqrt[m]{(N+\alpha)^m-x}})},\]
where $\theta_1$, $\theta_2$, $\cdots$, $\theta_m(\triangleq1)$
are all the pairwise different roots of  $Z^m=1$ in $\mathbb{C}$.
\end{theorem}
\subsection*{Acknowledgements}
The authors would like to thank the referee for helpful
suggestions.




























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