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\begin{center}
\vskip 1cm{\LARGE\bf Integrals and Polygamma Representations for \\
\vskip .12in
Binomial Sums
}
\vskip 1cm
\large
Anthony Sofo\\
School of Engineering and Science \\
Victoria University\\
PO Box 14428 \\
Melbourne City, VIC 8001 \\
Australia\\
\href{mailto:anthony.sofo@vu.edu.au}{\tt anthony.sofo@vu.edu.au}
\end{center}

\vskip .2 in

\begin{abstract}
We consider sums involving the product of reciprocal binomial coefficient
and polynomial terms and develop some double integral identities. In
particular cases it is possible to express the sums in closed form, give
some general results, recover some known results in Coffey and
produce new identities.
\end{abstract}


\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{defin}[theorem]{Definition}
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\def\QATOP#1#2{{#1 \atop #2}}

\section{Introduction}

In a recent paper Coffey \cite{10} considers summations over
digamma and polygamma functions and develops many results, namely two of his
propositions, in terms of the Riemann zeta function $\zeta \left( \cdot
\right) $, are respectively equations (59) and (66a)%
\begin{equation}
\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n}\ }{n^{p+1}\left( n+1\right) }%
=\left( -1\right) ^{p}\left( 1-2\ln 2\right) +\sum_{m=1}^{p}\left( -1\right)
^{p+m}~\left( 2^{-m}-1\right) \zeta \left( m+1\right)  \label{001}
\end{equation}%
and 
\begin{equation}
\sum_{n=1}^{\infty }\frac{1\ }{n^{p+1}\left( n+1\right) }=\left( -1\right)
^{p}+\sum_{m=1}^{p}\left( -1\right) ^{p+m}~\zeta \left( m+1\right) .
\label{002}
\end{equation}%
Coffey \cite{10} also constructs new integral representations for these
sums. The major aim of this paper is to investigate general binomial sums
with various parameters that then enables one to give more general
representations of (\ref{001}) and (\ref{002}), thereby generalizing the
propositions of Coffey, both in closed form in terms of zeta functions and
digamma functions at possible rational values of the argument, and in double
integral form. The following definitions will be useful. The Psi, or digamma
function $\psi \left( z\right) ,$ is defined by%
\begin{equation*}
\psi \left( z\right) =\frac{d}{dz}\log \Gamma \left( z\right) =\frac{\Gamma
^{\prime }\left( z\right) }{\Gamma \left( z\right) }=\sum\limits_{n=0}^{%
\infty }\left( \frac{1}{n+1}-\frac{1}{n+z}\right) -\gamma
\end{equation*}%
where $\gamma $ denotes the Euler-Mascheroni constant and $\Gamma \left(
z\right) $ is the Gamma function. Similarly%
\begin{equation}
\psi \left( z+1\right) =\sum\limits_{n=1}^{\infty }\left( \frac{1}{n}-\frac{1%
}{n+z}\right) -\gamma ,  \label{0.1}
\end{equation}%
and 
\begin{equation}
2\psi \left( 2z\right) =\psi \left( z\right) +\psi \left( z+\frac{1}{2}%
\right) +2\ln 2.  \label{0.2}
\end{equation}

Sums of reciprocals of binomial coefficients appear in the calculation of
massive Feynman diagrams \cite{11.3} within several different approaches:
for instance, as solutions of differential equations for Feynman amplitudes,
through a naive $\varepsilon $-expansion of hypergeometric functions within
Mellin-Barnes technique or in the framework of recently proposed algebraic
approach \cite{11.2}. \ There has recently been a renewed interest in the
study of series involving binomial coefficients and a number of authors have
obtained either closed form representation or integral representation for
some particular cases of these series. The interested reader is referred to 
\cite{1,alzer1,3,4,9,9.1,11,11.1,11.35,11.4,15,16,17,19,sof1,19.5,20,20.5,21}.

\section{The main results}

The following Lemma and well-known definition will be useful in the proof of
the main theorem.

\begin{definition}
\label{lemma1} Let $\left\vert z\right\vert \leq 1,~m\geq 1$ and $q\neq
-1,-2,-3,....$ Then%
\begin{equation}
\sum_{r=1}^{\infty }\frac{z^{r}\ }{\left( q+r\right) ^{m}}=\frac{\left(
-1\right) ^{m-1}}{\left( m-1\right) !}\int\limits_{0}^{1}\frac{%
z~y^{q~}\left( \ln \left( y\right) \right) ^{m-1}}{1-zy}dy.  \label{1.1}
\end{equation}
\end{definition}

\bigskip The next Lemma deals with two infinite sums.

\begin{lemma}
\label{lemma2} Let $a$ and $r$ be positive real numbers. Then%
\begin{equation}
\ \sum_{n=1}^{\infty }\frac{1}{n\left( an+r\right) }=\frac{H_{\frac{r}{a}%
}^{\left( 1\right) }}{r}\ \ \text{and }  \label{4.1}
\end{equation}%
\begin{equation}
\ \sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n}}{n\left( an+r\right) }=%
\frac{1}{r}\left\{ H_{\frac{r}{2a}}^{\left( 1\right) }-H_{\frac{r}{a}%
}^{\left( 1\right) }\right\} .  \label{4.2}
\end{equation}
\end{lemma}

\begin{proof}
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{1}{n\left( an+r\right) } &amp;=&amp;\frac{1}{r}%
\sum_{n=1}^{\infty }\left( \frac{1}{n}-\frac{1}{n+\frac{r}{a}}\right) \text{
and from (\ref{0.1})} \\
&amp;=&amp;\frac{1}{r}\left[ \gamma +\psi \left( \frac{r}{a}+1\right) \right] \\
&amp;=&amp;\frac{H_{\frac{r}{a}}^{\left( 1\right) }}{r}\ \text{; hence (\ref{4.1})
is attained.}
\end{eqnarray*}%
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n}}{n\left( an+r\right) } &amp;=&amp;%
\frac{1}{2r}\left[ \sum_{n=1}^{\infty }\left( \frac{1}{n}-\frac{1}{n+\frac{r%
}{2a}}\right) -\sum_{n=1}^{\infty }\left( \frac{1}{n}-\frac{1}{n+\frac{r-a}{%
2a}}\right) -\sum_{n=1}^{\infty }\frac{1}{n\left( 2n-1\right) }\right] \\
&amp;=&amp;\frac{1}{2r}\left[ -\gamma +\psi \left( \frac{r}{2a}+1\right) +\gamma
-\psi \left( \frac{r}{2a}+\frac{1}{2}\right) -2\ln 2\right] \\
&amp;=&amp;\frac{1}{2r}\left[ \psi \left( \frac{r}{2a}+1\right) -\psi \left( \frac{r%
}{2a}+\frac{1}{2}\right) -2\ln 2\right] ,
\end{eqnarray*}%
from the definition (\ref{0.2}) 
\begin{equation*}
\psi \left( \frac{r}{2a}+\frac{1}{2}\right) =2\psi \left( \frac{r}{a}%
+1\right) -\psi \left( \frac{r}{2a}+1\right) -2\ln 2;\ \text{hence}
\end{equation*}%
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n}}{n\left( an+r\right) } &amp;=&amp;%
\frac{1}{2r}\left[ 2\psi \left( \frac{r}{2a}+1\right) -2\psi \left( \frac{r}{%
a}+1\right) \right] \\
&amp;=&amp;\frac{1}{r}\left[ H_{\frac{r}{2a}}^{\left( 1\right) }-\gamma -H_{\frac{r}{%
a}}^{\left( 1\right) }+\gamma \right] \text{, therefore (\ref{4.2}) follows.}
\end{eqnarray*}
\end{proof}

\begin{remark}
In the following Corollaries and remarks we encounter harmonic numbers at
possible rational values of the argument, of the form $H_{\frac{r}{a}%
}^{\left( \alpha \right) }$ where $r=1,2,3,...,k,\ \alpha =1,2,3,...$ and $%
k\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
.$ \ The polygamma function $\psi ^{\left( \alpha \right) }\left( z\right) $
is defined as: 
\begin{equation*}
\psi ^{\left( \alpha \right) }\left( z\right) =\frac{d^{\alpha +1}}{%
dz^{\alpha +1}}\left[ \log \Gamma \left( z\right) \right] =\frac{d^{\alpha }%
}{dz^{\alpha }}\left[ \psi \left( z\right) \right] ,~z\neq \left\{
0,-1,-2,-3,...\right\} .
\end{equation*}%
To evaluate $H_{\frac{r}{a}}^{\left( \alpha \right) }$ we have available a
relation in terms of the polygamma function $\psi ^{\left( \alpha \right)
}\left( z\right) ,$ for rational arguments $z,$%
\begin{equation*}
H_{\frac{r}{a}}^{\left( \alpha +1\right) }=\zeta \left( \alpha +1\right) +%
\frac{\left( -1\right) ^{\alpha }}{\alpha !}\psi ^{\left( \alpha \right)
}\left( \frac{r}{a}+1\right)
\end{equation*}%
we also define 
\begin{equation*}
H_{\frac{r}{a}}^{\left( 1\right) }=\gamma +\psi \left( \frac{r}{a}+1\right) 
\text{, and }H_{0}^{\left( \alpha \right) }=0.
\end{equation*}%
The evaluation of the polygamma function $\psi ^{\left( \alpha \right)
}\left( \frac{r}{a}\right) $ at rational values of the argument can be
explicitly done via a formula as given by K\"{o}lbig \cite{kolbig2}, (see
also \cite{kolbig1}), or Choi and Cvijovic \cite{Choi1} in terms of the
polylogarithmic or other special functions. Some specific values are given
as, many others are listed in the book \cite{srichoibook}: 
\begin{equation*}
\psi ^{\left( n\right) }\left( \frac{1}{2}\right) =\left( -1\right)
^{n}n!\left( 2^{n+1}-1\right) \zeta \left( n+1\right)
\end{equation*}%
\begin{equation*}
H_{\frac{1}{4}}^{\left( 1\right) }=4-\frac{\pi }{2}-3\ln \left( 2\right) ,~\
H_{\frac{3}{4}}^{\left( 1\right) }=\frac{4}{3}+\frac{\pi }{2}-3\ln \left(
2\right) ,
\end{equation*}%
\begin{equation*}
H_{\frac{1}{3}}^{\left( 1\right) }=\frac{3}{2}-\frac{\pi }{2\sqrt{3}}-\frac{%
3\ln 3}{2},~\text{and }\ H_{\frac{5}{6}}^{\left( 1\right) }=\frac{6}{5}+%
\frac{\sqrt{3}\pi }{2}-\frac{3\ln \left( 3\right) }{2}-2\ln \left( 2\right) .
\end{equation*}
\end{remark}

We now state the following theorem.

\begin{theorem}
\label{t2} Let $a$ be a positive real number, $\left\vert t\right\vert \leq 1
$ $,$ $j\geq 0,$ and $k\in 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
\cup \left\{ 0\right\} .$ Then%
\begin{equation}
S_{k+1}\left( a,~j,~t\right) =\sum_{n=1}^{\infty }\frac{t^{n}\ }{%
n^{k+1}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) }  \label{6}
\end{equation}%
\begin{equation*}
=\left\{ 
\begin{array}{c}
\frac{\left( j+1\right) t\left( -1\right) ^{k}}{k!}\int\limits_{0}^{1}\int%
\limits_{0}^{1}\frac{\left( 1-x\right) ^{j}~x^{a}\left( \ln \left( y\right)
\right) ^{k}}{1-tx^{a}y}dxdy,~\text{for }k\geq 1 \\ 
at\int\limits_{0}^{1}\frac{\left( 1-x\right) ^{j+1}~x^{a-1}~}{1-tx^{a}}dx,%
\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ for }k=0%
\end{array}%
\right. .
\end{equation*}%
\begin{equation}
=T_{0\ \ a+k+1}F_{a+k}\left[ \left. 
\begin{array}{c}
\overset{\left( k+2\right) -terms}{\overbrace{1,1,.....,1}},\overset{\left(
a-1\right) -terms}{\overbrace{\frac{a+1}{a},..\dots .,\frac{2a-1}{a}}} \\%
[5pt]
\underset{k-terms}{\underbrace{2,2,.....,2}},\underset{a-terms}{\underbrace{%
\frac{a+j+2}{a},\dots ...,\frac{a+j+a+1}{a}}}%
\end{array}%
\right\vert t\right]   \label{6.01}
\end{equation}%
where $_{\ \ p}F_{q}\left[ \cdot \right] $ is the generalized hypergeometric
function, 
\begin{equation*}
T_{0}=t\left( j+1\right) \ B\left( j+1,a+1\right) ,
\end{equation*}%
and $B\left( \cdot ,\cdot \right) $ is the beta function.
\end{theorem}

\begin{proof}
Consider 
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{t^{n}\ }{n^{k+1}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) } &amp;=&amp;\sum_{n=1}^{\infty }\frac{\left( j+1\right) t^{n}\ ~\Gamma
\left( j+1\right) \Gamma \left( an+1\right) }{n^{k+1}~\Gamma \left(
an+j+2\right) } \\
&amp;=&amp;\left( j+1\right) \sum_{n=1}^{\infty }\frac{t^{n}\ ~}{n^{k+1}~}B\left(
an+1,j+1\right)
\end{eqnarray*}%
now replacing the beta function with its integral representation, we have$\ $
\begin{equation*}
\left( j+1\right) \sum_{n=1}^{\infty }\frac{t^{n}\ ~}{n^{k+1}~}B\left(
an+1,j+1\right) =\left( j+1\right) \sum_{n=1}^{\infty }\frac{t^{n}\ ~}{%
n^{k+1}~}\int\limits_{0}^{1}x^{an}\left( 1-x\right) ^{j}dx.
\end{equation*}%
By a justified changing the order of integration and summation we have,%
\begin{equation*}
\sum_{n=1}^{\infty }\frac{t^{n}\ }{n^{k+1}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) }=\left( j+1\right) \int\limits_{0}^{1}\left( 1-x\right)
^{j}\sum_{n=1}^{\infty }\frac{\left( tx^{a}\right) ^{n}\ ~}{n^{k+1}~}dx
\end{equation*}%
\begin{equation*}
=\frac{\left( j+1\right) t\left( -1\right) ^{k}}{k!}\int\limits_{0}^{1}\int%
\limits_{0}^{1}\frac{\left( 1-x\right) ^{j}~x^{a}\left( \ln \left( y\right)
\right) ^{k}}{1-tx^{a}y}dxdy,~\text{for }k\geq 1
\end{equation*}%
upon utilizing Definition \ref{lemma1}. The case of $k=0$ follows in a
similar way so that%
\begin{eqnarray*}
S_{1}\left( a,~j,~t\right) &amp;=&amp;\sum_{n=1}^{\infty }\frac{t^{n}\ }{n\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) } \\
&amp;=&amp;at\int\limits_{0}^{1}\frac{\left( 1-x\right) ^{j+1}~x^{a-1}}{1-tx^{a}}dx;
\end{eqnarray*}%
hence the integrals in (\ref{6}) are attained. By the consideration of the
ratio of successive terms $\frac{U_{n+1}}{U_{n}}$ where 
\begin{equation*}
U_{n}=\frac{t^{n}\ }{n^{k+1}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) }
\end{equation*}%
we obtain the result (\ref{6.01}).
\end{proof}

\bigskip The following interesting corollaries follow from Theorem \ref{t2}.

\begin{corollary}
\label{c1} Let $t=1$ and $a&gt;0.~$ Also let $j\geq 0$ and $k\geq 1$ be
integers. Then%
\begin{align}
S_{k+1}\left( a,~j,~1\right) &amp; =\sum_{n=1}^{\infty }\frac{1\ }{n^{k+1}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) }  \notag \\
&amp; =\frac{\left( j+1\right) \left( -1\right) ^{k}}{k!}\int\limits_{0}^{1}\int%
\limits_{0}^{1}\frac{\left( 1-x\right) ^{j}~x^{a}\left( \ln \left( y\right)
\right) ^{k}}{1-x^{a}y}dxdy,~\text{for }k\geq 1  \label{5} \\
&amp;  \notag \\
&amp; =\sum_{s=0}^{k-1}A_{s}\left( j+1\right) !\zeta \left( k+1-s\right)
+\sum_{r=1}^{j+1}\left( -1\right) ^{r+k+1}\left( \frac{a}{r}\right)
^{k}\left( \QATOP{j+1}{r}\right) H_{\frac{r}{a}}^{\left( 1\right) }
\label{5.1}
\end{align}%
where 
\begin{equation}
A_{s}=\lim_{n\rightarrow 0}\left[ \frac{1}{s!}\frac{d^{s}}{dn^{s}}\left\{ 
\frac{n^{k}}{n^{k}~\prod_{r=1}^{j+1}\left( an+r\right) }\right\} \right]
,~s=0,1,2,...k-1.  \label{5.11}
\end{equation}
\end{corollary}

\begin{proof}
By expansion,%
\begin{align*}
\sum_{n=1}^{\infty }\frac{1\ }{n^{k+1}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) }&amp; =\sum_{n=1}^{\infty }\frac{\left( j+1\right) !\ }{n^{k+1}\left(
an+1\right) _{j+2}}=\sum_{n=1}^{\infty }\frac{\left( j+1\right) !\ }{%
n^{k+1}\prod_{r=1}^{j+1}\left( an+r\right) } \\
&amp; =\sum_{n=1}^{\infty }\frac{\left( j+1\right) !}{n}\left[ \sum_{s=0}^{k-1}%
\frac{A_{s}}{n^{k-s}}+\sum_{r=1}^{j+1}\frac{B_{r}}{an+r}\right] ,
\end{align*}%
where%
\begin{equation*}
B_{r}=\lim_{n\rightarrow \left( -\frac{r}{a}\right) }\left\{ \frac{an+r}{%
\prod_{r=1}^{j+1}\left( an+r\right) }\right\} =\frac{\left( -1\right)
^{r+k+1}r}{\left( j+1\right) !}\left( \frac{a}{r}\right) ^{k}\left( \QATOP{%
j+1}{r}\right) ,
\end{equation*}%
$A_{s}$ is defined by (\ref{5.11}). Hence, after interchanging the sums, we
have%
\begin{align*}
&amp; \sum_{n=1}^{\infty }\frac{1\ }{n^{k+1}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) } \\
&amp; =\left( j+1\right) !\left[ \sum_{s=0}^{k-1}A_{s}\sum_{n=1}^{\infty }\frac{1%
}{n^{k+1-s}}+\sum_{r=1}^{j+1}B_{r}\sum_{n=1}^{\infty }\frac{1}{n\left(
an+r\right) }\right] \\
&amp; =\left( j+1\right) !\sum_{s=0}^{k-1}A_{s}~\zeta \left( k+1-s\right)
+\sum_{r=1}^{j+1}\left( -1\right) ^{r+k+1}\left( \frac{a}{r}\right)
^{k}\left( \QATOP{j+1}{r}\right) H_{\frac{r}{a}}^{\left( 1\right) }
\end{align*}%
upon utilizing Lemma \ref{lemma2}, which is the result (\ref{5.1}). The
degenerate case, for $j=-1,$ gives the known result%
\begin{equation*}
\sum_{n=1}^{\infty }\frac{1}{n^{k+1}}=\zeta \left( k+1\right) .
\end{equation*}%
The integral (\ref{5}) follows from the integral in (\ref{6}).
\end{proof}

A similar result is evident for the case $t=-1.$

\begin{corollary}
\label{c1A} Let $t=-1$ and $a&gt;0.~$ Also let $j\geq 0$ and $k\geq 1$ be
integers. Then%
\begin{align}
S_{k+1}\left( a,~j,~-1\right) &amp; =\sum_{n=1}^{\infty }\frac{\left( -1\right)
^{n}\ }{n^{k+1}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) }  \notag \\
&amp; =\frac{\left( j+1\right) \left( -1\right) ^{k+1}}{k!}\int\limits_{0}^{1}%
\int\limits_{0}^{1}\frac{\left( 1-x\right) ^{j}~x^{a}\left( \ln \left(
y\right) \right) ^{k}}{1+x^{a}y}dxdy,~\text{for }k\geq 1  \notag \\
&amp;  \notag \\
&amp; =\sum_{s=0}^{k-1}A_{s}\left( j+1\right) !\left( 2^{s-k}-1\right) \zeta
\left( k+1-s\right)  \label{5.2} \\
&amp; +\sum_{r=1}^{j+1}\left( -1\right) ^{r+k+1}\left( \frac{a}{r}\right)
^{k}\left( \QATOP{j+1}{r}\right) \left( H_{\frac{r}{a}}^{\left( 1\right)
}-H_{\frac{r}{2a}}^{\left( 1\right) }\right)  \notag
\end{align}
\end{corollary}

\begin{proof}
The proof, uses (\ref{4.2}) and follows the same details as that of
Corollary \ref{c1}, and will not be given here.
\end{proof}

The addition and subtraction of (\ref{5.1}) and (\ref{5.2}) gives us the
following representations.

\begin{remark}
Let $a&gt;0$ and let $j\geq 0$ and $k\geq 1$ be integers. Then%
\begin{eqnarray*}
&amp;&amp;2^{1-k}\sum_{n=1}^{\infty }\frac{1\ }{n^{k+1}\left( 
\begin{array}{c}
2an+j+1 \\ 
j+1%
\end{array}%
\right) } \\
&amp;=&amp;\sum_{s=0}^{k-1}A_{s}\left( j+1\right) !~2^{s-k}~\zeta \left(
k+1-s\right) +\sum_{r=1}^{j+1}\left( -1\right) ^{r+k+1}\left( \frac{a}{r}%
\right) ^{k}\left( \QATOP{j+1}{r}\right) \left( 2H_{\frac{r}{a}}^{\left(
1\right) }-H_{\frac{r}{2a}}^{\left( 1\right) }\right)
\end{eqnarray*}%
and 
\begin{eqnarray*}
&amp;&amp;\sum_{n=1}^{\infty }\frac{1\ }{\left( 2n-1\right) ^{k+1}\left( 
\begin{array}{c}
2an-a+j+1 \\ 
j+1%
\end{array}%
\right) } \\
&amp;=&amp;\sum_{s=0}^{k-1}A_{s}\left( j+1\right) !~\left( 2-2^{s-k}\right) ~\zeta
\left( k+1-s\right) +\sum_{r=1}^{j+1}\left( -1\right) ^{r+k+1}\left( \frac{a%
}{r}\right) ^{k}\left( \QATOP{j+1}{r}\right) H_{\frac{r}{2a}}^{\left(
1\right) }.
\end{eqnarray*}
\end{remark}

We give the following example to illustrate some of the above identities.

\begin{example}
Let $k=4,$ from (\ref{5.11})%
\begin{eqnarray*}
A_{0} &amp;=&amp;\frac{1}{\left( j+1\right) !},~~A_{1}=-\frac{aH_{j+1}^{\left(
1\right) }}{\left( j+1\right) !},~~A_{2}=\frac{a^{2}\left( \left(
H_{j+1}^{\left( 1\right) }\right) ^{2}+H_{j+1}^{\left( 2\right) }\right) }{%
2\left( j+1\right) !} \\
A_{3} &amp;=&amp;-\frac{a^{3}\left( \left( H_{j+1}^{\left( 1\right) }\right)
^{3}+3H_{j+1}^{\left( 1\right) }H_{j+1}^{\left( 2\right) }+2H_{j+1}^{\left(
3\right) }\right) }{6\left( j+1\right) !}
\end{eqnarray*}%
therefore%
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{1\ }{n^{5}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) } &amp;=&amp;\left( j+1\right) !\left[ A_{0}\zeta \left( 5\right)
+A_{1}\zeta \left( 4\right) +A_{2}\zeta \left( 3\right) +A_{3}\zeta \left(
2\right) \right] \\
&amp;&amp;+\sum_{r=1}^{j+1}\left( -1\right) ^{r+1}\left( \frac{a}{r}\right)
^{4}\left( \QATOP{j+1}{r}\right) H_{\frac{r}{a}}^{\left( 1\right) }
\end{eqnarray*}%
and 
\begin{eqnarray*}
\sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n}\ }{n^{5}\left( 
\begin{array}{c}
an+j+1 \\ 
j+1%
\end{array}%
\right) } &amp;=&amp;\left( j+1\right) !\left[ -\frac{15A_{0}}{16}\zeta \left(
5\right) -\frac{7A_{1}}{8}\zeta \left( 4\right) -\frac{3A_{2}}{4}\zeta
\left( 3\right) -\frac{A_{3}}{2}\zeta \left( 2\right) \right] \\
&amp;&amp;+\sum_{r=1}^{j+1}\left( -1\right) ^{r}\left( \frac{a}{r}\right) ^{4}\left( 
\QATOP{j+1}{r}\right) \left( H_{\frac{r}{a}}^{\left( 1\right) }-H_{\frac{r}{%
2a}}^{\left( 1\right) }\right) .
\end{eqnarray*}
\end{example}

\begin{remark}
The very special case of $a=1$ and $j=0$ allows one to evaluate 
\begin{eqnarray*}
A_{s} &amp;=&amp;\lim_{n\rightarrow 0}\left[ \frac{1}{s!}\frac{d^{s}}{dn^{s}}\left\{ 
\frac{n^{k}}{n^{k}~\left( n+1\right) }\right\} \right] ,~s=0,1,2,...k-1 \\
&amp;=&amp;\left( -1\right) ^{s},
\end{eqnarray*}%
and from (\ref{5.1}) and (\ref{5.2}) we can easily obtain (\ref{002}) and (%
\ref{001}).
\end{remark}

\bigskip

\bigskip A recurrence relation for a degenerate case, $j=0,$ of Theorem \ref%
{t2} is embodied in the following corollary.

\begin{corollary}
Let the conditions of Theorem \ref{t2} hold with $j=0$ and put%
\begin{eqnarray*}
S_{k+1}^{0} &amp;:&amp;=S_{k+1}^{0}\left( a,t\right) =\sum_{n=1}^{\infty }\frac{%
t^{n}\ }{n^{k+1}\left( an+1\right) } \\
&amp;=&amp;\frac{t}{a+1}~_{\ \ k+3}F_{k+2}\left[ \left. 
\begin{array}{c}
\overset{\left( k+2\right) -terms}{\overbrace{1,1,.....,1}},\frac{a+1}{a} \\%
[5pt] 
\underset{\left( k+1\right) -terms}{\underbrace{2,2,.....,2}},\frac{2a+1}{a}%
\end{array}%
\right\vert t\right] ,
\end{eqnarray*}%
then 
\begin{equation*}
S_{k+1}^{0}+aS_{k}^{0}=\mathrm{Li}_{k+1}\left( t\right) ,\text{ for }k\geq 1
\end{equation*}%
with solution 
\begin{eqnarray*}
S_{k+1}^{0} &amp;=&amp;\left( -a\right) ^{k}S_{1}^{0}+\sum_{r=1}^{k}\left( -a\right)
^{r-1}t~\Phi \left( t,k+2-r,1\right) \\
&amp;=&amp;\left( -a\right) ^{k}S_{1}^{0}+\sum_{r=1}^{k}\left( -a\right) ^{r-1}%
\mathrm{Li}_{k+2-r}\left( t\right)
\end{eqnarray*}%
where 
\begin{equation*}
S_{1}^{0}\left( a,t\right) =\sum_{n=1}^{\infty }\frac{t^{n}\ }{n\left(
an+1\right) }=\frac{t}{a+1}~_{\ \ 3}F_{2}\left[ \left. 
\begin{array}{c}
1,1,\frac{a+1}{a} \\[5pt] 
2,\frac{2a+1}{a}%
\end{array}%
\right\vert t\right]
\end{equation*}%
and $\Phi ,~\mathrm{Li}$ \ are the Lerch transcendent and the polylogarithm
respectively.
\end{corollary}

\begin{proof}
We notice that 
\begin{equation*}
S_{k+1}^{0}+aS_{k}^{0}=\sum_{n=1}^{\infty }\frac{t^{n}\ }{n^{k+1}}=\mathrm{Li%
}_{k+1}\left( t\right),
\end{equation*}
and hence the solution follows by iteration.
\end{proof}

\bigskip 
Related results may be seen in Coffey \cite[Lemmas 1 and 2]{11.01}.

Some examples are as follows:

$\ \bullet $ For $t=1$, we know that $\mathrm{Li}_{k+2-r}\left( 1\right)
=\zeta \left( k+2-r\right) .$ Hence

\begin{equation*}
S_{k+1}^{0}\left( a,1\right) =\left( -a\right) ^{k}S_{1}^{0}\left(
a,1\right) +\sum_{r=1}^{k}\left( -a\right) ^{r-1}\zeta \left( k+2-r\right) ,%
\text{ for }k\geq 1.
\end{equation*}%
When $a=1,$ we obtain Coffey's \cite{10} result, by noting that, from (\ref%
{4.1}), $S_{1}^{0}\left( 1,1\right) =1$ 
\begin{equation*}
S_{k+1}^{0}\left( 1,1\right) =\left( -1\right) ^{k}+\sum_{r=1}^{k}\left(
-1\right) ^{r-1}~\zeta \left( k+2-r\right) .
\end{equation*}%
When $a=\frac{1}{2},$%
\begin{equation*}
S_{k+1}^{0}\left( \frac{1}{2},1\right) =\frac{3\left( -1\right) ^{k}}{2^{k+1}%
}+\sum_{r=1}^{k}\frac{\left( -1\right) ^{r-1}}{2^{r-1}}\zeta \left(
k+2-r\right) .
\end{equation*}%
Similarly 
\begin{eqnarray*}
S_{k+1}^{0}\left( 8,1\right) &amp;=&amp;\left( -1\right) ^{k}2^{3\left( k+1\right)
}-\left( -1\right) ^{k}2^{3k-1}\pi \sqrt{3+2\sqrt{2}}-\left( -1\right)
^{k}2^{3k+2}\ln \left( 2\right) \\
&amp;&amp;-\left( -1\right) ^{k}2^{3k-3/2}\ln \left( 3+2\sqrt{2}\right)
+\sum_{r=1}^{k}\left( -2\right) ^{3\left( r-1\right) }\zeta \left(
k+2-r\right) .
\end{eqnarray*}

$\bullet $ For $t=-1$, $\mathrm{Li}_{k+2-r}\left( -1\right) =\eta \left(
k+2-r\right) =$ $\left( 2^{r-k-1}-1\right) \zeta \left( k+2-r\right) ,$
where $\eta \left( \cdot \right) $ is the Dirichlet Eta function. Hence

\begin{equation*}
S_{k+1}^{0}\left( a,-1\right) =\left( -a\right) ^{k}S_{1}^{0}\left(
a,-1\right) +\sum_{r=1}^{k}\left( -a\right) ^{r-1}\left( 2^{r-k-1}-1\right)
\zeta \left( k+2-r\right) ,\text{ for }k\geq 1.
\end{equation*}%
When $a=1,$ we obtain Coffey's \cite{10} result, by noting that 
\begin{equation*}
S_{k+1}^{0}\left( 1,-1\right) =\left( -1\right) ^{k}\left( 1-2\ln 2\right)
+\sum_{r=1}^{k}\left( -1\right) ^{r-1}~\left( 2^{r-k-1}-1\right) \zeta
\left( k+2-r\right) .
\end{equation*}%
When $a=4,$%
\begin{eqnarray*}
S_{k+1}^{0}\left( 4,-1\right) &amp;=&amp;\left( -2\right) ^{k}\left( 4-\sqrt{2}\ln
\left( 1+\sqrt{2}\right) -\frac{\pi }{2}-\ln 2\right) \\
&amp;&amp;+\sum_{r=1}^{k}\left( -1\right) ^{r-1}~2^{2\left( r-1\right) }\left(
2^{r-k-1}-1\right) \zeta \left( k+2-r\right) ,
\end{eqnarray*}%
and similarly 
\begin{eqnarray*}
S_{k+1}^{0}\left( \frac{1}{8},-1\right) &amp;=&amp;-\left( -1\right) ^{k}\frac{533}{%
840}2^{-3k}+\sum_{r=1}^{k}\frac{\left( -1\right) ^{r-1}}{8^{r-1}}~\left(
2^{r-k-1}-1\right) \zeta \left( k+2-r\right) \\
&amp;=&amp;-\frac{8}{9}~_{\ \ k+3}F_{k+2}\left[ \left. 
\begin{array}{c}
\overset{\left( k+2\right) -terms}{\overbrace{1,1,.....,1}},9 \\[5pt] 
\underset{\left( k+1\right) -terms}{\underbrace{2,2,.....,2}},10%
\end{array}%
\right\vert -1\right] .
\end{eqnarray*}

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\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 05A10; Secondary 11B65, 05A19, 33C20.

\noindent \emph{Keywords: } 
double integral, combinatorial identity, harmonic number,
polygamma function, recurrence.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  December 27 2009;
revised version received  February 16 2010.
Published in {\it Journal of Integer Sequences}, February 23 2010.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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