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\begin{document}

\title{Computer-free evaluation of an infinite double sum via Euler sums}

\author[A.~Panholzer]{Alois Panholzer}
\address{Institut f\"ur Diskrete Mathematik und Geometrie\\ 
Technische Universit{\"a}t Wien\\ Wiedner Hauptstra{\ss}e 8--10/104\\ 
A-1040 Wien\\ Austria.}
\email{Alois.Panholzer@tuwien.ac.at}

\author[H.~Prodinger]{Helmut Prodinger}
\address{Helmut Prodinger\\
Mathematics Department\\ Stellenbosch University \\
7602 Stellenbosch\\ South Africa.}
\email{hproding@sun.ac.za}

\date{}

\begin{abstract} A short and computer-free proof (using Euler sums and multiple zeta functions) is provided for a double sum that was recently computed by Pemantle and Schneider using the computer software \textsf{Sigma}.
\end{abstract}

\maketitle

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\font\rms=cmr8
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\markright{\its S\'eminaire Lotharingien de
Combinatoire \bfs 55 \rms (2005), Article~B55a\hfill}
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\section{Introduction}

The evaluation of the double sum
\begin{equation}
   \label{eqn1}
   S := \sum_{j, k =1}^{\infty} \frac{H_{j} (H_{k+1}-1)}{j k (k+1) (j+k)}=
-\zeta(2)-2\zeta(3)+4\zeta(2)\zeta(3)+2\zeta(5),
\end{equation}
(where $H_{n} := \sum_{k=1}^{n}\frac{1}{k}$ denote  harmonic numbers)
appears in \cite{PemSch04} and was obtained using Carsten Schneider's software \textsf{Sigma}.
Here, we will give  a short proof that is completely computer-free. It uses Euler sums
and multiple zeta functions.

\medskip

The second order harmonic numbers which appear in the sequel are denoted by 
$H_{n}^{(2)} := \sum_{k=1}^{n}\frac{1}{k^{2}}$.

We split the sum $S$ and  apply partial fraction decomposition:
\begin{equation*}
   S = \sum_{k \ge 1} \frac{H_{k+1}-1}{k (k+1)} \sum_{j \ge 1} \frac{H_{j}}{j (j+k)}=
   \sum_{k \ge 1} \frac{H_{k+1}-1}{k^2 (k+1)} 
 \sum_{j \ge 1} 
   H_{j} \Big(\frac{1}{j} - \frac{1}{j+k}\Big).
\end{equation*}
The inner sum is simplified as follows:
\begin{equation*}
 \sum_{j \ge 1}    H_{j} \Big(\frac{1}{j} - \frac{1}{j+k}\Big) =  \sum_{j \ge 1} 
   \Big(\frac{1}{j} - \frac{1}{j+k}\Big) \sum_{l=1}^{j} \frac{1}{l}
   = \sum_{l \ge 1} \frac{1}{l} \sum_{j \ge l} 
   \Big(\frac{1}{j} - \frac{1}{j+k}\Big).
\end{equation*}
The second sum here telescopes, and  only $k$ summands remain:
\begin{align*}
    \sum_{l \ge 1}& \frac{1}{l} \sum_{j \ge l} 
   \Big(\frac{1}{j} - \frac{1}{j+k}\Big) 
   = \sum_{l \ge 1} \frac{1}{l} \sum_{j=0}^{k-1} \frac{1}{l+j} 
     =  \sum_{l \ge 1} \frac{1}{l^{2}} + \sum_{j=1}^{k-1} \sum_{l \ge 1}
   \frac{1}{l(l+j)}\\& = \zeta(2)+\sum_{j=1}^{k-1}\frac{1}{j}
   \sum_{l \ge 1} \Big(\frac{1}{l} - \frac{1}{l+j}\Big)\qquad\text{[partial fraction 
decomposition]}\\
   &   = \zeta(2) + \sum_{j=1}^{k-1} 
   \frac{H_{j}}{j}\qquad\qquad\qquad\qquad\text{[again by telescoping]}\\
   &=\zeta(2) + \frac{H_{k}^{2} + H_{k}^{(2)}}{2} - \frac{H_{k}}{k} .
\end{align*}

Thus the task reduces to evaluate the following single sum:
\begin{equation}
   S = \sum_{k=1}^{\infty} \frac{H_{k+1}-1}{k^{2} (k+1)} 
   \Big(\zeta(2) + \frac{H_{k}^{2}+H_{k}^{(2)}}{2} - \frac{H_{k}}{k}\Big).
\end{equation}

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\markboth{\SMALL ALOIS PANHOLZER AND HELMUT PRODINGER}{\SMALL COMPUTER-FREE EVALUATION OF AN INFINITE DOUBLE SUM}
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After partial fraction expansion (and shifting the index if necessary),
sum $S$ can be written as follows:
\begin{equation*}
   S = - 2 \zeta(2) + \frac{1}{2} \sum_{k \ge 1} \frac{H_{k} H_{k}^{(2)}}{k^{2}} 
   + \frac{1}{2} \sum_{k \ge 1} \frac{H_{k}^{3}}{k^{2}}
   - \sum_{k \ge 1} \frac{H_{k}^{2}}{k^{3}} + (\zeta(2)-1) \sum_{k \ge 1} \frac{H_{k}}{k^{2}}
   - 2 \sum_{k \ge 1} \frac{1}{k^{2}}.
\end{equation*}

 For the final computation of $S$ we require the following evaluations of Euler sums via values of the Riemann
zeta function:
\begin{subequations}
\begin{align}
   \sum_{k \ge 1} \frac{H_{k} H_{k}^{(2)}}{k^{2}} & = \zeta(5) + \zeta(2) \zeta(3), 
   \label{eqna1} \\
   \sum_{k \ge 1} \frac{H_{k}^{3}}{k^{2}} & = 10 \zeta(5) + \zeta(2) \zeta(3), \label{eqna2} \\
   \sum_{k \ge 1} \frac{H_{k}^{2}}{k^{3}} & = \frac{7}{2} \zeta(5) - \zeta(2) \zeta(3), 
   \label{eqna3} \\
   \sum_{k \ge 1} \frac{H_{k}}{k^{2}} & = 2 \zeta(3), \label{eqna4}
\end{align}
\end{subequations}
from which equation~\eqref{eqn1} then follows.

Equations~\eqref{eqna3} and \eqref{eqna4} can be found explicitly in \cite{FlaSal98}. 
In \cite{FlaSal98} one also finds the identities
\begin{subequations}
\begin{align}
   \sum_{k \ge 1} \frac{H_{k}^{3}}{(k+1)^{2}} & = \frac{15}{2} \zeta(5) + \zeta(2) \zeta(3),
   \label{eqnb1} \\
   \sum_{k \ge 1} \frac{H_{k}}{k^{4}} & = 3 \zeta(5) - \zeta(2) \zeta(3),
   \label{eqnb2}
\end{align}
\end{subequations}
and due to
\begin{equation*}
   \sum_{k \ge 1} \frac{H_{k}^{3}}{k^{2}} = \sum_{k \ge 1} \frac{H_{k}^{3}}{(k+1)^{2}}
   + 3 \sum_{k \ge 1} \frac{H_{k}^{2}}{k^{3}} - 3 \sum_{k \ge 1} \frac{H_{k}}{k^{4}}
   + \sum_{k \ge 1} \frac{1}{k^{5}},
\end{equation*}
we obtain equation \eqref{eqna2} as well,  using \eqref{eqnb1}, \eqref{eqnb2} and \eqref{eqna3}.

To show \eqref{eqna1} we will apply Theorem~$2$ of \cite{BorGir96}, which gives
\begin{equation*}
   \sum_{k \ge 1} \frac{H_{k-1} H_{k-1}^{(2)}}{k^{2}} = \zeta(2,1,2) + \zeta(2,2,1) 
   + \zeta(2,3),
\end{equation*}
where the \emph{multiple zeta functions} are defined  by
\begin{equation*}
   \zeta(a_{1}, a_{2}, \dots, a_{m}) = \sum_{k_{1} > k_{2} > \cdots > k_{m} \ge 1}
   \frac{1}{k_{1}^{a_{1}} k_{2}^{a_{2}} \cdots k_{m}^{a_{m}}}.
\end{equation*}

Since
\begin{align*}
   \sum_{k \ge 1} \frac{H_{k} H_{k}^{(2)}}{k^{2}} & = \sum_{k \ge 1} 
   \frac{H_{k-1} H_{k-1}^{(2)}}{k^{2}} + \sum_{k \ge 1} \frac{H_{k-1}^{(2)}}{k^{3}}
   + \sum_{k \ge 1} \frac{H_{k-1}}{k^{4}} + \sum_{k \ge 1} \frac{1}{k^{5}} \\
   & = \sum_{k \ge 1} \frac{H_{k-1} H_{k-1}^{(2)}}{k^{2}} + \sum_{k > l \ge 1} 
   \frac{1}{k^{3} l^{2}} + \sum_{k > l \ge 1} \frac{1}{k^{4}l} + \sum_{k \ge 1} 
   \frac{1}{k^{5}},
\end{align*}
we obtain
\begin{equation*}
   \sum_{k \ge 1} \frac{H_{k} H_{k}^{(2)}}{k^{2}} =
   \zeta(2,1,2) + \zeta(2,2,1) + \zeta(2,3) + \zeta(3,2) + \zeta(4,1) + \zeta(5).
\end{equation*}
Using the following evaluations of the multiple zeta function given in \cite{BorBorGir95}
resp. \cite{BorGir96}:
\begin{align*}
   \zeta(2,1,2) & = \zeta(2,3) = \frac{9}{2} \zeta(5) - 2 \zeta(2) \zeta(3), \\
   \zeta(2,2,1) & = \zeta(3,2) = - \frac{11}{2} + 3 \zeta(2) \zeta(3), \\
   \zeta(4,1) & = 2 \zeta(5) - \zeta(2) \zeta(3),
\end{align*}
equation~\eqref{eqna1} follows.

\bibliographystyle{plain}
\begin{thebibliography}{1}

\bibitem{BorBorGir95}
D.~Borwein, J.~Borwein and R.~Girgensohn, Explicit evaluation of Euler sums,
\emph{Proceedings of the Edinburgh Mathematical Society} 38, 277--294, 1995.

\bibitem{BorGir96}
J.~Borwein and R.~Girgensohn, Evaluation of triple Euler sums,
\emph{Electronic Journal of Combinatorics} 3, research paper \#23, 1996.

\bibitem{FlaSal98}
P.~Flajolet and B.~Salvy, Euler sums and contour integral representations,
\emph{Experimental Mathematics} 7, 15--35, 1998.

\bibitem{PemSch04}
R.~Pemantle and C.~Schneider, When is $0.999\dots$ equal to $1$?,
preprint, 2004.

\end{thebibliography}

\end{document}

