\documentclass[a4paper,reqno,12pt]{amsart} \usepackage{epsfig} %\renewcommand{\baselinestretch}{1.5} \setlength{\textwidth}{6in} \setlength{\textheight}{9in} \setlength{\topmargin}{0cm} \setlength{\oddsidemargin}{5mm} \setlength{\evensidemargin}{5mm} \newtheorem{theorem}{Theorem}%[section] \newtheorem{proposition}[theorem]{Proposition}%[section] \newtheorem{lemma}[theorem]{Lemma}%[section] \newtheorem{example}[theorem]{Example}%[section] \newtheorem{cor}[theorem]{Corollary} \sloppy \allowdisplaybreaks \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 %First page headline in AmS-LaTeX for S\'eminaire Lotharingien de Combinatoire %--first part \thispagestyle{myheadings} \font\rms=cmr8 \font\its=cmti8 \font\bfs=cmbx8 \markright{\its S\'eminaire Lotharingien de Combinatoire \bfs 55 \rms (2005), Article~B55a\hfill} \def\thepage{} % % \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} %First page headline in AmS-LaTeX for S\'eminaire Lotharingien de Combinatoire %--restoring the headers and pagenumbering \pagenumbering{arabic} \addtocounter{page}{1} \markboth{\SMALL ALOIS PANHOLZER AND HELMUT PRODINGER}{\SMALL COMPUTER-FREE EVALUATION OF AN INFINITE DOUBLE SUM} % % 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}