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\begin{center}
\vskip 1cm{\LARGE\bf On the Summation of Certain Iterated Series}
\vskip 1cm {\large
Ovidiu Furdui\\
Campia Turzii \\
405100 Cluj \\
Romania\\ 
\href{mailto:ofurdui@yahoo.com}{\tt ofurdui@yahoo.com}  \\
\ \\
Tiberiu Trif\\
Department of Mathematics\\
Babe\c{s}-Bolyai University \\
Cluj-Napoca \\
Romania\\
\href{mailto:ttrif@math.ubbcluj.ro}{\tt ttrif@math.ubbcluj.ro} \\
}
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\vskip .2in

\begin{abstract}
The paper gives a unified treatment of the summation of certain iterated series of the form
 $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}a_{n+m},$
where $(a_{n})_{n\in \mathbb{N}}$ is a sequence of real numbers. We prove that, under certain conditions, the double iterated series equals the difference of two single series.
\end{abstract}


                                                                                


\section{Introduction}
The goal of this paper is to provide a unified treatment of the
summation of a special class of double series that have appeared recently.
Below, we collect some problems that led to the motivation of this article.
\medskip

\textbf{Problem 1} (\cite{FURDUI1})
Find the sum
\begin{equation*}\label{Furdui1}
    \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(-1)^{n+m}}
    {\big(\lfloor\sqrt{n+m}\rfloor\big)^3}\,,
\end{equation*}
where $\lfloor{a}\rfloor$ denotes the greatest integer less than or
equal to $a$.
\medskip

\textbf{Problem 2} (\cite{F})
Find
\begin{equation}\label{alternate-harmonic-double-series}
    \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(-1)^{n+m}}
    {n+m}\,.
\end{equation}
\medskip

\textbf{Problem 3} (\cite{EFTHIMIOU})
Compute the sum
\begin{equation}\label{Efthimiou}
    \sum_{n=0}^\infty \sum_{m=1}^\infty (-1)^{n+m}
    \frac{\ln(n+m)}{n+m}\,.
\end{equation}
\medskip

\textbf{Problem 4} (\cite{FURDUI2})
Find
\begin{equation}\label{Furdui2}
    \sum_{n=1}^\infty \sum_{m=1}^\infty (-1)^{n+m}
    \frac{H_{n+m}}{n+m}\,,
\end{equation}
where $H_n$ denotes the $n$th harmonic number.
\medskip

Although the problems can be solved by various techniques, in this paper we give a general method for summing the iterated series
\begin{equation}\label{iterated-series}
    \sum_{n=1}^\infty\sum_{m=1}^\infty a_{n+m},
\end{equation}
where $(a_n)_{n\in \mathbb{N}}$ is a sequence of real numbers, and we show that, under certain conditions, the double iterated series equals the difference of two single series.
The organization of the paper is as follows: in Section~\ref{sect2} we give the first two main results of the paper, which are about the evaluation of the double iterated series (\ref{iterated-series}), and in Section~\ref{sect3} we give the closed form evaluation of multiple series of the form $\sum_{n_{1}=1}^{\infty}\cdots \sum_{n_{k}=1}^{\infty}a_{n_{1}+n_{2}+\cdots+n_{k}}.$
\section{Double iterated series} \label{sect2}
The first main result of this section is the following theorem.
\begin{theorem} \label{thu}
Suppose that both series
\begin{equation}\label{single-series}
    \sum_{k=1}^\infty a_k \qquad\mbox{and}\qquad \sum_{k=1}^\infty
    ka_k
\end{equation}
 converge and let $\sigma$ and $\widetilde{\sigma}$ denote
their sums, respectively. Then the iterated series {\rm
(\ref{iterated-series})} converges and its sum $s$ equals
$\widetilde{\sigma}-\sigma$.
\end{theorem}
\begin{proof} For positive integers $\nu$ and $n$ we let
\begin{equation*}
    A_\nu=\sum_{m=1}^\infty a_{\nu+m} \qquad\mbox{and}\qquad
    s_n=\sum_{\nu=1}^n A_\nu.
\end{equation*}
Likewise, for every positive integer $n$ we let
\begin{equation*}
    \sigma_n=\sum_{k=1}^n a_k \qquad\mbox{and}\qquad
    \widetilde{\sigma}_n=\sum_{k=1}^n ka_k.
\end{equation*}
Since $A_\nu=\sigma-\sigma_\nu$, it follows that
\begin{equation}\label{formula-s-n}
    s_n = n\sigma-\sum_{\nu=1}^n \sigma_\nu = n\sigma-\sum_{k=1}^n
    (n+1-k)a_k=n\sigma+\widetilde{\sigma}_n-(n+1)\sigma_n=\frac{u_{n}}{v_{n}},
\end{equation}
 where
\begin{equation*}
    u_n=\sigma-\sigma_n+\frac{\widetilde{\sigma}_n-\sigma_n}{n}
    \qquad\mbox{and}\qquad v_n=\frac{1}{n}\,.
\end{equation*}
On the other hand, it is straight-forward to show that
\begin{equation*}
    \frac{u_{n+1}-u_n}{v_{n+1}-v_n}=\widetilde{\sigma}_{n+1}-
    \sigma_{n+1}.
\end{equation*}
Since
\begin{equation*}
    \lim_{n\to\infty}\frac{u_{n+1}-u_n}{v_{n+1}-v_n}=
    \widetilde{\sigma}-\sigma,
\end{equation*}
an application of Ces\'aro-Stolz theorem, the $\left(\frac{0}{0}\right)$ case,
implies that
\begin{equation*}
    s=\lim_{n\to\infty}s_n=\lim_{n\to\infty}\frac{u_n}{v_n}=\widetilde{\sigma}-\sigma,
\end{equation*}
and the theorem is proved.
\end{proof}

{\bf Solution to Problem 1.} Let $a_k=(-1)^k/\big(\lfloor\sqrt{k}\rfloor\big)^3$, and note that both series (\ref{single-series})
\begin{equation*}
    \sum_{k=1}^\infty a_k=\sum_{k=1}^\infty (-1)^k\frac{1}
    {\big(\lfloor\sqrt{k}\rfloor\big)^3} \qquad\mbox{and}\qquad
    \sum_{k=1}^\infty ka_k=\sum_{k=1}^\infty (-1)^k\frac{k}
    {\big(\lfloor\sqrt{k}\rfloor\big)^3},
\end{equation*}
converge based on the Leibniz test. Moreover,
\begin{equation*}
\begin{aligned}
  \widetilde{\sigma}&amp;=\sum_{k=1}^\infty ka_k=\sum_{k=1}^\infty \frac{(-1)^k k}{\big(\lfloor\sqrt{k}\rfloor\big)^3}=\sum_{N=1}^\infty\left(\sum_{k=N^2}^{N^2+2N}\frac{(-1)^k k}{\big(\lfloor\sqrt{k}\rfloor\big)^3}\right)=\sum_{N=1}^\infty\frac{1}{N^3}\left(\sum_{k=N^2}^{N^2+2N}(-1)^k k\right)\\
  &amp;=\sum_{N=1}^\infty\frac{(-1)^N(N^2+N)}{N^3}=\sum_{N=1}^\infty\frac{(-1)^N}{N}+\sum_{N=1}^\infty\frac{(-1)^N}{N^2}\\
  &amp;=-\ln 2-\frac{\pi^2}{12}.\\
  \end{aligned}
  \end{equation*}
On the other hand,
\begin{equation*}
  \sigma=\sum_{k=1}^\infty a_k=\sum_{N=1}^\infty\left(\sum_{k=N^2}^{N^2+2N}\frac{(-1)^k}{\big(\lfloor\sqrt{k}\rfloor\big)^3}\right)
  =\sum_{N=1}^\infty\frac{1}{N^3}\left(\sum_{k=N^2}^{N^2+2N}(-1)^k \right)=\sum_{N=1}^\infty\frac{(-1)^N}{N^3}=-\frac{3}{4}\,\zeta(3),
\end{equation*}
where $\zeta(3)=\sum_{k=1}^{\infty}1/k^{3}$ denotes the Ap\' ery constant. It follows, based on Theorem \ref{thu}, that
\begin{equation*}
    \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(-1)^{n+m}}
    {\big(\lfloor\sqrt{n+m}\rfloor\big)^3}
    =\frac{3}{4}\,\zeta(3)-\ln 2-\frac{\pi^2}{12},
\end{equation*}
and the problem is solved.

\begin{remark} It is worth mentioning that if the series $\sum_{k=2}^\infty (k-1)a_k,$
would be absolutely convergent, the conclusion of Theorem \ref{thu}
would follow by rearranging the terms of the iterated series
(\ref{iterated-series}) as follows
\begin{equation*}
 \sum_{k=2}^\infty (k-1)a_k=a_{1+1}+\big(a_{1+2}+a_{2+1}\big)+\cdots+\big(a_{1+(k-1)}+a_{2+(k-2)}+\cdots+a_{(k-1)+1}\big)+\cdots.
\end{equation*}
Unfortunately, when $a_k=(-1)^k/\lfloor\sqrt{k}\rfloor^3,$
the absolute convergence of the series $\sum_{k=2}^\infty(k-1)a_k$ fails to hold.
Also, the convergence of series
(\ref{single-series}) is sufficient but not necessary for the
convergence of the iterated series (\ref{iterated-series}). This is
shown by the series
(\ref{alternate-harmonic-double-series}), (\ref{Efthimiou}), and
(\ref{Furdui2}). Clearly, Theorem \ref{thu} does not apply to these series.
However, the next theorem, which can be applied for summing the series
(\ref{alternate-harmonic-double-series}), (\ref{Efthimiou}) and
(\ref{Furdui2}), is the second new result of this section.
\end{remark}
\begin{theorem} \label{thd}
With the same notation as in the statement and the proof of Theorem \ref{thu}, let us suppose
that the first series in {\rm (\ref{single-series})} converges, that
$\widetilde{\sigma}_{2n}=o(n)$, and that the limit
\begin{equation}\label{limit}
    \ell=\lim_{n\to\infty}\Big(na_{2n}+\widetilde{\sigma}_{2n-1}
    \Big)
\end{equation}
exists in $\mathbb{R}$. Then the iterated series {\rm
(\ref{iterated-series})} converges and its sum $s$ is given by
$s=\ell-\sigma$.
\end{theorem}
\begin{proof} Let $w_n=na_{2n}+
\widetilde{\sigma}_{2n-1}$. With the same notation as in the statement
and the proof of Theorem \ref{thu}, we have, based on (\ref{formula-s-n}), that
\begin{equation*}
    \sigma+s_{2n}=(2n+1)(\sigma-\sigma_{2n})+\widetilde{\sigma}_{2n}
    =\frac{x_n}{y_n}\,,
\end{equation*}
where $x_n=\sigma-\sigma_{2n}+\frac{1}{2n+1}\,
\widetilde{\sigma}_{2n}$ and $y_n=\frac{1}{2n+1}$. On the other hand,
\begin{equation*}
    \frac{x_{n+1}-x_n}{y_{n+1}-y_n}=\frac{2n+1}{2}\,a_{2n+2}+
    \widetilde{\sigma}_{2n+1}=w_{n+1}-\frac{1}{2}\,a_{2n+2}\,.
\end{equation*}
Since $a_n\to 0$ as $n\to\infty$, we have, based on the Ces\'aro-Stolz
theorem, the $\left(\frac{0}{0}\right)$ case, and (\ref{limit}), that $\lim_{n\to\infty}(\sigma+s_{2n}) =\ell$, and it follows that
$\lim_{n\to\infty}s_{2n}=\ell-\sigma$. On the other hand, since
\begin{equation*}
    s_{2n+1}=s_{2n}+A_{2n+1}=s_{2n}+\sigma-\sigma_{2n+1},
\end{equation*}
we get that $\lim_{n\to\infty}s_{2n+1}=\lim_{n\to\infty}s_{2n}
=\ell-\sigma$. Thus, $\lim_{n\to\infty}s_n=\ell-\sigma$, and the theorem is proved. 
\end{proof}
\smallskip
{\bf Solution to Problem 2.}
Let $a_k=(-1)^k/k$ and we note that
\begin{equation*}
    \sigma=\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty \frac{(-1)^k} {k}=-\ln 2.
\end{equation*}
On the other hand,
$\widetilde{\sigma}_{2n}=\sum_{k=1}^{2n}(-1)^k=0$, and
\begin{equation*}
    w_n=na_{2n}+\widetilde{\sigma}_{2n-1}
    =\frac{1}{2}+\sum_{k=1}^{2n-1}k\cdot \frac{(-1)^k}{k}=
    -\frac{1}{2},\quad\mbox{for all }n\in\mathbb{N}.
\end{equation*}
Thus, $\ell=-\frac{1}{2}$. An application of Theorem \ref{thd} shows
that
\begin{equation*}
    \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(-1)^{n+m}}
    {n+m}=\ln 2-\frac{1}{2}\,,
\end{equation*}
and the problem is solved.
\medskip

Before we give the solution to Problem 3 we need to introduce a notation. Recall that, the double factorial of a positive integer $n$ is defined by the following formula
\begin{equation*}
n!!=
\begin{cases}
\ n(n-2)\cdots 5\cdot 3\cdot 1, \quad \text{if} \,\, n \,\, \text{is odd;}\\
\ n(n-2)\cdots 6\cdot 4\cdot 2, \quad \text{if} \,\, n \,\, \text{is even.}\\
\end{cases}
\end{equation*}
\medskip

{\bf Solution to Problem 3.} Let $a_k=(-1)^k\frac{\ln
k}{k}$. It can be proved, see \cite{KLAMKIN1}, that
\begin{equation*}
    \sigma=\sum_{k=1}^\infty(-1)^k\frac{\ln k}{k}=\gamma\ln 2-
    \frac{1}{2}\,\ln^2 2,
\end{equation*}
where $\gamma$ denotes the Euler-Mascheroni constant. On the other hand,
\begin{eqnarray*}
  \widetilde{\sigma}_{2n}&amp;=&amp;\sum_{k=1}^{2n}(-1)^k\ln k=\ln\frac{(2n)!!}{(2n-1)!!}=\frac{1}{2}\ln\frac{\big((2n)!!\big)^2}
  {\big((2n-1)!!\big)^2(2n+1)}+\frac{\ln(2n+1)}{2}\,,
\end{eqnarray*}
and
\begin{eqnarray*}
  w_n&amp;=&amp;na_{2n}+\widetilde{\sigma}_{2n-1}=\frac{1}{2}\,\ln(2n)+\sum_{k=1}^{2n-1}(-1)^k\ln k=\frac{1}{2}\,\ln(2n)+\ln\frac{(2n-2)!!}{(2n-1)!!}\\
  &amp;=&amp; \frac{1}{2}\ln\frac{\big((2n)!!\big)^2}
  {\big((2n-1)!!\big)^2(2n+1)}+\frac{1}{2}\ln
  \frac{2n+1}{2n}\,.
\end{eqnarray*}
An application of the Wallis formula
\begin{equation*}
    \lim_{n\to\infty}\frac{\big((2n)!!\big)^2}
    {\big((2n-1)!!\big)^2(2n+1)}=\frac{\pi}{2}\,,
\end{equation*}
shows that $\widetilde{\sigma}_{2n}=o(n)$ and $\ell=\lim_{n\to\infty}w_n=\frac{1}{2}\ln\frac{\pi}{2}\,$.
 Theorem \ref{thd} implies that
\begin{equation*}
    \sum_{n=1}^\infty \sum_{m=1}^\infty
    (-1)^{n+m}\frac{\ln(n+m)}{n+m}=\frac{1}{2}\ln\frac{\pi}{2}-\sigma,
\end{equation*}
and hence
\begin{equation*}
    \sum_{n=0}^\infty \sum_{m=1}^\infty
    (-1)^{n+m}\frac{\ln(n+m)}{n+m}=\sigma+
    \sum_{n=1}^\infty \sum_{m=1}^\infty
    (-1)^{n+m}\frac{\ln(n+m)}{n+m}
    =\frac{1}{2}\ln\frac{\pi}{2}\,,
\end{equation*}
and the problem is solved.
\medskip

Before we give the solution to Problem 4 we need to mention the properties of a special function. Recall that, the dilogarithm function, (see \cite{L, S}), denoted by  ${\rm{Li}}_{2}(z)$, is the special function defined by
\begin{equation*}
{\rm{Li}}_{2}(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{2}}=-\int_{0}^{z}\frac{\ln(1-t)}{t}dt, \quad \quad |z|\leq 1.
\end{equation*}
A special Landen identity involving this function, \cite[p.\ 107]{S}, is given by
\begin{equation*}
{\rm{Li}}_{2}(z)+{\rm{Li}}_{2}(1-z)=\frac{\pi^{2}}{6}-\ln(z)\ln(1-z).
\end{equation*}
This implies, when $z=1/2$, that
\begin{equation}\label{doii}
{\rm{Li}}_{2}\left(\frac{1}{2}\right)=\frac{\pi^{2}}{12}-\frac{\ln^{2}2}{2}.
\end{equation}
Now we are ready to give the solution to Problem 4.
\medskip

{\bf Solution to Problem 4.} Let $a_k=(-1)^k\frac{H_k}{k}$. To calculate
\begin{equation*}
    \sigma=\sum_{k=1}^\infty (-1)^k\frac{H_k}{k},
\end{equation*}
we will be using a general result established by Kantor \cite{KANTOR}: given a sequence
$(\alpha_k)_{k\geq 1}$ of real numbers, we consider the power series
\begin{equation}\label{power-series}
    f(x)=\sum_{k=1}^\infty \alpha_kx^k \quad\mbox{and}\quad
    g(x)=\sum_{k=1}^\infty \alpha_kH_kx^k.
\end{equation}
By the ratio test, the two power series have the same radius of convergence. Kantor \cite{KANTOR} proved that for every real number $x$ for which the
first series in (\ref{power-series}) converges one has
\begin{equation*}
    g(x)=\int_0^1\frac{f(x)-f(tx)}{1-t}\,dt.
\end{equation*}
Let $\alpha_k=(-1)^k/k$, and note that
\begin{equation*}
    f(x)=\sum_{k=1}^\infty(-1)^k\frac{x^k}{k}=-\ln(1+x),
    \qquad x\in(-1,1].
\end{equation*}
It follows that
\begin{equation*}
    \sigma=g(1)=\int_0^1\frac{\ln(1+t)-\ln 2}{1-t}\,dt.
\end{equation*}
The substitution $x=1-t$ implies that
\begin{equation*}
    \sigma=\int_0^1\frac{1}{x}\,\ln\left(1-\frac{x}{2}\right) dx=\int_0^{1/2}\frac{\ln(1-y)}{y}\,dy=-{\rm{Li}}_{2}\left(\frac{1}{2}\right)=\frac{\ln^{2}2}{2}-\frac{\pi^{2}}{12}.
\end{equation*}
On the other hand,
\begin{equation*}
    \widetilde{\sigma}_{2n}=\sum_{k=1}^{2n}(-1)^kH_k=
    \sum_{k=1}^n\big(H_{2k}-H_{2k-1}\big)= \sum_{k=1}^n\frac{1}{2k}
    =\frac{1}{2}\,H_n,
\end{equation*}
and
\begin{equation*}
    w_n=na_{2n}+\widetilde{\sigma}_{2n-1}= \frac{1}{2}\,H_{2n}
    +\widetilde{\sigma}_{2n}-H_{2n}=\frac{1}{2}\, \big(
    H_n-H_{2n}\big).
\end{equation*}
Since $H_n=\ln n+\gamma+o(1)$, we have that all the hypotheses of Theorem \ref{thd} are satisfied and $\ell=\lim_{n\to\infty}w_n=-\frac{1}{2}\,\ln 2.$
Thus,
\begin{equation*}
    \sum_{n=1}^\infty \sum_{m=1}^\infty (-1)^{n+m}\frac{H_{n+m}}{n+m}=l-\sigma=-\frac{\ln 2}{2}-\frac{\ln^{2}2}{2}+\frac{\pi^{2}}{12},
\end{equation*}
and the problem is solved.
\section{Multiple series of a special form} \label{sect3}

In this section we study the multiple series of the form
\begin{equation} \label{unu}
\sum\limits_{n_{1},n_{2},\ldots,n_{k}=1}^{\infty}a_{n_{1}+n_{2}+\cdots+n_{k}},
\end{equation}
 where $(a_{n})_{n\in \mathbb{N}}$ is a sequence of real numbers that verifies a special condition. Before we give the main theorem of this section we collect some results we need in our analysis. Recall that, the Stirling numbers of the first kind, denoted by $s(n,k),$ are the special numbers defined by the generating function
  $$z(z-1)(z-2)\cdots(z-n+1)=\sum\limits_{k=0}^{n}s(n,k)z^{k}.$$ For recurrence relations as well as interesting properties satisfied by these numbers the reader is referred to the book by Srivastava and Choi \cite{S}.  The main result of this section is the following theorem.

 \begin{theorem} \label{prima}
 Let $k\geq 1$ be a natural number and let $(a_{n})_{n\in \mathbb{N}}$ be a sequence of real numbers such that $\sum_{n=1}^{\infty}a_{n}n^{k-1}$ converges absolutely. Then the multiple series (\ref{unu}) converges and
 \begin{equation*} \sum\limits_{n_{1},n_{2},\ldots,n_{k}=1}^{\infty}a_{n_{1}+n_{2}+\cdots+n_{k}}=\frac{1}{(k-1)!}\sum\limits_{i=1}^{k}s(k,i)\left(\sum\limits_{p=k}^{\infty}a_{p}p^{i-1}\right),
 \end{equation*}
 where $s(k,i)$ are the Stirling numbers of the first kind.
 \end{theorem}
 \begin{proof} We have
\begin{equation*}
\begin{aligned}
\sum\limits_{n_{1},n_{2},\ldots,n_{k}=1}^{\infty}a_{n_{1}+n_{2}+\cdots+n_{k}}&amp;=\sum\limits_{p=k}^{\infty}\left(\sum\limits_{n_{1}+n_{2}+\cdots+n_{k}=p}a_{n_{1}+n_{2}+\cdots+n_{k}}\right)\\
&amp;=\sum\limits_{p=k}^{\infty}a_{p}\left(\sum\limits_{n_{1}+n_{2}+\cdots+n_{k}=p}1\right)\\
&amp;=\sum\limits_{p=k}^{\infty}a_{p}\binom{p-1}{k-1}\\
&amp;=\frac{1}{(k-1)!}\sum\limits_{p=k}^{\infty}\frac{a_{p}}{p}\left(p(p-1)(p-2)\cdots (p-k+1)\right)\\
&amp;=\frac{1}{(k-1)!}\sum\limits_{p=k}^{\infty}\frac{a_{p}}{p}\left(\sum\limits_{i=0}^{k}s(k,i)p^{i}\right)\\
&amp;=\frac{1}{(k-1)!}\sum\limits_{i=0}^{k}s(k,i)\left(\sum\limits_{p=k}^{\infty}a_{p}p^{i-1}\right)\\
&amp;=\frac{1}{(k-1)!}\sum\limits_{i=1}^{k}s(k,i)\left(\sum\limits_{p=k}^{\infty}a_{p}p^{i-1}\right),\
\end{aligned}
\end{equation*}
since $s(k,0)=0$ for $k\in \mathbb{N}$. The theorem is proved.
\end{proof}
The next result, which is a consequence of Theorem \ref{prima}, is about the calculation of another multiple series of a special form.

\begin{corollary} \label{coroc}Let $1\leq i\leq k$ be fixed natural numbers and let $(a_{n})_{n\in \mathbb{N}}$ be a sequence of positive numbers such that $\sum_{n=1}^{\infty}a_{n}n^{k}$ converges. Then
 \begin{equation*}
 \sum\limits_{n_{1},n_{2},\ldots,n_{k}=1}^{\infty}n_{i}\cdot a_{n_{1}+n_{2}+\cdots+n_{k}}=\frac{1}{k!}\sum\limits_{i=1}^{k}s(k,i)\left(\sum\limits_{p=k}^{\infty}a_{p}p^{i}\right),
 \end{equation*}
 where $s(k,i)$ are the Stirling numbers of the first kind.
\end{corollary}
\begin{proof} We have, based on symmetry reasons, that for all $i,j=1,\ldots, k,$ one has that
\begin{equation*}
\sum\limits_{n_{1},n_{2},\ldots,n_{k}=1}^{\infty}n_{i}\cdot a_{n_{1}+n_{2}+\cdots+n_{k}}=\sum\limits_{n_{1},n_{2},\ldots,n_{k}=1}^{\infty}n_{j}\cdot a_{n_{1}+n_{2}+\cdots+n_{k}},
\end{equation*}
and hence
\begin{equation*}
\sum\limits_{n_{1},n_{2},\ldots,n_{k}=1}^{\infty}n_{i}\cdot a_{n_{1}+n_{2}+\cdots+n_{k}}=\frac{1}{k}\sum\limits_{n_{1},n_{2},\ldots,n_{k}=1}^{\infty}(n_{1}+n_{2}+\cdots+n_{k})\cdot a_{n_{1}+n_{2}+\cdots+n_{k}},
\end{equation*}
and the result follows based on Theorem \ref{prima} applied to the sequence $(na_{n})_{n\in \mathbb{N}}$.
\end{proof}
\begin{corollary} a) Let $k$ be a fixed positive integer and let $m&gt;k$. Then
\begin{multline*}
\sum\limits_{n_{1},n_{2},\ldots, n_{k}=1}^{\infty}\frac{1}{(n_{1}+n_{2}+\cdots+n_{k})^{m}}\\
=\frac{1}{(k-1)!}\sum\limits_{i=1}^{k}s(k,i)\left(\zeta(m+1-i)-1-\frac{1}{2^{m+1-i}}-\cdots-\frac{1}{(k-1)^{m+1-i}}\right),
\end{multline*}
where the parenthesis contains only the term $\zeta(m+1-i)$ when $k=1$.

b)  Let $k$ and $i$ be fixed positive integers such that $1\leq i\leq k,$ and let $m$ be such that $m-k&gt;1$. Then
\begin{multline*}
\sum\limits_{n_{1},n_{2},\ldots, n_{k}=1}^{\infty}\frac{n_{i}}{(n_{1}+n_{2}+\cdots+n_{k})^{m}}\\
=\frac{1}{k!}\sum_{i=1}^{k}s(k,i)\left(\zeta(m-i)-1-\frac{1}{2^{m-i}}-\cdots-\frac{1}{(k-1)^{m-i}}\right),
\end{multline*}
where the parenthesis contains only the term $\zeta(m-i)$ when $k=1$.
\end{corollary}
The next result specializes to the case when the sequence $a_{n}=1/n!$. Recall that, if $k\in \mathbb{N}$, the series $S_{k}=\sum_{n=1}^{\infty}\frac{n^{k}}{n!},$ also known as the Wolstenholme series, equals an integral multiple of $e$, i.e., $S_{k}=B_{k}e$, (see \cite[p.\ 197]{B}). The integer $B_{k}$ is known in the mathematical literature as the $k$th Bell number \cite{Comtet} and the equality $S_{k}=B_{k}e$ is known as Dobinski's formula \cite{Dob}. For example, $B_{1}=1$, $B_{2}=2$, $B_{3}=5$, $B_{4}=15$, $B_{5}=52$. One can prove that the sequence $(B_{k})_{k\in \mathbb{N}}$ verifies the recurrence formula
$$B_{k}=\sum\limits_{j=1}^{k-1}\binom{k-1}{j}B_{j}+1.$$
The next corollary gives a recurrence relation involving the Stirling numbers of the first kind and the Bell numbers. The formula is obtained by calculating, by two different methods, a multiple series involving a factorial term.
\begin{corollary} Let $k\geq 2$ and let $i$ be fixed positive integers such that $1\leq i\leq k$. Then
\begin{equation*}
a) \quad \sum\limits_{n_{1},n_{2},\ldots, n_{k}=1}^{\infty}\frac{n_{i}}{(n_{1}+n_{2}+\cdots+n_{k})!}=\frac{e}{k!}.
\end{equation*}
\begin{equation*}
b) \quad \sum\limits_{i=1}^{k}s(k,i)\left(B_{i}e-\sum\limits_{p=1}^{k-1}\frac{p^{i}}{p!}\right)=e.
\end{equation*}
\end{corollary}
\begin{proof} We have, based on Corollary \ref{coroc}, that
\begin{equation*}
\sum\limits_{n_{1},n_{2},\ldots, n_{k}=1}^{\infty}\frac{n_{i}}{(n_{1}+n_{2}+\cdots+n_{k})!}=\frac{1}{k!} \sum\limits_{i=1}^{k}s(k,i)\left(B_{i}e-\sum\limits_{p=1}^{k-1}\frac{p^{i}}{p!}\right).
\end{equation*}
On the other hand,
\begin{equation*}
\begin{aligned}
\sum\limits_{n_{1},n_{2},\ldots, n_{k}=1}^{\infty}\frac{n_{i}}{(n_{1}+n_{2}+\cdots+n_{k})!}&amp;=\frac{1}{k}\sum\limits_{n_{1},n_{2},\ldots, n_{k}=1}^{\infty}\frac{1}{(n_{1}+n_{2}+\cdots+n_{k}-1)!}\\
&amp;=\sum\limits_{p=k}^{\infty}\left(\sum\limits_{n_{1}+\cdots+n_{k}=p}\frac{1}{(n_{1}+n_{2}+\cdots+n_{k}-1)!}\right)\\
&amp;=\frac{1}{k}\sum\limits_{p=k}^{\infty}\frac{1}{(p-1)!}\binom{p-1}{k-1}\\
&amp;=\frac{1}{k!}\sum\limits_{p=k}^{\infty}\frac{1}{(p-k)!}\\
&amp;=\frac{e}{k!},
\end{aligned}
\end{equation*}
and the corollary is proved.
\end{proof}
The next result refers to the calculation of multiple Wolstenholme series.
\begin{corollary}
Let $m$ and $k$ be natural numbers. Then
\begin{equation*}
\sum\limits_{n_{1},\ldots, n_{k}=1}^{\infty}\frac{(n_{1}+n_{2}+\cdots+n_{k})^{m}}{(n_{1}+n_{2}+\cdots+n_{k})!}=\frac{1}{(k-1)!}\sum\limits_{i=1}^{k}s(k,i)\left(B_{m+i-1}e-\sum\limits_{p=1}^{k-1}\frac{p^{m+i-1}}{p!}\right),
\end{equation*}
where the parenthesis contains only the term $B_{m+i-1}e$ when $k=1$.
\end{corollary}


\bibliographystyle{amsplain}
\begin{thebibliography}{9}

\bibitem{B} T. J. J'A. Bromwich, {\it An Introduction to the Theory of Infinite Series}, AMS Chelsea Publishing, Third Edition, 1991.

\bibitem{Comtet}
L. Comtet, \textit{Advanced Combinatorics}, D. Reidel, 1974.

\bibitem{Dob}
G. Dobi\'nski, Summirung der Reihe $\sum n^m/n!$ f\"ur $m=1,2,$
$3,4,5,\dots$, \textit{Arch. f\"ur Mat. und Physik}
\textbf{61} (1877), 333--336.

\bibitem{EFTHIMIOU}  C. Efthimiou,
Problem 1838,  \textit{Math. Mag.} \textbf{83} (2010), 65.

\bibitem{F}  O. Furdui,
Problem 5013, \textit{School Science and Mathematics} \textbf{108} March (2008), 2.

\bibitem{FURDUI1}  O. Furdui,
Problem 1849, \textit{Math. Mag.} \textbf{83} (2010), 227.

\bibitem{FURDUI2}  O. Furdui,
Problem 11519, \textit{Amer. Math. Monthly} \textbf{117} (2010), 649.

\bibitem{KANTOR} W. H. M. Kantor,
Solution to problem 4946, \textit{Amer. Math. Monthly} \textbf{69}
(1962), 239.

\bibitem{KLAMKIN1}  M. S. Klamkin,
Problem 4592, \textit{Amer. Math. Monthly} \textbf{61} (1954), 350.

\bibitem{KLAMKIN2} M. S. Klamkin,
Problem 4946, \textit{Amer. Math. Monthly} \textbf{68} (1961), 67.

\bibitem{L} L. Lewin, {\it Polylogarithms and Associated Functions}, Elsevier, 1981.
\bibitem{S} H. M. Srivastava and J. Choi, {\it Series Associated with the Zeta and Related Functions}, Kluwer, 2001.

\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}: Primary
40A05; Secondary 40B05.

\noindent \emph{Keywords: } Bell numbers, iterated series,
double alternating harmonic series, Stirling numbers, Wolstenholme series.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000110} and
\seqnum{A094638}.)

\bigskip
\hrule
\bigskip

\noindent Received February 10 2011;
revised versions received May 9 2011.
Published in {\it Journal of Integer Sequences}, May 17 2011.

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