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\begin{center}
\vskip 1cm{\LARGE\bf 
Tornheim-Like Series, Harmonic Numbers, \\
\vskip .1in
and Zeta Values
}
\vskip 1cm
\large
Ilham A. Aliev and Ayhan Dil\\
Department of Mathematics\\
Akdeniz University\\
07070 Antalya\\
Turkey \\
\href{mailto:ialiev@akdeniz.edu.tr}{\tt ialiev@akdeniz.edu.tr} \\
\href{mailto:adil@akdeniz.edu.tr}{\tt adil@akdeniz.edu.tr}
\end{center}

\vskip .2 in

\begin{abstract}
We obtain explicit evaluations of the Tornheim-like multiple series
involving harmonic numbers. We give a new relationship between harmonic
numbers and $\zeta(  2)  $. We also present closed-form formulas of some
multiple series in terms of zeta values.
\end{abstract}

\section{Introduction}

The Riemann zeta function is defined by
\[
\zeta(  z)  =\sum_{k=1}^{\infty}\frac{1}{k^{z}},\text{ \ \ }(
\operatorname{Re}z&gt;1)  .
\]
For even positive integers, one has the well-known relationship between zeta
values and Bernoulli numbers:
\begin{equation}
\zeta(  2n)  =(  -1)  ^{n+1}\frac{(  2\pi)
^{2n}}{2(  2n)  !}B_{2n}. \label{zb}
\end{equation}
Here 
$$ B_{0}=1,\quad B_{2}=\frac{1}{6}, \quad B_{4}=-\frac{1}{30}, \quad
B_{6}=\frac {1}{42},\ldots \ .$$
This result was proved for the first time by Euler in 1740.

For odd positive
integers, no simple expression like (\ref{zb}) is known. Roger Ap\'{e}ry
\cite{AR} proved the irrationality of $\zeta(  3)  $, and after that
$\zeta(  3)  $ was named Ap\'{e}ry's constant. Rivoal
\cite{TR}\ showed that infinitely many of the numbers $\zeta(
2n+1)  $ must be irrational.  Zudilin \cite{WZ} showed that
at least one of the numbers $\zeta(  5)  $, $\zeta(  7)
$, $\zeta(  9)  $ and $\zeta(  11)  $ is irrational.

For
a positive integer $n$ and an integer $m$ the $n$th partial sum of
$\zeta(  m)  $ is called the $n$th generalized harmonic number of
order $m$, and is denoted by $H_{n}^{(  m)  }$, i.e.,
\[
H_{n}^{(  m)  }=\sum_{k=1}^{n}\frac{1}{k^{m}}.
\]

The special case $m=1$ is the classical harmonic number
\[
H_{n}=\sum_{k=1}^{n}\frac{1}{k},\text{ }H_{0}=0.
\]
It is well known that there are deep relationships between Tornheim-like
series, harmonic numbers and zeta values. The Tornheim double series \cite{Tor}
(or the so-called Witten zeta function \cite{DZ}) is defined by
\begin{equation}
S(  a,b,c)  :=\sum_{m,n=1}^{\infty}\frac{1}{m^{a}n^{b}(
m+n)  ^{c}}. \label{TDS}
\end{equation}
The following equation is a simple and nice example of a connection between
Tornheim-like series, harmonic numbers, and zeta values (see \cite{B, BB, E1,
E2, N}):
\begin{equation}
\sum_{n,m=1}^{\infty}\frac{1}{nm(  n+m)  }=\sum_{m=1}^{\infty}
\frac{H_{m}}{m^{2}}=2\zeta(  3)  . \label{hz}
\end{equation}
Tornheim-like series have attracted increasing attention in recent years and they have proven to be a powerful tool for finding many interesting relationships between various zeta values (\cite{BAB, BA, B, BB, BD, E1, E2, K, M, S2, XZ}).

Boyadzhiev \cite{KB1, KB2} described a simple method for evaluating double series
of the form (\ref{TDS}) in terms of zeta values. Kuba \cite{K} considered the
following general sum:
\[
V=\sum_{j,k=1}^{\infty}\frac{H_{j+k}^{(  u)  }}{j^{r}k^{s}(
j+k)  ^{t}}.
\]
This sum includes the Tornheim double series (\ref{TDS}) as a special case.
Kuba \cite{K} proved that whenever $w=r+s+t+u$ is even, for $r,s,t,w\in
\mathbb{N}$ , the series $V$ can be explicitly evaluated in terms of zeta functions.

On the other hand, Xu and Li \cite{XZ} used the Tornheim type series for the evaluation of non-linear Euler sums. Among other results they
obtained
\begin{equation}
\sum_{m=1}^{\infty}\frac{H_{m+k}}{m(  m+k)  }=\frac{H_{k}^{2}
+H_{k}^{(  2)  }}{k},k\in\mathbb{N}=\left\{  1,2,3,\ldots\right\}
. \label{XL}
\end{equation}
From (\ref{hz}) and (\ref{XL}) it is easy to see that the value of the series
\[
a(  k)  =\sum_{m=1}^{\infty}\frac{H_{m+k}}{m(  m+k)
},k\in\mathbb{N\cup}\left\{  0\right\}
\]
is irrational for $k=0$ and rational for every $k\in\mathbb{N}$. Hence the
following questions naturally arise: for integers $s\in\mathbb{N\cup
}\left\{  0\right\}  $, are the values of the double series
\[
\sum_{n,m=1}^{\infty}\frac{H_{n+m+s}}{nm(  n+m+s)  },
\]
and more generally, the multiple series
\[
A_{n}(  s)  =\sum_{k_{1}=1}^{\infty}\cdots\sum_{k_{n-1}=1}^{\infty
}\frac{H_{k_{1}+\cdots+k_{n-1}+s}}{k_{1}\cdots k_{n-1}(  k_{1}
+\cdots+k_{n-1}+s)  }
\]
rational or irrational numbers? This question is studied in the second
section. Namely, in the case when $n$ is odd, we have solved this question
exactly. If $n$ is even, we give a partial solution depending on the odd
zeta values.

In the third section, some new relationships between harmonic numbers and
$\zeta(  2)  $ are given and explicit evaluation formulas for some
double series via zeta values are established.

\section{Explicit evaluations of the Tornheim-like multiple series involving
harmonic numbers}

\begin{theorem}
\label{it}Consider the double series
\[
A(  s)  =\sum_{n,m=1}^{\infty}\frac{H_{n+m+s}}{nm(
n+m+s)  },\text{ \ \ }s\in\mathbb{N\cup}\left\{  0\right\}  .
\]
For any $s\in\mathbb{N}$ the value of $A(  s)  $ is rational but
$A(  0)  $ is irrational. More precisely,
\begin{displaymath}
A(  s)  = \begin{cases}
6\zeta(  4),  &amp; \text{if }s=0;\\
6\sum_{j=0}^{s-1}(  -1)  ^{j}\binom{s-1}{j}\frac{1}{(
j+1)  ^{4}}, &amp; \text{if }s\geq1 .
\end{cases}
\end{displaymath}

\end{theorem}

\begin{proof}
Using telescoping series, we have
\begin{align*}
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+m+s}  &amp;  =\sum_{k=1}^{\infty
}(  \frac{1}{k}-\frac{1}{k+n+m+s}) \\
&amp;  =(  n+m+s)  \sum_{k=1}^{\infty}\frac{1}{k(  k+n+m+s)
}.
\end{align*}
It then follows that
\begin{align*}
A(  s)   &amp;  =\sum_{n,m=1}^{\infty}\frac{1}{nm(  n+m+s)
}(  1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+m+s}) \\
&amp;  =\sum_{n,m,k=1}^{\infty}\frac{1}{nmk(  n+m+k+s)  }\\
&amp;  =\sum_{n,m,k=1}^{\infty}\left(  \int_{0}^{1}x^{n-1}dx\right) \left( 
\int_{0}^{1}y^{m-1}dy\right)  \left(  \int_{0}^{1}z^{k-1}dz\right) \left(
\int_{0}^{1}t^{n+m+k+s-1}dt\right) \\
&amp;  =\int_{0}^{1}t^{s+2}\left[  \int_{0}^{1}\left(  \sum_{n=1}^{\infty}(
xt)  ^{n-1}\right)  dx\int_{0}^{1}\left(  \sum_{m=1}^{\infty}(
yt)  ^{m-1}\right)  dy\int_{0}^{1}\left(  \sum_{k=1}^{\infty}(
zt)  ^{k-1}\right)  dz\right]  dt\\
&amp;  =\int_{0}^{1}t^{s+2}\left[  \int_{0}^{1}\frac{1}{1-xt}dx\int_{0}^{1}
\frac{1}{1-yt}dy\int_{0}^{1}\frac{1}{1-zt}dz\right]  dt.
\end{align*}
Since
\[
\int_{0}^{1}\frac{1}{1-ut}du=-\frac{1}{t}\ln(  1-t)  ,
\]
we have
\begin{equation}
A(  s)  =-\int_{0}^{1}t^{s-1}\ln^{3}(  1-t)
dt=-\int_{0}^{1}(  1-t)  ^{s-1}\ln^{3}t\, dt. \label{u1}
\end{equation}
Setting $s=0$, it follows that
\begin{align*}
A(  0)   &amp;  =-\int_{0}^{1}\frac{1}{1-t}\ln^{3}t\, dt=-\sum
_{j=0}^{\infty}\int_{0}^{1}t^{j}\ln^{3}t \, dt\\
&amp;  =-\sum_{j=0}^{\infty}(  -\frac{6}{(  j+1)  ^{4}})
=6\zeta(  4)  =\frac{\pi^{4}}{16}.
\end{align*}
On the other hand, if $s\geq1$, then utilizing the formulas
\[
(  1-t)  ^{s-1}=\sum_{j=0}^{s-1}(  -1)  ^{j}\binom
{s-1}{j}t^{j}
\]
and
\[
\int_{0}^{1}t^{j}\ln^{3}t\, dt=-\frac{3!}{(  j+1)  ^{4}},
\]
(\ref{u1}) can be computed explicitly as
\begin{align*}
A(  s)   &amp;  =-\int_{0}^{1}(  1-t)  ^{s-1}\ln^{3}t\, dt\\
&amp;  =3!\sum_{j=0}^{s-1}(  -1)  ^{j}\binom{s-1}{j}\frac{1}{(
j+1)  ^{4}}.
\end{align*}
This proves the stated result.
\end{proof}

In the same way as in Theorem \ref{it}, by making use of the formulas
\[
(  1-t)  ^{s-1}=\sum_{j=0}^{s-1}(  -1)  ^{j}\binom
{s-1}{j}t^{j}\text{ and }\int_{0}^{1}t^{j}\ln^{k}t\, dt=(  -1)
^{k}\frac{k!}{(  j+1)  ^{k+1}},
\]
one can prove the following more general result.

\begin{theorem}
\label{G} Let
\[
A_{n}(  s)  =\sum_{k_{1}=1}^{\infty}\cdots\sum_{k_{n-1}=1}^{\infty
}\frac{H_{k_{1}+\cdots+k_{n-1}+s}}{k_{1}\cdots k_{n-1}(  k_{1}
+\cdots+k_{n-1}+s)  },\text{ }s\in\mathbb{N\cup}\left\{  0\right\}
,n\geq2.
\]
Then
\begin{equation}
A_{n}(  s)  =
\begin{cases}
n!\zeta(  n+1),  &amp; \text{if }s=0;\\
n!\sum_{j=0}^{s-1}(  -1)  ^{j}\binom{s-1}{j}\frac{1}{(
j+1)  ^{n+1}}, &amp; \text{if }s\geq1.
\end{cases}
 \label{g}
\end{equation}

\end{theorem}

Two special cases of the theorem are as follows:
\[
A_{2}(  s)  =\sum_{k=1}^{\infty}\frac{H_{k+s}}{k(  k+s)
}=
\begin{cases}
2!\zeta(  3),  &amp; \text{if }s=0;\\
2!\sum_{j=0}^{s-1}(  -1)  ^{j}\binom{s-1}{j}\frac{1}{(
j+1)  ^{3}}, &amp; \text{if }s\geq1,
\end{cases}
\]
and
\[
A_{4}(  s)  =\sum_{k,m,n=1}^{\infty}\frac{H_{k+m+n+s}}{kmn(
k+m+n+s)  }=
\begin{cases}
4!\zeta(  5),  &amp; \text{if }s=0;\\
4!\sum_{j=0}^{s-1}(  -1)  ^{j}\binom{s-1}{j}\frac{1}{(
j+1)  ^{5}}, &amp; \text{if }s\geq1.
\end{cases}
\]


\begin{remark}
It can be easily seen from (\ref{g}) that the expression $A_{n}(
s)  $ is a rational number for all $s\geq1$ and $n\geq2$. However
$A_{2}(  0)  =2!\zeta(  3)  $ is irrational (Ap\'{e}ry).
If $n\geq4$ and even, it is not known whether the numbers $A_{n}(
0)  =n!\zeta(  n+1)  $ are irrational or not. On the other
hand, for any odd $n\in\mathbb{N}$ we have $A_{n}(  0)
=n!\zeta(  n+1)  =r_{n}\pi^{n+1}$ (see (\ref{zb})), which is irrational
because $r_{n}$ is rational and $\pi^{n+1}$ is irrational. Notice that, as is
well known, the irrationality of $\pi^{n}$ is a consequence of the
transcendence of $\pi$.
\end{remark}

\begin{remark}
There is an interesting connection between the multiple series $A_{n}(
s)  $\ and the Bell polynomials $b_{n}(  x)  $. For a given
sequence $x=(  x_{1},x_{2},\ldots)  $, the Bell polynomials
$b_{n}(  x)  =b_{n}(  x_{1},x_{2},\ldots)  $ are defined
by the generating function (see \cite{C} or \cite{Mb})
\[
\sum_{n=0}^{\infty}b_{n}(  x)  \frac{t^{n}}{n!}=\exp\left\{
\sum_{k=1}^{\infty}x_{k}\frac{t^{k}}{k!}\right\}  .
\]
As a result of this definition one has the following explicit expression
\cite{C}:
\[
b_{n}(  x)  =\sum_{\sigma(  n)  }\frac{n!}{a_{1}
!a_{2}!\cdots a_{n}!}\left(  \frac{x_{1}}{1}\right)  ^{a_{1}}\left(
\frac{x_{2}}{2}\right)  ^{a_{2}}\cdots\left(  \frac{x_{n}}{n}\right)  ^{a_{n}
},
\]
where the summation ranges over the set $\sigma(  n)  $ of all
partitions of $n$. Using this explicit expression we can give a list of the
first few Bell polynomials as
\begin{align}
b_{0}(  x)  &amp;=1, &amp; b_{1}(  x)  &amp;=x_{1}, \label{t}\\
b_{2}(  x)  &amp;=x_{1}^{2}+x_{2}, &amp; b_{3}(  x)  &amp;=x_{1} ^{3}+3x_{1}x_{2}+2x_{3}.
\nonumber
\end{align}
Considering the well-known harmonic number identity (see \cite[Eq. (3.56)]{WJ}
or \cite[Corollary 2.2]{KL})
\begin{align*}
&amp;  i!m\dbinom{m+n}{n}\sum_{k=0}^{n}(  -1)  ^{k}\binom{n}{k}\frac
{1}{(  m+k)  ^{i+1}}\\
&amp;  =b_{i}(  H_{m+n}-H_{m-1},H_{m+n}^{(  2)  }-H_{m-1}^{(
2)  },\ldots ,H_{m+n}^{(  i)  }-H_{m-1}^{(  i)
})
\end{align*}
we have
\begin{equation}
n!s\sum_{j=0}^{s-1}(  -1)  ^{j}\binom{s-1}{j}\frac{1}{(
j+1)  ^{n+1}}=b_{n}(  H_{s},H_{s}^{(  2)  }
,\ldots,H_{s}^{(  n)  })  \label{s},
\end{equation}
which corresponds to (\ref{g}) for $s\geq1$. Hence we can restate
$A_{n}(  s)  $ in terms of the Bell polynomials as
\begin{displaymath}
A_{n}(  s)  =
\begin{cases}
n!\zeta(  n+1),  &amp; \text{if }s=0;\\
\frac{1}{s}b_{n}(  H_{s},H_{s}^{(  2)  },\ldots,H_{s}^{(
n)  }),  &amp; \text{if }s\geq1.
\end{cases}
\end{displaymath}
Thanks to this formula, considering (\ref{t}) we can write $A_{n}(
s)  $ as a finite combination of the harmonic and generalized harmonic
numbers. For instance, for $n=2$ we have
\[
A_{2}(  s)  =
\begin{cases}
2\zeta(  3),  &amp; \text{if }s=0;\\
\frac{1}{s}(  (  H_{s})  ^{2}+H_{s}^{(  2)
}),  &amp; \text{if }s\geq1,
\end{cases}
\]
which coincides with (\ref{XL}). For $n=3$ we have
\[
A_{3}(  s)  =
\begin{cases}
6\zeta(  4),  &amp; \text{if }s=0;\\
\frac{1}{s}(  (  H_{s})  ^{3}+3H_{s}H_{s}^{(  2)
}+2H_{s}^{(  3)  }),  &amp; \text{if }s\geq1.
\end{cases}
\]

\end{remark}

\section{Explicit evaluations of some double series in zeta values}

The next theorem gives a new relationship between harmonic numbers and
$\zeta(  2)  $.

\begin{theorem}
Let $O_{m}=\sum_{k=1}^{m}\frac{1}{2k-1}=H_{2m}-\frac{1}{2}H_{m}$. Then
\begin{equation}
\sum_{m=1}^{\infty}\frac{O_{m}}{2m(  2m+1)  }=\frac{1}{4}
\zeta(  2)  .\label{on}
\end{equation}
\end{theorem}

\begin{proof}
Let
\[
A=\sum_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+1)  }
\]
and
\[
B=\sum_{m,n=1}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+1)  }.
\]
From the equation
\[
\sum_{k=0}^{\infty}\frac{1}{(  2k+1)  ^{2}}=\frac{\pi^{2}}{8},
\]
we have
\begin{equation}
A=\frac{\pi^{2}}{4}-1+B.\label{2}
\end{equation}
Further, using telescoping series, we have
\begin{align*}
B &amp;  =\sum_{m=1}^{\infty}\frac{1}{2m+1}\frac{1}{2m}\sum_{n=1}^{\infty}(
\frac{1}{2n+1}-\frac{1}{2n+1+2m})  \\
&amp;  =\sum_{m=1}^{\infty}\frac{1}{2m(  2m+1)  }(  \frac{1}
{3}+\frac{1}{5}+\cdots+\frac{1}{2m+1})  \\
&amp;  =\sum_{m=1}^{\infty}\frac{O_{m}-\frac{2m}{2m+1}}{2m(  2m+1)  }\\
&amp;  =\sum_{m=1}^{\infty}\frac{O_{m}}{2m(  2m+1)  }-\sum
_{m=1}^{\infty}\frac{1}{(  2m+1)  ^{2}}\\
&amp;  =\sum_{m=1}^{\infty}\frac{O_{m}}{2m(  2m+1)  }-\frac{\pi^{2}}
{8}+1.
\end{align*}
Hence we obtain that
\begin{equation}
B=\sum_{m=1}^{\infty}\frac{O_{m}}{2m(  2m+1)  }+1-\frac{3}{4}
\zeta(  2)  .\label{s1}
\end{equation}
Now let us evaluate $A$.
\begin{align*}
A &amp;  =\sum_{m,n=0}^{\infty}\left(  \int_{0}^{1}x^{2m}dx\right)  \left( 
\int_{0}^{1}y^{2n}dy\right)  \left(  \int_{0}^{1}t^{2m+2n}dt\right)  \\
&amp;  =\int_{0}^{1}\left(  \int_{0}^{1}\sum_{m=0}^{\infty}(  xt)
^{2m}dx\int_{0}^{1}\sum_{n=0}^{\infty}(  yt)  ^{2n}dy\right)  dt\\
&amp;  =\int_{0}^{1}\left(  \int_{0}^{1}\frac{1}{1-(  xt)  ^{2}}
dx\int_{0}^{1}\frac{1}{1-(  yt)  ^{2}}dy\right)  dt\\
&amp;  =\frac{1}{4}\int_{0}^{1}\frac{1}{t^{2}}\ln^{2}\left(  \frac{1+t}{1-t}\right)  dt.
\end{align*}
The substitution $\frac{1+t}{1-t}=u$ immediately leads to the following
equality:
\[
A=\frac{1}{2}\int_{1}^{\infty}\frac{1}{(  1-u)  ^{2}}\ln^{2}udu.
\]
Integration by parts gives
\begin{align}
A &amp;  =\int_{1}^{\infty}\frac{1}{u(  u-1)  }\ln u\, du=\int_{1}
^{\infty}\frac{1}{1-\frac{1}{u}}\frac{\ln u}{u^{2}}du\nonumber\\
&amp;  =\sum_{k=0}^{\infty}\int_{1}^{\infty}u^{-k-2}\ln u \,du=\sum_{k=0}^{\infty
}\frac{1}{(  k+1)  ^{2}}=\zeta(  2)  =\frac{\pi^{2}}
{6}.\label{4}
\end{align}
Similarly, from (\ref{2}), (\ref{s1}) and (\ref{4}) we have
\[
\frac{\pi^{2}}{6}=\frac{\pi^{2}}{4}-1+\sum_{m=1}^{\infty}\frac{O_{m}
}{2m(  2m+1)  }-\frac{3}{4}\zeta(  2)  +1
\]
and as a result
\[
\sum_{m=1}^{\infty}\frac{O_{m}}{2m(  2m+1)  }=\frac{1}{4}
\zeta(  2)  .
\]

\end{proof}

\begin{remark}
Considering (\ref{s1}) we have
\begin{equation}
B=\sum_{m=1}^{\infty}\frac{H_{2m+1}-1-\frac{1}{2}H_{m}}{2m(  2m+1)
}.\label{r1}
\end{equation}
Now we use (\ref{2}) and (\ref{4}) to conclude that
\begin{equation}
\sum_{m=1}^{\infty}\frac{2H_{2m+1}-H_{m}}{2m(  2m+1)  }=2(
2-\ln2)  -\zeta(  2)  .\label{ln}
\end{equation}

\end{remark}

In the following theorem we obtain interesting relationships between some
special double series and zeta values $\zeta(  2)  $ and
$\zeta(  3)  $.

\begin{theorem}
We have the following series evaluations:
\begin{itemize}
\item[(a)]
$\sum_{m,n=0}^{\infty}\frac
{1}{(  m+\frac{1}{2})  (  n+\frac{1}{2})  (
m+n+\frac{1}{2})  (  m+n+1)  }=16\zeta(  2)
-14\zeta(  3)  
$.

\item[(b)] $\sum_{m,n=0}^{\infty}\frac
{1}{(  m+\frac{1}{2})  (  n+\frac{1}{2})  (
m+n+1)  (  m+n+\frac{3}{2})  }=14\zeta(  3)
-8\zeta(  2)  $ .

\item[(c)]
$\sum_{m,n=0}^{\infty}\frac{1}{(  m+\frac{1}{2})
(  n+\frac{1}{2})  (  m+n+\frac{1}{2})  (
m+n+1)  (  m+n+\frac{3}{2})  }=24\zeta(  2)
-28\zeta(  3)  $.
\end{itemize}
\end{theorem}

\begin{proof}
\begin{itemize}
\item[(a)]
We have
\begin{align*}
&amp;  \sum_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+2)  }\\
&amp;  =\sum_{m,n=0}^{\infty}\left(  \int_{0}^{1}x^{2m}dx\right)  \left(  \int
_{0}^{1}y^{2n}dy\right)  \left(  \int_{0}^{1}t^{2m+2n+1}dt\right) \\
&amp;  =\int_{0}^{1}\left(  \int_{0}^{1}\sum_{m=0}^{\infty}(  xt)
^{2m}dx\int_{0}^{1}\sum_{n=0}^{\infty}(  yt)  ^{2n}dy\right)  t\, dt\\
&amp;  =\frac{1}{4}\int_{0}^{1}\frac{1}{t}\ln^{2}\left(  \frac{1+t}{1-t}\right)
dt.
\end{align*}
Here the substitution $\frac{1+t}{1-t}=u$ leads to the following equality:
\begin{align*}
\frac{1}{4}\int_{0}^{1}\frac{1}{t}\ln^{2}\left(  \frac{1+t}{1-t}\right)  dt
&amp;  =\frac{1}{2}\int_{1}^{\infty}\frac{1}{(  u^{2}-1)  }\ln
^{2}udu=\frac{1}{2}\int_{1}^{\infty}\frac{1}{u^{2}}\frac{1}{(  1-\frac
{1}{u^{2}})  }\ln^{2}udu\\
&amp;  =\frac{1}{2}\sum_{k=0}^{\infty}\int_{1}^{\infty}u^{-2k-2}\ln u \,du.
\end{align*}
After integration by parts we get
\[
\frac{1}{2}\sum_{k=0}^{\infty}\int_{1}^{\infty}u^{-2k-2}\ln u \,du=\sum
_{k=0}^{\infty}\frac{1}{(  2k+1)  ^{3}}=\frac{7}{8}\zeta(
3)  .
\]
Hence
\begin{equation}
\sum_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+2)  }=\frac{7}{8}\zeta(  3)  . \label{6}
\end{equation}
On the other hand, according to the formula (\ref{4}),
\begin{equation}
A=\sum_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+1)  }=\zeta(  2)  . \label{7}
\end{equation}
Now, from (\ref{6}) and (\ref{7}) it follows that
\begin{align*}
\zeta(  2)  -\frac{7}{8}\zeta(  3)   &amp;  =\sum
_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)  (
2m+2n+1)  }\\
&amp;  -\sum_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+2)  }\\
&amp;  =\sum_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+1)  (  2m+2n+2)  }
\end{align*}
and this proves ($a$).

\item[(b)]
By the same method in the proof of (a), we have
\begin{align*}
&amp;  \sum_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+3)  }\\
&amp;  =\sum_{m,n=0}^{\infty}\left( \int_{0}^{1}x^{2m}dx\right)  \left(  \int
_{0}^{1}y^{2n}dy\right)  \left(  \int_{0}^{1}t^{2m+2n+2}dt\right) \\
&amp;  =\int_{0}^{1}\left(  \int_{0}^{1}\frac{1}{1-(  xt)  ^{2}
}dx\int_{0}^{1}\frac{1}{1-(  yt)  ^{2}}dy\right)t^{2}   dt\\
&amp;  =\frac{1}{4}\int_{0}^{1}\ln^{2}\left(  \frac{1+t}{1-t}\right)  dt\\
&amp;  =\frac{1}{2}\int_{1}^{\infty}\frac{1}{(  u+1)  ^{2}}\ln
^{2}udu=-\frac{1}{2}\int_{1}^{\infty}\ln^{2}ud(  \frac{1}{u+1}) \\
&amp;  =\int_{1}^{\infty}\frac{1}{u(  u+1)  }\ln u \, du=\sum_{k=2}
^{\infty}(  -1)  ^{k}\int_{1}^{\infty}u^{-k}\ln u \, du\\
&amp;  =\sum_{k=2}^{\infty}(  -1)  ^{k}\frac{1}{(  k-1)
^{2}}=\frac{\pi^{2}}{12}.
\end{align*}
Thus
\begin{equation}
\sum_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+3)  }=\frac{\pi^{2}}{12}=\frac{1}{2}\zeta(  2)
. \label{b}
\end{equation}
Now, from (\ref{6}) and (\ref{b}) we have
\[
\frac{7}{8}\zeta(  3)  -\frac{1}{2}\zeta(  2)
=\sum_{m,n=0}^{\infty}\frac{1}{(  2m+1)  (  2n+1)
(  2m+2n+2)  (  2m+2n+3)  }
\]
and this proves (b).

\item[(c)]
Finally, formula (c) can be obtained by subtracting the formula (b) from
the formula (a).
\end{itemize}
\end{proof}

\section{Acknowledgments}
The authors gratefully acknowledge the many helpful suggestions of the
anonymous referee.

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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary 11M32;
Secondary 40B05.

\noindent \emph{Keywords: } Tornheim series, harmonic number, Riemann
zeta value.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000110},
\seqnum{A001008}, and
\seqnum{A263633}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received December 6 2021;
revised versions received  December 9 2021; April 5 2022; April 6 2022.
Published in {\it Journal of Integer Sequences}, June 20 2022.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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