TITLE: Double series convergent to $2\zeta(4)$?
QUESTION [15 upvotes]: Using a computer I found the double sum
$$S(n)= \sum_{j=1}^n\sum_{k=1}^n \frac{j^2 + jk + k^2}{j^2(j+k)^2k^2}$$
has values
$$S(10) \quad\quad= 1.881427206538142 \\ S(1000) \quad= 2.161366028875634 \\S(100000) =  2.164613524212465\\$$
As a guess I compared with fractions $\pi^p/q$ where $p,q$ are positive integers and it appears 
$$\lim_{n \to \infty} S(n) = \frac{\pi^4}{45} =  2\zeta(4) \approx 2.164646467422276 $$
I'd be interested in seeing a proof if true.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[#ffd,10px]{\ds{\sum_{j = 1}^{\infty}\sum_{k = 1}^{\infty}
{j^{2} + jk + k^{2} \over j^{2}\pars{j + k}^{2}k^{2}}}} =
\sum_{j = 1}^{\infty}\sum_{k = 1}^{\infty}
{j^{2} + jk + k^{2} \over j^{2}k^{2}}\
\overbrace{\bracks{-\int_{0}^{1}\ln\pars{x}x^{j + k - 1}\,\dd x}}
^{\ds{1 \over \pars{j + k}^{2}}}
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{x}
\pars{\sum_{j = 1}^{\infty}x^{j}
\sum_{k = 1}^{\infty}{x^{k} \over k^{2}} +
\sum_{j = 1}^{\infty}{x^{j} \over j}
\sum_{k = 1}^{\infty}{x^{k} \over k} +
\sum_{j = 1}^{\infty}{x^{j} \over j^{2}}
\sum_{k = 1}^{\infty}x^{k}}\,{\dd x \over x}
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{x}
\bracks{2\sum_{j = 1}^{\infty}x^{j}
\sum_{k = 1}^{\infty}{x^{k} \over k^{2}} +
\pars{\sum_{j = 1}^{\infty}{x^{j} \over j}}^{2}}\,{\dd x \over x}
\label{1}\tag{1}
\end{align}

Note that $\ds{\sum_{\ell = 1}^{\infty}{x^{\ell} \over \ell^{s}} =
\,\mrm{Li}_{s}\pars{x}}$ where $\ds{\mrm{Li}_{s}}$ is the Polylogarithm Function. Moreover,
  $\ds{\mrm{Li}_{1}\pars{x} = -\ln\pars{1- x}}$,
  $\ds{\mrm{Li}_{s + 1}\pars{z} = \int_{0}^{z}{\mrm{Li}_{s}\pars{t} \over t}\,\dd t}$ and
  $\ds{\sum_{j = 1}^{\infty}x^{j} = {x \over 1 - x}}$.

\eqref{1} becomes
\begin{align}
&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\bbox[#ffd,10px]{\ds{\sum_{j = 1}^{\infty}\sum_{k = 1}^{\infty}
{j^{2} + jk + k^{2} \over j^{2}\pars{j + k}^{2}k^{2}}}} =
-2\int_{0}^{1}{\ln\pars{x}\,\mrm{Li}_{2}\pars{x} \over 1 - x}\,\dd x -
\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x
\\[5mm] = &\
-2\int_{0}^{1}\overbrace{\ln\pars{1 - x} \over x}^{\ds{-\mrm{Li}_{2}'\pars{x}}}\,\
\mrm{Li}_{2}\pars{1 - x} \,\dd x -
\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x
\label{2}\tag{2}
\end{align}
where we set $\ds{x \mapsto 1 - x}$ in the first integral. With Euler Reflection Formula
$\ds{\mrm{Li}_{2}\pars{1 - x} = -\mrm{Li}_{2}\pars{x} + {\pi^{2} \over 6} -\ln\pars{x}\ln\pars{1 - x}}$. \eqref{2} becomes:
\begin{align}
&\bbox[#ffd,10px]{\ds{\sum_{j = 1}^{\infty}\sum_{k = 1}^{\infty}
{j^{2} + jk + k^{2} \over j^{2}\pars{j + k}^{2}k^{2}}}}
\\ = &\ -\
\overbrace{\int_{0}^{1}\totald{\mrm{Li}_{2}^{2}\pars{x}}{x} \,\dd x}
^{\ds{\pi^{4} \over 36}}\ +\
{\pi^{2} \over 3}\
\overbrace{\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\,\dd x}
^{\ds{\pi^{2} \over 6}}\ +\
\overbrace{\int_{0}^{1}{\ln\pars{x}\ln^{2}\pars{1 - x} \over x}\,\dd x}^{\ds{-\,{\pi^{4} \over 180}}}
\\[5mm] = &\
\bbx{\large{\pi^{4} \over 45}}
\end{align}