Method of proof of $\sum\limits_{n=1}^{\infty}\tfrac{\coth n\pi}{n^7}=\tfrac{19}{56700}\pi^7$ The following formula was stated by Ramanujan:
$$\sum\limits_{n=1}^{\infty}\frac{\coth n\pi}{n^7}=\frac{19\pi^7}{56700}$$
Does anybody know the method of proof of this formula? I know that typically Ramanujan used extensively methods of divergent series, but I cannot see how to attempt a proof of this result. It looks somehow like a relatively simple result, but I can't see what methods might be used to obtain it.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

There is a straightforward way to evaluate the
  $\texttt{initial (k,n)-sum}$ which is the starting point of 
  @robjohn fine answer:

\begin{align}
\sum_{n = 1}^{\infty}{\pi\coth\pars{\pi n} \over n^{7}} & =
\sum_{n = 1}^{\infty}{1 \over n^{8}} +
2\sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}{1 \over n^{6}\pars{n^{2} + k^{2}}}
\\[5mm] & =
\zeta\pars{8} +
\sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}\bracks{%
{1 \over n^{6}\pars{n^{2} + k^{2}}} + {1 \over k^{6}\pars{k^{2} + n^{2}}}}
\\[5mm] & =
\zeta\pars{8} +
\sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}
{k^{6} + n^{6} \over k^{6}n^{6}\pars{n^{2} + k^{2}}} =
\zeta\pars{8} +
\sum_{n = 1}^{\infty}\sum_{k = 1}^{\infty}
{k^{4} - k^{2}n^{2} + n^{4} \over k^{6}n^{6}}
\\[5mm] & =
\zeta\pars{8} +
\sum_{n = 1}^{\infty}{1 \over n^{6}}\sum_{k = 1}^{\infty}{1 \over k^{2}} -
\sum_{n = 1}^{\infty}{1 \over n^{4}}\sum_{k = 1}^{\infty}{1 \over k^{4}} +
\sum_{n = 1}^{\infty}{1 \over n^{2}}\sum_{k = 1}^{\infty}{1 \over k^{6}}
\\[5mm] & =
\zeta\pars{8} + 2\zeta\pars{6}\zeta\pars{2} - \zeta^{2}\pars{4}
\end{align}
