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.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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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}
A: In the following a general expression for
$$
\sum_{n=1}^\infty\frac{\coth(n\pi)}{n^{4K-1}}
$$
with $K\in\mathbb Z_+$ will be derived.
Consider the following function:
$$
f_k(z)=\frac{\cot z\coth z}{z^{k}}.\tag1
$$
The function has the pole of order $k+2$ at $z=0$ and simple poles at the points $z=n\pi$ and $z=i n\pi$, with $n\in\mathbb Z$, $n\ne0$.
Let us integrate the function along a square contour connecting the following points of the complex plane:
$$\gamma_\nu:\;
(-\nu,-\nu)\to(\nu,-\nu)\to(\nu,\nu)\to(-\nu,\nu)\to   
$$
with $\nu\in\mathbb R,\nu\not\in\mathbb Z,\nu>0$.
By Cauchy's theorem the exact choice of $\nu\in(n\pi,n\pi+\pi)$ does not matter for the value of the integral. To simplify its estimate for large $n$ 
let $\nu=\left(n+\frac14\right)\pi$. In this case:
$$
|\cot(x\pm i\nu)\coth(x\pm i\nu)|^2=\frac{1+\cot^2x\coth^2\nu}{\cot^2x+\coth^2\nu}\cdot
\underbrace{\frac{1+\coth^2x\cot^2\nu}{\coth^2x+\cot^2\nu}}_{=1}
=\frac{\tan^2x+\coth^2\nu}{1+\tan^2x\coth^2\nu}\le\coth^2\nu.
$$
Due to the symmetry the same estimate is valid for $|\cot(i x\pm \nu)\coth(i x\pm \nu)|^2$. Therefore:
$$
\lim_{n\to\infty}\oint_{\gamma_{\nu}}f_k(z)dz=0.
$$
Thus by the residue theorem we have:
$$
\operatorname{Res}(f_k,0)+\sum_{n=1}^\infty\left[\operatorname{Res}(f_k,n\pi)+\operatorname{Res}(f_k,-n\pi)+\operatorname{Res}(f_k,in\pi)+\operatorname{Res}(f_k,-in\pi)\right]=0\tag2
$$
The pole at $z=0$ can be calculated as follows. Recall that:
$$
\cot z=\frac1z\sum_{n=0}^\infty\frac{(-1)^nB_{2n}}{(2n)!}(2z)^{2n};\quad
\coth z=\frac1z\sum_{n=0}^\infty\frac{B_{2n}}{(2n)!}(2z)^{2n},
$$
where $B_n$ are Bernoulli numbers.
From this one obtains:
$$
\cot z\coth z=\frac1{z^2}\sum_{n=0}^\infty\frac{A_{n}}{n!}(2z)^{n},\text{ with } 
A_{2n+1}=0,\;A_{2n}=\sum_{k=0}^n(-1)^k\binom{2n}{2k}B_{2k}B_{2n-2k}.\tag3
$$
It follows
$$
\operatorname{Res}(f_k,0)=\frac{2^{k+1}A_{k+1}}{(k+1)!}\equiv A^*_{k+1}.\tag4
$$
Next we compute the residues at the simple poles:
$$
\operatorname{Res}(f_k,n\pi)=\lim_{z\to n\pi}(z-n\pi)\frac{\cot z\coth z}{z^{k}}=\lim_{\zeta\to0}\zeta\frac{\cot (\zeta)\coth(\zeta+n\pi)}{(\zeta+n\pi)^{k}}=\frac{\coth(n\pi)}{(n\pi)^{k}},\tag{5a}
$$
and
$$
\operatorname{Res}(f_k,i n\pi)=\lim_{z\to in\pi}(z-in\pi)\frac{\cot z\coth z}{z^{k}}=\lim_{\zeta\to0}\zeta\frac{\cot (\zeta+in\pi)\coth(\zeta)}{(\zeta+i n\pi)^{k}}=\frac{\coth(n\pi)}{i(in\pi)^{k}},\tag{5b}
$$
where we used 
$$\begin{align}
&\lim_{x\to0}x\cot x=\lim_{x\to0}x\coth x=1,\\ 
&\cot(x+n\pi)=\cot(x),\\ 
&\coth(x+in\pi)=\coth(x),\\ 
&\cot(i x)=-i\coth(x).
\end{align}
$$
From $(5)$ one obtains:
$$
\operatorname{Res}(f_k,n\pi)+\operatorname{Res}(f_k,-n\pi)+\operatorname{Res}(f_k,in\pi)+\operatorname{Res}(f_k,-in\pi)
=\begin{cases}
4\dfrac{\coth(n\pi)}{(n\pi)^k},&k=-1\text{ mod }4,\\
0,&k\ne -1\text{ mod }4.\\
\end{cases}\tag6
$$
Combining $(2)$, $(4)$ and $(6)$ one draws two conclusions:
$$A_k=0\text{ for } k\ne 0\text{ mod }4.\tag7$$
$$4\sum_{n=1}^\infty\dfrac{\coth(n\pi)}{(n\pi)^{4K-1}}=-A^*_{4K}.\tag8$$
Observe that (7) is non-trivial only for $k=2\text{ mod }4$. No closed-form expression is known for $A_{4K}$ but its calculation using $(3)$ represents no difficulty. The first four values of $A^*_{4K}$ are:
$$
A^*_0=1,\quad A^*_4=-\frac7{45},\quad A^*_8=-\frac{19}{14175},\quad A^*_{12}=-\frac{2906}{212837625}.
$$
Particularly, for $K=2$ one obtains:
$$\sum_{n=1}^\infty\dfrac{\coth(n\pi)}{n^{7}}=-\dfrac{A^*_{8}\pi^7}4
=\dfrac{19\pi^7}{56700}.$$

Using for $n\ne0$ the relation:
$$
B_{2n}=\frac{(-1)^{n+1}(2n)!}{(2\pi)^{2n}}2\zeta(2n),
$$
where $\zeta(z)$ is the Riemann zeta-function, the equation $(8)$ can be rewritten as:
$$
\pi\sum_{n=1}^\infty\dfrac{\coth(n\pi)}{n^{4K-1}}=
\zeta(4K)-\sum_{k=1}^{2K-1}(-1)^k\zeta(2k)\zeta(4K-2k),
$$
in agreement with previous results for $K=2$.
A: Since $(7)$ from this answer is valid for any $z\in\mathbb{C}$, we have
$$
\begin{align}
\pi\coth(\pi n)
&=\sum_{k\in\mathbb{Z}}\frac1{n+ik}\\
&=\frac1n+2n\sum_{k=1}^\infty\frac1{n^2+k^2}\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{n=1}^\infty\frac{\pi\coth(\pi n)}{n^7}
&=\sum_{n=1}^\infty\frac1{n^8}+2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n^6(n^2+k^2)}\\
&=\zeta(8)+2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2}\left(\frac1{n^6}-\frac1{n^4(n^2+k^2)}\right)\\
&=\zeta(8)+2\zeta(2)\zeta(6)-2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2}\frac1{n^4(n^2+k^2)}\tag{2}\\
&=\zeta(8)+2\zeta(2)\zeta(6)-2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^4}\left(\frac1{n^4}-\frac1{n^2(n^2+k^2)}\right)\\
&=\zeta(8)+2\zeta(2)\zeta(6)-2\zeta(4)\zeta(4)+2\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^4n^2(n^2+k^2)}\tag{3}\\[6pt]
&=\zeta(8)+2\zeta(2)\zeta(6)-\zeta(4)\zeta(4)\tag{4}\\[12pt]
&=\frac{19\pi^8}{56700}\tag{5}
\end{align}
$$
where $(4)$ is the average of $(2)$ and $(3)$. Also. we've used the values of $\zeta(2k)$ computed in this answer. Thus,
$$
\sum_{n=1}^\infty\frac{\coth(\pi n)}{n^7}=\frac{19\pi^7}{56700}\tag{6}
$$
