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.
Best Answer
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, $$ \sum_{n=1}^\infty\frac{\coth(\pi n)}{n^7}=\frac{19\pi^7}{56700}\tag{6} $$