Ramanujan series Type $\sum _{k=1}^{\infty } \frac{\sinh (2 \pi k)}{2 \sqrt{2} \pi ^9 k^{11} (1-\cosh (2 \pi k))}$ In Ramanujan lost notebook I see series like
$$\sum _{k=1}^{\infty } \frac{343 \sinh \left(\sqrt{2} \pi  k\right)}{32 \pi ^{11} k^{13} \left(\cos \left(\sqrt{2} \pi  k\right)-\cosh \left(\sqrt{2} \pi  k\right)\right)}=\frac{721 (-1)^{3/4} \pi ^2}{277992000}-\frac{721 \sqrt[4]{-1} \pi ^2}{277992000}$$ Mathematica verifies its correctness numerically, but other series cannot be dealt with Mathematica, for example:
$$\sum _{k=1}^{\infty } \frac{\sinh (2 \pi  k)}{2 \sqrt{2} \pi ^9 k^{11} (1-\cosh (2 \pi  k))}=-\frac{1453 \pi ^2}{851350500 \sqrt{2}}$$
How can we prove this identity? Thank for helping.
 A: Both of your sums can be brought under same context. Let $\zeta \notin \mathbb{R}$, integrate $$f(z)=\frac{{\cot \pi \zeta z\cot \pi z}}{{{z^n}}}$$ around a big circle centered at origin. When $n\geq 2$, the integral around big circle $\to 0$. $f(z)$ has poles at $z=k, k\zeta^{-1}$ for $k\in \mathbb{Z}$, if $n$ is moreover odd, then
$$\tag{*}\sum_{k = 1}^\infty  {\left( {\frac{{\cot \pi \zeta k}}{{{k^n}}} + {\zeta ^{n - 1}}\frac{{\cot \pi {\zeta ^{ - 1}}k}}{{{k^n}}}} \right)}  =  - \frac{\pi }{2}{\mathop{\rm Res}\nolimits} [\frac{{\cot \pi \zeta z\cot \pi z}}{{{z^n}}},z = 0]$$
This is essentially the functional equation here.


*

*When $\zeta = i$, $n\equiv 1\pmod{4}$, LHS of $(*)$ becomes
$$-2i\sum_{k = 1}^\infty  \frac{\coth \pi k}{k^n}$$ so this sum
can be explicitly calculated. As pointed out by a comment, when $n=11$, this is your first sum.


*When $\zeta = e^{\pi i /4}$, $n\equiv 5\pmod{8}$, LHS of $(*)$
becomes $$2i\sum\limits_{k = 1}^\infty  {\frac{{\sinh \sqrt 2 \pi
   k}}{{\cos (\sqrt 2 \pi k) - \cosh (\sqrt 2 \pi k)}}\frac{1}{{{k^n}}}}
   $$ this is your first sum when $n=13$.


*When $\zeta = e^{\pi i /4}$, $n\equiv 1\pmod{8}$, LHS of $(*)$
becomes $$-2\sum\limits_{k = 1}^\infty  {\frac{{\sin \sqrt 2 \pi
   k}}{{\cos (\sqrt 2 \pi k) - \cosh (\sqrt 2 \pi k)}}\frac{1}{{{k^n}}}}
   $$ for example when $n=17$, this equals $-\frac{41 \pi
   ^{17}}{181976169375 \sqrt{2}}$.


*When $\zeta=e^{\pi i/3}$, $\cot \pi {\zeta ^{ \pm 1}}k = \frac{{ \pm
   i\sinh \sqrt 3 \pi k}}{{{{( - 1)}^k} - \cosh \sqrt 3 \pi k}}$, so LHS
of $(*)$ becomes $$i(1 - {\zeta ^{n - 1}})\sum\limits_{k = 1}^\infty 
   {\frac{{\sinh \sqrt 3 \pi k}}{{{{( - 1)}^k} - \cosh \sqrt 3 \pi
   k}}\frac{1}{{{k^n}}}} $$ For instance, when $n=11$, we have
$$\sum\limits_{k = 1}^\infty  {\frac{{\sinh \sqrt 3 \pi k}}{{{{( -
   1)}^k} - \cosh \sqrt 3 \pi k}}\frac{1}{{{k^{11}}}}}  =  -
   \frac{{7457{\pi ^{11}}}}{{1277025750\sqrt 3 }}$$


*There are also formulas of comparable simplicity when $\zeta = e^{\pi i/6}$.
