What I do so far
\begin{align*}
\text{Show that} \quad &\sum_{n=1}^{+\infty}\frac{1}{(n\cdot\sinh(n\pi))^2} = \frac{2}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{(2n-1)^2} – \frac{11\pi^2}{180} \\
\text{Lemma 1 } &\sum_{n = – \infty }^\infty \frac{1}{{z + n}} = \frac{\pi }{{\tan (\pi z)}} \\
\text{Lemma 2 } &\frac{1}{{\sinh^2(\pi z)}} = \frac{1}{{\pi^2 z^2}} + \frac{4z^2}{{\pi^2}}\sum_{k=1}^\infty \frac{1}{{(z^2 + k^2)^2}} – \frac{2}{{\pi^2}}\sum_{k=1}^\infty \frac{1}{{z^2 + k^2}} \\
&\text{Because:} \nonumber \\
&\frac{\pi }{{\tan (\pi z)}} = \sum_{k = – \infty }^\infty \frac{1}{{z + k}} \Rightarrow \left( \frac{\pi }{{\tan (\pi z)}} \right)' = -\sum_{k = – \infty }^\infty \frac{1}{{(z + k)^2}} \nonumber \\
&\Rightarrow \boxed{\frac{\pi^2}{\sin^2(\pi z)}} = \sum_{k = – \infty }^\infty \frac{1}{{(z + k)^2}} \Rightarrow \nonumber \\
&\Rightarrow \frac{\pi^2}{\sin^2(\pi iz)} = \sum_{k = – \infty }^\infty \frac{1}{{(iz + k)^2}} \Rightarrow \frac{\pi^2}{\sinh^2(\pi z)} = \sum_{k = – \infty }^\infty \frac{1}{{(iz + k)^2}} \Rightarrow \boxed{\frac{\pi^2}{\sinh^2(\pi z)} = \sum_{k = – \infty }^\infty \frac{1}{{(z + ik)^2}} = \sum_{k = – \infty }^\infty \frac{1}{{(z – ik)^2}}} \nonumber \\
&\text{then} \nonumber \\
&\frac{1}{{\sinh^2(\pi z)}} = \frac{1}{{2\pi^2}}\sum_{k = – \infty }^\infty \left( \frac{1}{{(z + ik)^2}} + \frac{1}{{(z – ik)^2}} \right) \nonumber \\
&= \frac{1}{{\pi^2}}\sum_{k = – \infty }^\infty \frac{z^2 – k^2}{{(z^2 + k^2)^2}} = \frac{1}{{\pi^2}}\sum_{k = – \infty }^\infty \frac{2z^2 – (z^2 + k^2)}{{(z^2 + k^2)^2}} \nonumber \\
&= \frac{1}{{\pi^2}} \left( 2z^2\sum_{k = – \infty }^\infty \frac{1}{{(z^2 + k^2)^2}} – \frac{1}{{\pi^2}}\sum_{k = – \infty }^\infty \frac{1}{{z^2 + k^2}} \right) \nonumber \\
&\Rightarrow \boxed{\frac{1}{{\sinh^2(\pi z)}} = \frac{1}{{\pi^2 z^2}} + \frac{4z^2}{{\pi^2}}\sum_{k=1}^\infty \frac{1}{{(z^2 + k^2)^2}} – \frac{2}{{\pi^2}}\sum_{k=1}^\infty \frac{1}{{z^2 + k^2}}}
\end{align*}
We replace $z$ with $n$ and sum, so it results:
\begin{align*}
&\sum_{n=1}^\infty \frac{1}{n^2\sinh^2(\pi n)} \\
&= \frac{1}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^4} + \frac{4}{\pi^2}\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(n^2+k^2)^2} – \frac{2}{\pi^2}\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^2(n^2+k^2)}
\end{align*}
But it is known that
$$
\frac{1}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^4} = \frac{1}{\pi^2}\cdot \zeta(4) = \frac{\pi^2}{90}
$$
for the sum
$$
\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^2(n^2+k^2)}
$$
we have:
\begin{align*}
S &= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n^2(n^2+k^2)} \\
&= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k^2}\left(\frac{1}{n^2} – \frac{1}{n^2+k^2}\right) \\
&= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k^2}\frac{1}{n^2} – \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k^2(n^2+k^2)} \\
&\Rightarrow S = \left(\sum_{n=1}^\infty \frac{1}{n^2}\right)\left(\sum_{k=1}^\infty \frac{1}{k^2}\right) – S \\
&\Rightarrow 2S = \left(\frac{\pi^2}{6}\right)^2 \\
&\Rightarrow S = \frac{\pi^4}{72}
\end{align*}
So finally
$$
\boxed{\sum_{n=1}^\infty \frac{1}{n^2\sinh^2(\pi n)} = \frac{4}{\pi^2}\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(n^2+k^2)^2} – \frac{\pi^2}{60}}
$$
The final double sum is related to functions $\zeta(2)$ & $\zeta(4)$ that are missing all add-ins of natural ones that are not written as sum of two squares. From Fermat's theorem (proved by Euler) we know that if some natural number has a prime factor of form $p = 4k + 3$ raised to odd force, then it is not written as the sum of two squares.
I read that
$$
\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(n^2+k^2)^2} = \zeta(2) \cdot \sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} – \zeta(4)
$$
somewhere, but I have no proof of this, nor have I been able to discover any.
Given the above:
\begin{align*}
&\sum_{n=1}^\infty \frac{1}{n^2\sinh^2(\pi n)} \\
&= \frac{4}{\pi^2}\left(\zeta(2) \cdot \sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} – \zeta(4)\right) – \frac{\pi^2}{60} \\
&= \frac{4}{\pi^2}\left(\frac{\pi^2}{6} \cdot \sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} – \frac{\pi^4}{90}\right) – \frac{\pi^2}{60} \\
&= \frac{2}{3}\sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} – \left(\frac{2\pi^2}{45} + \frac{\pi^2}{60}\right) \\
&= \frac{2}{3}\sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} – \frac{11\pi^2}{180}
\end{align*}
I would be very interested to see a proof of the relationship
$$
\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(n^2+k^2)^2} = \zeta(2) \cdot \sum_{n=1}^\infty (-1)^{n-1}\frac{1}{(2n-1)^2} – \zeta(4)
$$
which is true (tested with handmade program in Visual Basic).
Best Answer
First we have, for $s>1$ and $\beta(s)$ the Dirichlet beta function, $$\DeclareMathOperator{\csch}{csch} \sum_{(x,y)\in \mathbb{Z}^2, (x,y)\neq (0,0)} \frac{1}{(x^2+y^2)^s} = 4\zeta(s)\beta(s)$$ this can be proved in many ways, the most elegant way is probably to realize LHS as $4\zeta_{\mathbb{Q}(\sqrt{-1})}(s)$ as Dedekind zeta function of a number field (which is a PID and $4$ = number of units), $\zeta_{\mathbb{Q}(\sqrt{-1})}(s)$ factors into Dirichlet L-function because the field is abelian.
Now let $s=2$ and using $$\sum_{x\in \mathbb{Z}} \frac{1}{(x^2+y^2)^2} = \frac{\pi \coth (\pi y)+\pi ^2 y \text{csch}^2(\pi y)}{2 y^3}$$ we have (with $G = \beta(2)$), $$\begin{aligned} 4\zeta(2)G &= 2\sum_{y\geq 1, x\in \mathbb{Z}} \frac{1}{(x^2+y^2)^2} + 2\zeta(4) \\ &= 2\sum_{y\geq 1} \frac{\pi \coth (\pi y)+\pi ^2 y \text{csch}^2(\pi y)}{2 y^3} + 2\zeta(4) \end{aligned}$$ the desired evaluation of $\sum_{n\geq 1} \frac{\csch^2(n\pi)}{n^2}$ follows immediately from $\sum_{n=1}^\infty \frac{\coth(n\pi)}{n^3} = \frac{7\pi^3}{180}$