A Ramanujan sum involving $\sinh$ Today, in a personal communication, I was asked to prove the classical result
$$\boxed{ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\sinh(\pi n)} = \frac{\pi^3}{360}}\tag{CR} $$
which I believe is due to Ramanujan. My proof can be found here and it is based on the closed form for $\sum_{n\geq 1}\frac{1}{m^2+n^2}$ and double counting. I would use this question for collecting alternative proofs. I am already aware that standard techniques for tackling series with a similar structure are


*

*The Poisson summation formula;

*The residue theorem coupled with the Laplace transform of $\text{Li}_k$ (I used this approach due to Simon Plouffe here, for instance);

*Eisenstein series related to Gaussian integers;

*Identities involving Dirichlet's series, since $$r_2(N)=\left|\{(a,b)\in\mathbb{Z}^2:a^2+b^2=N\}\right|=4 \sum_{d\mid N}\chi_4(d)$$
(this can be regarded as an analytic-combinatorial equivalent of the statement "$\mathbb{Z}[i]$ is a UFD"). 


I also know that a cornerstone is given by Zucker's The Summation of Series of Hyperbolic Functions, 1979. Let us see if we can devise a very short proof of $(\text{CR})$ through these ingredients or other ones.
 A: I realized my original proof can be shortened by exploiting symmetry.
The first three lines are unchanged:
$$\frac{1}{\sinh z}=\frac{1}{z}+\sum_{m\geq 1}\left(\frac{1}{z-m\pi i}+\frac{1}{z+m\pi i}\right)(-1)^m $$
$$\frac{1}{\sinh(\pi n)}=\frac{1}{\pi n}+\frac{1}{\pi}\sum_{m\geq 1}\left(\frac{1}{n-mi}+\frac{1}{n+mi}\right)(-1)^m $$
$$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\sinh(\pi n)}=\frac{\eta(4)}{\pi}+\frac{2}{\pi}\sum_{m\geq 1}\sum_{n\geq 1}\frac{(-1)^{n+m+1}}{n^2(n^2+m^2)}$$
then by using $2\sum_{m,n\geq 1}f(m,n) = \sum_{m,n\geq 1}f(m,n)+f(n,m)$ we immediately get
$$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\sinh(\pi n)}=\frac{\eta(4)-\eta(2)^2}{\pi}=\color{red}{\frac{\pi^3}{360}}$$
where $\eta(s)=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s}=(1-2^{1-s})\,\zeta(s)$ for any $s>1$.
A: This can be derived very simply using contour integration in the complex plane.  Consider the contour integral
$$\pi \oint_C \frac{dz}{z^3 \sinh{(\pi z)} \sin{(\pi z)}} $$
where $C$ is a square centered at the origin of side $2 N+1$.  As $N \to \infty$, one may show that the contour integral approaches zero.  (Consider the magnitude of the integrand over the sides of the square.)
This means that the sum of the residues at the poles $z=\pm n$ and $z=\pm i n$ vanishes.  Note that for $n \ne 0$, the residues at $z=n$ and $z=i n$ are equal to $(-1)^n/(\pi n^3 \sinh{(\pi n)})$. 
 Also, the summand is even in $n$.  Thus we have
$$4 \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3 \sinh{(\pi n)}} + \operatorname*{Res}_{z=0} \frac{\pi}{z^3 \sinh{(\pi z)} \sin{(\pi z)}} = 0$$
Because
$$\operatorname*{Res}_{z=0} \frac{\pi}{z^3 \sinh{(\pi z)} \sin{(\pi z)}} = \frac{\pi^3}{90} $$
the stated result follows.  (NB that last residue at $z=0$ is best done using a Laurent series expansion, as the pole is of order 5.)
