Conjecture $\sum_{n=1}^\infty\frac{n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}=\frac{1}{12\pi^2}$ Wolfram Alpha numerical calculation  shows that the quantity
$$
\sum_{n=1}^\infty\frac{12\pi^2n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}
$$
is 1 with high accuracy. Can anybody prove the resulting conjecture:

$$
\sum_{n=1}^\infty\frac{n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}\overset{?}{=}\frac{1}{12\pi^2}
$$

 A: Take the function
$$
f(z) = \frac{z^2}{\sin (\pi  z) \left(\cosh \left(\sqrt{3} \pi  z\right)-\cos (\pi  z) \right)}.
$$
Observe that $\sin (\pi  z) \left(\cosh \left(\sqrt{3} \pi  z\right)-\cos (\pi  z) \right)=2\sin (\pi  z) \sin (\pi  z \omega) \sin (\pi  z\omega^2) $, where $\omega=e^{2\pi i/3}$. This means that $f(z)dz=\frac{z^3}{\sin (\pi  z) \left(\cosh \left(\sqrt{3} \pi  z\right)-\cos (\pi  z) \right)}\frac{dz}{z}$ is symmetric under $z\to z\omega$.
The rest is a routine procedure in application of residue theorem. Consider a closed contour $C$ composed of 2 rays, $z=x$, and $z=x\omega$ $(x>0)$ closed by a circle of large radius $R$. The poles inside the contour are $z=ne^{\pi i/3}$. Note that the integral along the circle vanishes in the limit $R\to\infty$, and the integrals along straight lines cancel each other out. Now one carefully calculates the contribution of residues at $z=n$ and $z=n\omega$ and the contribution of the residue at $z=0$:
$$
\lim_{R\to\infty}\int_Cf(z)dz=-\pi i~\sum_{n=1}^\infty\underset{z=n}{\text{res}}f(z)\cdot 2-\frac{2\pi i}{3}\underset{z=0}{\text{res}}f(z)=2\pi i~\sum_{n=1}^\infty\underset{z=ne^{\pi i/3}}{\text{res}}f(z).
$$
As a result 
$$
2\sum_{n=1}^\infty\frac{n^2}{\pi(-1)^n \left(\cosh(\pi n\sqrt{3})-(-1)^n\right)}=-\frac{1}{3}\frac{1}{2\pi^3}
$$
Finally
$$
\sum_{n=1}^\infty\frac{n^2}{1-(-1)^n \cosh(\pi n\sqrt{3})}{=}\frac{1}{12\pi^2}
$$
