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:
$$ \sum_{n=1}^\infty\frac{n^2}{(-1)^n \cosh(\pi n\sqrt{3})-1}\overset{?}{=}\frac{1}{12\pi^2} $$
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 $$ \sum_{n=1}^\infty\frac{n^2}{1-(-1)^n \cosh(\pi n\sqrt{3})}{=}\frac{1}{12\pi^2} $$
If we put $q=e^{-\pi\sqrt{3}}$ then the desired sum in question is given by $F(-q) $ where $$F(q) =2\sum_{n=1}^{\infty}\frac{n^2q^n}{(1-q^n)^2}$$ The above function is connected to well known functions $P, Q, R$ of Ramanujan : \begin{align} P(q) &=1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}\tag{1}\\ Q(q) &=1+240\sum_{n\geq 1}\frac{n^3q^n}{1-q^n}\tag{2}\\ R(q) &=1-504\sum_{n\geq 1}\frac{n^5q^n}{1-q^n}\tag{3} \end{align} We have the following relation $$q\frac{dP(q)} {dq} =-12F(q)=\frac{P^2(q)-Q(q)}{12}\tag{4}$$ The first equality above is a matter of differentiating the series for $P(q) $ but the second identity is deep and was proved by Ramanujan.
Thus the desired sum $F(-q)$ is given by $$\frac{Q(-q)-P^2(-q)}{144}$$ The calculation of $P(-q), Q(-q) $ can be done via their link with elliptic integrals $K, E$ with modulus $k$ corresponding to nome $q$. We have \begin{align} P(-q) &=\left(\frac{2K}{\pi}\right) ^2\left(\frac{6E}{K}+4k^2-5\right)\tag{5}\\ Q(-q) &=\left(\frac{2K}{\pi}\right) ^4(1-16k^2k'^2)\tag{6} \end{align} Let us also note that the modulus $k$ corresponding to $q$ is given by $$k=\frac{\sqrt{3}-1}{2\sqrt{2}}$$ This value of $k$ is obtained from the modular equation of degree $3$ $$\sqrt{kl} +\sqrt{k'l'} =1$$ by putting $l=k', l'=k$. Using the value of $k$ in $(6)$ we get that $Q(-q) =0$ and it remains to evaluate $P(-q) $. It can be proved with some effort that $P(-q) =2\sqrt{3}/\pi$ which gives the sum of series in question as $-1/12\pi^2$.
The value of $P(-q) $ is obtained via the value of $P(q^2)$. The formula for $P(q^2)$ is $$P(q^2)=\left(\frac{2K}{\pi}\right)^2\left(\frac{3E}{K}+k^2-2\right)\tag{7}$$ Using $(5),(7)$ we get $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)\tag{8}$$ Ramanujan gave a technique to evaluate the value of $P(q^2)$ using the identity $$3P(q^6)-P(q^2)=\frac{4KL}{\pi^2}(1+kl+k'l')$$ where $l, l', L$ are related to nome $q^3$. Putting $q=e^{-\pi/\sqrt{3}},k=(\sqrt{3}+1)/2\sqrt{2},l=(\sqrt {3}-1)/2\sqrt{2}$ and noting that $l=k'$, $K/L=L'/L=\sqrt{3}$ we get $$3P(e^{-2\pi\sqrt {3}})-P(e^{-2\pi/\sqrt {3}})=\frac{3\sqrt {3}}{2}\cdot\frac{4L^2}{\pi^2}\tag{9}$$ Further using logarithmic differentiation of transformation formula for Dedekind eta function one can prove that $$nP(e^{-2 \pi\sqrt {n}}) +P(e^{-2\pi/\sqrt {n}}) =\frac{6\sqrt{n}}{\pi}\tag{10}$$ Putting $n=3$ in above equation and adding it to $(9)$ we get $$P(e^{-2\pi\sqrt{3}})=\frac{\sqrt{3}}{4}\cdot\frac{4L^2}{\pi^2}+\frac{\sqrt{3}} {\pi}$$ Changing notation a bit so that $q=e^{-\pi\sqrt{3}}$ and then $L$ can be replaced by $K$ and we have $$P(q^2)=\frac{\sqrt{3}}{4}\left(\frac{2K}{\pi}\right) ^2+\frac{\sqrt{3}}{\pi}$$ Using this value in $(8) $ and $k=(\sqrt{3}-1)/2\sqrt{2}$ we get $P(-q) =2\sqrt {3}/\pi$.
Alternatively we can use the notation $q=\exp(2\pi i\tau) $ where $\tau$ has positive imaginary part and define $$E_2(\tau)=P(q), E_4(\tau)=Q(q) $$ and use the complex version of the identity $(10)$ as $$E_2(-1/\tau)=\tau^2E_2(\tau)-\frac {6i\tau}{\pi}\tag{11}$$ (the identity $(10)$ is arrived at by putting $\tau=i\sqrt{n} $). The identity for $E_4$ is $$E_4(-1/\tau)=\tau^4E_4(\tau)$$ If we put $\tau=(-1+i\sqrt{3})/2$ so that $\tau$ is the cube root of unity with positive imaginary part then we have $-1/\tau=\tau+1$ and hence $$\tau^4E_4(\tau)=E_4(-1/\tau)=E_4(\tau+1)=E_4(\tau)$$ so that $E_4(\tau)=0$ or $Q(-e^{-\pi\sqrt{3}})=0$.
Further we have $$E_2(-1/\tau)=E_2(\tau+1)=E_2(\tau)$$ and from equation $(11)$ we get $$E_2(\tau)=\frac{6i\tau}{(\tau^2-1)\pi}=\frac{2\sqrt{3}}{\pi}$$ as the value of $P(-e^{-\pi\sqrt{3}})$. This avoids the difficult but more general technique of Ramanujan described earlier.
