Tricky Sum 1.0

Firstly, note that $\sin^2(\omega\pi n) \;= \; \left\{ \begin{array}{lll} -\sinh^2\left(\tfrac12\sqrt{3}\pi n\right) &\hspace{1cm} & n\; \mathrm{even} \\[1ex] \cosh^2\left(\tfrac12\sqrt{3}\pi n\right) & & n\; \mathrm{odd} \end{array}\right.$ and hence $\sum_{n=1}^\infty \csc^2(\omega \pi n) \; = \; \sum_{n=1}^\infty \mathrm{sech}^2\big(\sqrt{3}\pi (n-\tfrac12)\big) - \sum_{n=1}^\infty \mathrm{cosech}^2\big(\sqrt{3}\pi n\big) \; = \; S_1 - S_2$ To evaluate $S_1$ and $S_2$ , we need to handle so-called singular values of the complete elliptic integral of the first kind $\mathsf{K}(k) \; = \; \int_0^{\frac12\pi}\frac{d\theta}{\sqrt{1 - k^2\sin^2\theta}} \hspace{2cm} 0 < k < 1$ and its complementary function $\mathsf{K}'(k) \; = \; \mathsf{K}\big(\sqrt{1-k^2}\big)$ A singular value $k_r$ of the elliptic integral is one for which $\frac{\mathsf{K}'(k_r)}{\mathsf{K}(k_r)} \; = \; \sqrt{r} \hspace{2cm} r \in \mathbb{N}$ We are interested in $k_3 =\tfrac1{2\sqrt{2}}(\sqrt{3}-1)$ , and we have an expression for the complete elliptic integral of the second kind $\mathsf{E}(k_3) \; = \; \frac{\pi}{4\sqrt{3}\mathsf{K}(k_3)} + \frac{\sqrt{3}+1}{2\sqrt{3}}\mathsf{K}(k_3)$ given the value of the elliptic alpha function $\alpha(3) = \tfrac12(\sqrt{3}-1)$ . In fact it is possible to express $\mathsf{K}(k_3)$ exactly in terms of Gamma functions, $\mathsf{K}(k_3) \; = \; \frac{3^{\frac14}\Gamma^3(\frac13)}{2^{\frac73}\pi}$ but this is not needed here.

Using the results of this paper and the above observations (note that this paper writes elliptic integral expressions in term of $m = k^2$ instead of $k$ ), we obtain $\begin{aligned} S_1 & = \; \frac{2\mathsf{K}(k_3)\mathsf{E}(k_3)}{\pi^2} - 2(1 - k_3^2)\frac{\mathsf{K}(k_3)^2}{\pi^2} \; = \; \frac{1}{2\pi\sqrt{3}} - \frac{1}{2\sqrt{3}}\,\frac{\mathsf{K}(k_3)^2}{\pi^2} \\ S_2 & = \; \tfrac16+ \tfrac23(2 - k_3^2)\frac{\mathsf{K}(k_3)^2}{\pi^2} - \frac{2\mathsf{K}(k_3)\mathsf{E}(k_3)}{\pi^2} \; = \; \frac16 - \frac{1}{2\pi\sqrt{3}} - \frac{1}{2\sqrt{3}}\,\frac{\mathsf{K}(k_3)^2}{\pi^2} \end{aligned}$ so that $\sum_{n=1}^\infty \csc^2(\omega \pi n) \; = \; S_1 - S_2 \; = \; \frac{1}{\pi\sqrt{3}} - \frac16$ Thus $A = \frac{1}{\sqrt{3}}$ and $B = -\tfrac16$ , so that $\tfrac{A^2}{B^2} =\boxed{12}$ .