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This is some what related to a sum I previously posted a week or so ago.

Using complex analysis, is there a way to show:

$\displaystyle \sum_{n=1}^{\infty}\frac{\coth^{2}(\pi n)}{n^{2}}=\frac{2}{3}K+\frac{19{\pi}^{2}}{180}$, where K is the Catalan constant.

I have tried, but failed to see how the Catalan constant could be incorporated in the closed form solution.

The residue at z=0 is $\frac{-8{\pi}^{2}}{45}$.

The residue at $z=n$ is $\displaystyle \lim_{z\to n}\frac{(z-n)\pi \cos(\pi z)\coth^{2}(\pi z)}{z^{2}\sinh^{2}(\pi z)}=\frac{\coth^{2}(\pi n)}{n^{2}}$

The residue at $z=ni$ (the zeroes of $\sinh^{2}(\pi z)$) are where I hit a snag.

$\displaystyle \lim_{z\to ni}\frac{(z-ni)\pi \cot(\pi z)\cosh^{2}(\pi z)}{z^{2}\sinh^{2}(\pi z)}$

I used L'Hopital and arrived at an undefined result.

Apparently, this is more involved than the typical sum using $\pi\cot(\pi z)$ to sum a series. Especially, some how incorporating the Catalan constant.

Thanks.

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