# Series $\sum\limits_{n=1}^\infty \frac{1}{\cosh(\pi n)}= \frac{1}{2} \left(\frac{\sqrt{\pi}}{\Gamma^2 \left( \frac{3}{4}\right)}-1\right)$

I was playing around with Mathematica and found that

$$\sum_{n=1}^\infty\frac1{\cosh(\pi n)} = \frac12\left(\frac{\sqrt{\pi}}{\Gamma \left(\tfrac34\right)^2}-1\right)$$

Does anybody know how to prove this manually?

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Here's a short sketch which I might further expand on later.

Consider the equivalent Lambert series

$$2\sum_{k=1}^\infty \frac{q^k}{1+q^{2k}}$$

with $q=\exp(-\pi)$.

This sum, through the use an appropriate geometric series, can also be expressed as

$$2\sum_{j=0}^\infty (-1)^j\frac{q^{2j+1}}{1-q^{2j+1}}$$

This sum can be expressed in terms of the Jacobi theta function $\vartheta_3(0,q)$. Using formula 57 in the previously given link, we have the expression

$$\frac{\vartheta_3(0,q)^2-1}{2}\tag{1}$$

Formula 45 in that same link gives the relation

$$\vartheta_3(0,\exp(-\pi))=\frac{\sqrt[4]\pi}{\Gamma(3/4)}\tag{2}$$

Substituting $(2)$ into $(1)$ gives the identity in the OP.

If you're interested in hyperbolic sums like these, you will want to see these three papers.

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