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

Question

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?

Answer 1

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.

Answer 2

This is by no means obvious. First, see the MathWorld page on Hyperbolic Secant.

There you will see that this identity is due to Ramanujan.

Namely, it follows from the Ramanujan cos/cosh identity.

To prove this manually, then, you may wish to read up on how this latter identity is established.

Answer 3

it possible to gives $$\sum _{k=1}^{\infty } \frac{1}{2 \cosh \left(\pi ^2 k\right)}=\frac{i \psi _e^{(0)}\left(1+\frac{i \pi }{2}\right)}{2 \pi }-\frac{i \psi _e^{(0)}\left(1-\frac{i \pi }{2}\right)}{2 \pi }+\frac{1}{4}+\frac{1}{4 \pi }$$