Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

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


Formula 45 in that same link gives the relation


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.

share|cite|improve this answer

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.

share|cite|improve this answer
Thanks B.D! I had heard about Ramanujan cos/cosh identity but did not know it could be used to prove this series. – Shobhit Apr 8 '13 at 11:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.