Summing series containing $e^{an} \pm 1$ term in the denominator I was looking over Ramanujan's first letter to Hardy and came across several series of a similar form:
$$ \sum_{n=1}^{\infty} \frac{n^{13}}{e^{2\pi n}-1} = \frac{1}{24} $$
$$ \sum_{n=1}^{\infty} \frac{\coth(n\pi)}{n^7} = \frac{19 \pi^7}{56700} $$
$$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5 \cosh \left(\frac{(2n+1)\pi}{2} \right)} = \frac{\pi^5}{768} $$
Does anyone know what method(s) Ramanujan used or is likely to have used to compute these, or where I can find this information? More generally I am wondering what are some techniques of summing series in which the denominator of the general term contains an exponential $\pm 1$, with perhaps another exponential or rational function in the numerator? Series of the form
$$ \sum_{n=1}^{\infty} \frac{p(n)}{q(n)} \frac{1}{e^{an} \pm 1} \hspace{0.5cm} \text{or} \hspace{0.5cm} \sum_{n=1}^{\infty} \frac{p(n)}{q(n)} \frac{1}{\cosh(an)} \hspace{0.5cm} \text{or} \hspace{0.5cm} \sum_{n=1}^{\infty} \frac{p(n)}{q(n)} \coth(an) $$
where $p$ and $q$ are polynomials and $a>0$ is a constant.
 A: The first identity is (from my point of view) a really hard one. It was already many times discussed on MSE. Particular interest for you can present this discussion, where also a reference concerning the method  probably used by Ramanujan was given.
The others two sums can be evaluated by elementary means of residue calculus. The general method is described here. The same method can be applied whenever a polynomial is present in the denominator of the expression. Let us demonstrate this on the example of the series
$$
\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^{4K+1} \cosh \left(\frac{(2n+1)\pi}{2} \right)}
$$
For this we consider the integral:
$$
\oint_{\Gamma_\nu}\frac{dz}{z^k\cos z\cosh z}
$$
where $\Gamma_\nu$ is a square contour 
$$
(-\nu,-\nu)\to(\nu,-\nu)\to(\nu,\nu)\to(-\nu,\nu)\to(-\nu,-\nu)
$$
avoiding the poles of the denominator situated at $(n+\frac12)\pi$ and $(n+\frac12)\pi i$ ($n\in\mathbb Z$). Exactly the same reasoning as in the cited answer will result in the identity:
$$
4\sum_{n=0}^{\infty} \frac{(-1)^n}{\left(\frac{(2n+1)\pi}{2} \right)^{4K+1} \cosh \left(\frac{(2n+1)\pi}{2} \right)}=A_{4K}
$$
with $A_k$ defined by the series expansion
$$
\frac1{\cos z\cosh z}=\sum_{k=0}^\infty A_kz^k.
$$
As, particularly, $A_4=\frac16$ one obtains for $K=1$ the value of the sum cited in your question.
