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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.

Dave
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    Ramanujan used various techniques to sum various series. The easiest results were based on algebraic manipulation. The more difficult ones (the first formula here) were based on his theory of elliptic and theta functions. Other techniques were based on partial fractions for analytic functions and his own equivalent of mellin transform. The true surprise is that his methods were inherent based on calculus and algebraic manipulation and avoided complex analysis completely. – Paramanand Singh May 04 '20 at 16:35
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    You should get hold of Collected Papers of Ramanujan to get an insight into his methods. It contains some of very ingenious and economic proofs of Ramanujan. Unfortunately he does not provide computational details needed for proofs of his most difficult results. As if the papers were written for someone who had almost magical powers of algebraic manipulation like Ramanujan. – Paramanand Singh May 04 '20 at 16:40

1 Answers1

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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.

user
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