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In Ramanujan's Notebooks Volume 2 by B.C. Berndt I came across the formula $$ \sum_{n=1}^{\infty} \frac{1}{n(e^{2\pi n}-1)} = \log 2 + \frac{3}{4} \log \pi - \frac{\pi}{12} -\log \Gamma \left(\frac{1}{4} \right). $$ The proof provided uses some special values from the theory of elliptic functions, but I am unfamiliar with the subject. Does there exist a different evaluation of the series? I tried contour integration, cotangent partial fraction, and applying Poisson summation or converting to an integral but so far no success. Any help is appreciated!

Dave
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    Using the formula searching tool https://approach0.xyz/ I found this – Jean Marie Apr 25 '21 at 05:32
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    See this :https://math.stackexchange.com/q/938123/72031 – Paramanand Singh Apr 25 '21 at 05:38
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    See as well the generalization here – Jean Marie Apr 25 '21 at 05:39
  • Theta functions and elliptic integrals are the key here, but maybe some Mellin transform expert can also help here. – Paramanand Singh Apr 25 '21 at 05:40
  • Thanks for the links everyone but all of these methods use the theory of elliptic/theta functions. For example the value $\theta(e^{-\pi}) = \frac{2}{\pi} K(\frac{1}{2}) $ is used, or converting the series to a Dedekind eta function evaluation, etc. Is there any way to evaluate the sum without using elliptic/theta functions? (I am not familiar with that theory) – Dave Apr 25 '21 at 05:52

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$$ \frac{1}{n(e^{2\pi n}-1)}=\sum_{k=0}^\infty \frac{e^{-2 \pi (k+1) n}}{n}$$ $$\sum_{n=1}^\infty \frac{e^{-2 \pi (k+1) n}}{n}=\log \left(1-e^{-2 \pi (k+1)}\right)$$ So, now, you need to compute $$\sum_{k=0}^\infty \log \left(1-e^{-2 \pi (k+1)}\right)$$ and this is explained in the link provided by @Jean-Marie.

Claude Leibovici
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