Evaluate the series $ \sum_{n=1}^{\infty} \frac{1}{n(e^{2\pi n}-1)} $.

Question

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!

Answer 1

$$ \frac{1}{n(e^{2\pi n}-1)}=\sum_{k=0}^\infty \frac{e^{-2 \pi (k+1) n}}{n}$$ $$\sum_{n=1}^\infty \frac{e^{-2 \pi (k+1) n}}{n}=\log \left(1-e^{-2 \pi (k+1)}\right)$$ 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.