Ramanujan-Sums... How do they do?

Question

Prove that:

$\displaystyle \frac{1}{e^{2\pi}-1}+\frac{2}{e^{4\pi}-1}+\frac{3}{e^{6\pi}-1}+...=\frac{1}{24}-\frac{1}{8\pi}$

I would like to see different ways of solving this beautiful sum, whoever is encouraged? :)

Thanks to all.

Answer 1

In your case, consider the function

$$ \frac{x}{e^{2\pi x}-1} $$

which has the Mellin transform

$$ \left( 2\,\pi \right) ^{-s-1}\Gamma \left( s+1 \right) \zeta \left( s+1 \right). $$

Then just follow the technique used in this similar problem.

Answer 2

This problem is already solved in Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$ and has a nice solution there.

But then the simplest proof comes from Ramanujan. He uses the transformation formula for "eta" function and logarithmic differentiation. If we define $\eta(q)$ by $$\eta(q) = q^{1/12}(1 - q^{2})(1 - q^{4})(1 - q^{6})\cdots$$ then we have $$\frac{\eta(e^{-\pi\sqrt{n}})}{\eta(e^{-\pi/\sqrt{n}})} = n^{-1/4}$$ or in other words $$n^{1/4}e^{-\pi\sqrt{n}/12}(1 - e^{-2\pi\sqrt{n}})(1 - e^{-4\pi\sqrt{n}})\cdots = e^{-\pi/(12\sqrt{n})}(1 - e^{-2\pi/\sqrt{n}})(1 - e^{-4\pi/\sqrt{n}})\cdots$$

Then if we differentiate the above equation logarithmically with respect to $n$ we get $$n\left\{1 - 24\left(\frac{1}{e^{2\pi\sqrt{n}} - 1} + \frac{2}{e^{4\pi\sqrt{n}} - 1}+ \cdots\right)\right\} + \left\{1 - 24\left(\frac{1}{e^{2\pi/\sqrt{n}} - 1} + \frac{2}{e^{4\pi/\sqrt{n}} - 1}+ \cdots\right)\right\} = \frac{6\sqrt{n}}{\pi}$$

Putting $n = 1$ in above equation we get $$1 - 24\left(\frac{1}{e^{2\pi} - 1} + \frac{2}{e^{4\pi} - 1} + \frac{3}{e^{6\pi} - 1}\right) = \frac{3}{\pi}$$ and thus the proof is completed. The above derivation is coming straight from Ramanujan's most famous and amazing paper "Modular Equations and Approximations to $\pi$".