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My friend, Mr. Barreto, did post a problem on Art of Problem Solving (AOPS) about infinite sum and asked me to post the question here in order to have various approaches and answers. I am also curious about this problem as well. Here is his post:

"When trying to prove the following Ramanujan identity: $$\alpha\sum_{k=1}^\infty\frac{k}{e^{2\alpha k}-1}+\beta\sum_{k=1}^\infty\frac{k}{e^{2\beta k}-1}=\frac{\alpha+\beta}{24}-\frac{1}{4}\quad.............(1)$$ initially the method consists of using $\frac{1}{e^{x}-1}=\frac12\left(\coth\left(\frac x2\right)-1\right)$ and then (1) is : $$\frac12\sum_{k=1}^\infty k\left\{\alpha \coth(\alpha k)+\beta\coth(\beta k)\right\}-\frac{\alpha+\beta}{2}\sum_{k=1}^\infty k\quad ............(2)$$ We know that these sums are diverging, but one way to escape the obvious in this way is to use a little of the Theory of Divergent Series by the author of the identity (Ramanujan). So, for divergent series, we have: $$\sum_{n=1}^\infty f(n)=-\frac{f(0)}{2}+i\int_0^\infty\frac{f(it)-f(-it)}{e^{2\pi t}-1}\mathrm dt$$ Here is the point of doubt, I am concluding through the formula above, that (2) is: $$(\ast)\ \sum_{k=1}^\infty k=-\frac{1}{12},\qquad (\ast\ast)\ \sum_{k=1}^\infty k\left\{\alpha \coth(\alpha k)+\beta\coth(\beta k)\right\}=-1,$$ that way we get: $\frac{\alpha+\beta}{24}-\frac12$ and not $\frac{\alpha+\beta}{24}-\frac14$. I don't know if the theory used is really applicable, I just found it convincing to use the Ramanujan Summation to prove Ramanujan's Identity (1). I hope that only the algebraic part is wrong, because I found the path simple and elegant."

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  • $\begingroup$ Methods of summing divergent series may be dubious (or, more precisely, reliable only within their limit of applicability), but the series in the LHS of your "identity" (1) are most definitely convergent for $\alpha,\beta>0$, and have a positive sum. The RHS is negative for $\alpha+\beta<6$. So you (you, not Mr. Barreto!) may want to explain how that could be an identity. $\endgroup$
    – user436658
    Oct 31, 2020 at 20:18
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    $\begingroup$ This appears to be the just the logarithm of transformation formula of Dedekind Eta function. $\endgroup$
    – Paramanand Singh
    Nov 9, 2020 at 15:05
  • $\begingroup$ Is there another way to evaluate it if one might ask, sir? $\endgroup$ Nov 9, 2020 at 15:08
  • $\begingroup$ The sums in the identity are intimately connected to the theory of theta functions and elliptic integrals. Perhaps there might be some approach via contour integration but I am not even a novice in that area. I am posting a solution based on eta functions. $\endgroup$
    – Paramanand Singh
    Nov 9, 2020 at 15:33

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Let us define the Dedekind Eta function as $$\eta(q) =q^{1/12}\prod_{n\geq 1}(1-q^{2n})\tag{1}$$ It is well known that it satisfies the transformation formula $$\alpha^{1/4}\eta(e^{-\alpha}) =\beta^{1/4} \eta(e^{-\beta})\tag{2}$$ where $\alpha, \beta$ are positive with $\alpha\beta=\pi^2 $.

We can observe that $$\frac{d\beta} {d\alpha} =-\frac{\beta} {\alpha} $$ and logarithmic differentiation of equation $(2)$ gives us $$\alpha\sum_{n\geq 1}\frac{n}{e^{2n\alpha}-1}+\beta\sum_{n\geq 1}\frac {n} {e^{2n\beta}-1}=\frac{\alpha +\beta}{24}-\frac{1}{4}$$ with some amount of algebra.


The transformation formula $(2)$ is almost obvious if one writes the expression of eta function in terms of corresponding elliptic modulus and integrals. But including such details might not be possible here. There are approaches based on complex analysis which I find more complicated.

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