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Prove that $$\sum_{k=1}^{\infty} \frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$$

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$\large\frac{2\pi k}{\text{e}^{2\pi k}-1}=k\pi (\coth(k\pi)-1)$ –  B. S. Jan 8 '13 at 17:09
    
I can transform the sum to $\sum_{m=1}^{\infty} (-1)^{m+1} \mathrm{csch}^2{(\pi m)}$. Unfortunately, I have to get back to work. I will come back if nobody has solved by then. –  Ron Gordon Jan 8 '13 at 17:20
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@Chris'ssister Have you tried using the Laplace transform as a tool to find the limit of this series using this paper? See mathdl.maa.org/images/cms_upload/A_Laplace_Transform18380.pdf. –  Elias Jan 8 '13 at 18:25
    
@Elias: Thank you for that link I find it interesting. I'm going through it right now. –  Chris's sis Jan 8 '13 at 18:29
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Interestingly, if one blindly applies the Plana summation formula$$ \sum_{k=1}^\infty f(k) = -\frac{1}{2} f(0) + \int_0^\infty f(t) \mathrm{d}t + \int_0^\infty i \frac{f(i t)-f(-i t)}{\\exp(2 \pi t)-1} \mathrm{d}t$$ although it is not applicable because $f(t)$ is not bounded in the right half complex plane, one gets $\frac{1}{12} - \frac{1}{4 \pi}$ which twice the exact answer. This hints that the sum may be amendable to complex analysis techniques. Here $f(k)$ refers to the summand. –  user40314 Feb 9 '13 at 13:49
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1 Answer

up vote 16 down vote accepted

Rewrite $$\frac{1}{e^{2\pi k} -1} = \sum_{n=1}^\infty e^{-2\pi k n}.$$

So we need to evaluate $$\sum_{n,k=1}^\infty k e^{-2\pi k n}.$$

Summing first over $k$, we have $$ \sum_{k=1}^\infty k e^{-2\pi k n} = \frac1{2\pi}\frac{\partial}{\partial n} \sum_{k=1}^\infty e^{-2\pi k n} = \frac1{2\pi}\frac{\partial}{\partial n} \frac{1}{e^{2\pi n} -1} =\frac{e^{2\pi n}}{(e^{2\pi n}-1)^2} = \frac{1}{4 \sinh^2(\pi n)} .$$

The sum $$\sum_{n=1}^\infty \frac{1}{\sinh^2(\pi n)} =\frac{1}{6} - \frac{1}{2\pi} $$ is evaluated here, see also page 3 here, and the quoted result follows.

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beautiful answer! –  nbubis Jan 8 '13 at 18:38
    
Better yet, here's a link to the paper in which the result is derived. math.uiuc.edu/~berndt/articles/venkalatex.pdf –  Ron Gordon Jan 8 '13 at 18:40
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@nbubis: thanks, I was trying to evaluate $\sum_n \text{csch}^2 (\pi n)$ from first principles but it seems not so easy. The second reference needs two pages to prove it (ok, he has a more general result). –  Fabian Jan 8 '13 at 18:40
    
@fabian: apologies, didn't see that. Very nontrivial exercise indeed. –  Ron Gordon Jan 8 '13 at 18:41
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@Chris'ssister: I wondered a long time myself (tried it very hard). However, the statement which was evidently first proved by T. S. Nanjundiah [11] in 1951 in the paper made me decide that there is none. –  Fabian Jan 8 '13 at 19:15
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