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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    $\begingroup$ $\large\frac{2\pi k}{\text{e}^{2\pi k}-1}=k\pi (\coth(k\pi)-1)$ $\endgroup$
    – Mikasa
    Jan 8 '13 at 17:09
  • $\begingroup$ 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. $\endgroup$
    – Ron Gordon
    Jan 8 '13 at 17:20
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    $\begingroup$ @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. $\endgroup$ Jan 8 '13 at 18:25
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    $\begingroup$ 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. $\endgroup$
    – user40314
    Feb 9 '13 at 13:49
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    $\begingroup$ It is amusing since, "essentially", the sum is the energy of a photon gas with discrete wave numbers. I guess @Fabian answer exploits this fact. $\endgroup$ May 3 '14 at 23:02

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.

  • $\begingroup$ beautiful answer! $\endgroup$
    – nbubis
    Jan 8 '13 at 18:38
  • $\begingroup$ Better yet, here's a link to the paper in which the result is derived. math.uiuc.edu/~berndt/articles/venkalatex.pdf $\endgroup$
    – Ron Gordon
    Jan 8 '13 at 18:40
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    $\begingroup$ @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). $\endgroup$
    – Fabian
    Jan 8 '13 at 18:40
  • $\begingroup$ @fabian: apologies, didn't see that. Very nontrivial exercise indeed. $\endgroup$
    – Ron Gordon
    Jan 8 '13 at 18:41
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    $\begingroup$ @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. $\endgroup$
    – Fabian
    Jan 8 '13 at 19:15

Consider the function complex valued function

$$f(z)=\frac{\cot(\pi z)}{\sinh^2(z \pi)}$$

and integrate it along a quadratic contour $\mathcal{C}$ with verticies $\{( N/2,i N/2),(- N/2,i N/2),(- N/2,-i N/2),( N/2,-i N/2)\}$ where $N$ is a (big) odd natural number. Note that the choice of such of kind verticies is necessary to bypass the poles of $f(z)$.

The result is (we travel counter-clockwise)

$$ \oint_{\mathcal{C}}f(z)dz=\int_{N/2}^{-N/2}dxf(x+i N/2)+i\int_{N/2}^{-N/2}dyf(i y+ N/2)+\\\int_{-N/2}^{N/2}dxf(x-i N/2)+i\int_{-N/2}^{N/2}dyf(i y+ N/2) $$

we now have (this is an exercise for the reader)

$\lim_{N\rightarrow\infty}f(x\pm i N/2)=\frac{\mp i}{\cosh^2(\pi x)}$ and that $\lim_{N\rightarrow\infty}f(i y\pm N/2)=0$.

Furthermore it is elementary to show (the integrand is the derivative of $\tanh$) that $\int_{\mathbb{R}}\frac{1}{\cosh^2(\pi x)}dx=\frac{2}{\pi}$ which means that

$$ \oint_{\mathcal{C}}f(z)dz=-2\frac{2i}{\pi}+2\cdot0=-\frac{4i}{\pi} \quad (\spadesuit) $$

in the limit $N\rightarrow \infty$. On the other hand we know by the residue theorem that (note that $(N-1)/2$ is an integer by construction)

$$ \oint_{\mathcal{C}}f(z)dz=2 \pi i\sum_{-(N-1)/2\leq k\leq (N-1)/2}\text{Res}(f(z),z=k)+\text{Res}(f(z),z=i k) $$

it is now straightforward to show that

$$ \text{Res}(f(z),z=k)=\text{Res}(f(z),z=i k) = \frac{1}{\pi\sinh^2(\pi k)} \quad \text{if}\,\,k\neq0 $$

$$ \text{Res}(f(z),z=0) =-\frac{2}{3 \pi} $$

which allows us to write that

$$ \oint_{\mathcal{C}} f(z)dz = 8 i\sum_{k=1}^{\infty}\frac{1}{\sinh^2(\pi k)}-\frac{4 i}{3} \quad (\heartsuit) $$

In the limit of $N\rightarrow\infty$.

Now putting togehter $(\heartsuit)$ and $(\spadesuit)$ it follows that

$$ \sum_{k=1}^{\infty}\frac{1}{\sinh^2(\pi k)}=\frac{1}{6}-\frac{1}{2\pi} \quad\blacksquare $$

this proves the result used by @Fabian


if we set $f_a(z)=\frac{\cot(\pi z)}{\sinh^2(a z \pi)}$ and follow the same steps as above we can show that on the one hand the sum of residues equals

$$ \oint_{\mathcal{C}} f_a(z)dz=4i\sum_{k\geq1}\frac{1}{\sinh^2(k \pi a)}+\frac{4i}{a^2}\sum_{k\geq1}\frac{1}{\sinh^2(k \pi/ a)}-\frac {2 i}{3}\frac{1+a^2}{a^2}\quad (\heartsuit \heartsuit) $$

and on the other hand, by parametrizing the contour of integration as shown above, we get that $$ \oint_{\mathcal{C}} f_a(z)dz=-\frac{4i}{\pi a} \quad (\spadesuit \spadesuit) $$

Now putting togehter $(\heartsuit\heartsuit)$ and $(\spadesuit\spadesuit)$ it follows that

$$ \sum_{k\geq1}\frac{1}{\sinh^2(k \pi a)}+\frac{1}{a^2}\sum_{k\geq1}\frac{1}{\sinh^2(k \pi/ a)}=\frac{1+a^2}{6 a^2}-\frac{1}{\pi a}\quad\blacksquare\quad\blacksquare $$

in accordance with Berndt et al.

  • $\begingroup$ @r9m thanks, that is very kind of you $\endgroup$
    – tired
    May 21 '17 at 16:56
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    $\begingroup$ It is interesting to prove that some parts of the contour don't vanish. $\endgroup$ May 22 '17 at 13:03
  • $\begingroup$ Nice answer. For some reasons I simply missed this post. $\endgroup$ Apr 29 '18 at 20:41
  • $\begingroup$ $+1$. Wow !!!.${}{}{}{}{}$. $\endgroup$ Jan 31 at 18:51

If we define Dedekind eta function $\eta(q)$ via the equation $$\eta(q) = q^{1/12}\prod_{i = 1}^{\infty}(1 - q^{2i})\tag{1}$$ then it can be proved via functional equation of Dedekind eta function that $$n^{1/4}\eta(e^{-\pi\sqrt{n}}) = \eta(e^{-\pi/\sqrt{n}})\tag{2}$$ Let's define function $P(q)$ as $$P(q) = 1 - 24\sum_{i = 1}^{\infty}\frac{iq^{2i}}{1 - q^{2i}} = 12q\frac{d}{dq}\log\eta(q)\tag{3}$$ Logarithmic differentiation of equation $(2)$ with respect to $n$ gives us $$nP(e^{-\pi\sqrt{n}}) + P(e^{-\pi/\sqrt{n}}) = \frac{6\sqrt{n}}{\pi}\tag{4}$$ and putting $n = 1$ we get $$P(e^{-\pi}) = \frac{3}{\pi}\tag{5}$$ and looking at the definition of $P(q)$ we see that the above equation is exactly what is asked in the question.

  • $\begingroup$ this made short work of it (+1) $\endgroup$
    – tired
    May 17 '17 at 21:59
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    $\begingroup$ @tired: Ramanujan dealt with such sums of hyperbolic functions in great detail and I learnt most of his technique from his collected papers. The beauty of his approach is the use of elementary tools (algebra and calculus of single real variable) and it shows that these basic tools (available to any high schooler) can prove very deep results. $\endgroup$ May 17 '17 at 22:02
  • $\begingroup$ it is somehow logical that Ramanujan chosed this appproach because his lack of a formal mathematical education (as Euler did in a time where formal mathematics wasn't developed). And it is absolutely incredible how far he came... $\endgroup$
    – tired
    May 17 '17 at 22:12
  • $\begingroup$ @tired : fully agree. By the way in case you have not noticed the last equation in your answer dealing with variable $a$ is same as the equation $(4)$ of my answer (just put $a=\sqrt{n} $). $\endgroup$ May 18 '17 at 3:28
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    $\begingroup$ @tired: there are identities to handle the powers of hyperbolic functions also. I will extract some info and post after sometime. $\endgroup$ May 18 '17 at 17:54

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