# Closed form for $\sum_{n=1}^{\infty}\frac{1}{\sinh^2\!\pi n}$ conjectured

By trial and error I have found numerically

$$\sum_{n=1}^{\infty}\frac{1}{\sinh^2\!\pi n}=\frac{1}{6}-\frac{1}{2\pi}$$

How can this result be derived analytically?

• Try using the fact that $\sinh x = - i\cdot \sin (ix)$ Jul 21, 2016 at 14:21
• It has the opposite sign (it must be positive, being a series of squares). Jul 21, 2016 at 14:26
• It depends on properties of the Dedekind eta function (en.wikipedia.org/wiki/Dedekind_eta_function). Following Zucker, THE SUMMATION OF SERIES OF HYPERBOLIC FUNCTIONS, your series is $I_1(1)$. It can be tackled through the Laplace/Mellin transform. Jul 21, 2016 at 14:45
• Evaluating numerically is not what "trial and error" means.
– anon
Jul 21, 2016 at 15:45
• related: math.stackexchange.com/q/346713 Aug 3, 2016 at 16:36

Hint. This is a normally convergent series, then one may interchange the two summations, \begin{align} \sum_{n=1}^{\infty}\frac{1}{\sinh^2\!\pi n}&=4\sum_{n=1}^{\infty}\frac{e^{-2\pi n}}{(1-e^{-2\pi n})^2} \\\\&=4\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}k\:e^{-2\pi n k} \\\\&=4\sum_{k=1}^{\infty}k\left(\sum_{n=1}^{\infty}e^{-2\pi n k}\right) \\\\&=4\sum_{k=1}^{\infty}\frac{k}{e^{2\pi k}-1} \\\\&=\frac16-\frac1{2\pi} \end{align} where we have concluded using a classic evaluation.

• Quelle élégance ! Jul 21, 2016 at 19:29

By considering the logarithmic derivative of the Weierstrass product $$\frac{\sinh(\pi x)}{\pi x}=\prod_{m\geq 1}\left(1+\frac{x^2}{m^2}\right) \tag{1}$$ we have: $$\sum_{m\in\mathbb{Z}}\frac{x}{x^2+m^2}=\frac{1}{2}\sum_{m\in\mathbb{Z}}\left(\frac{1}{x-mi}+\frac{1}{x+mi}\right)=\pi \coth(\pi x) \tag{2}$$ and by differentiating again: $$\frac{\pi^2}{\sinh^2(\pi x)}=\frac{1}{2}\sum_{m\in\mathbb{Z}}\left(\frac{1}{(x-mi)^2}+\frac{1}{(x+mi)^2}\right)\tag{3}$$ so we have: $$\sum_{n\geq 1}\frac{1}{\sinh^2(\pi n)}=\frac{1}{2\pi^2}\sum_{n\geq 1}\sum_{m\in\mathbb{Z}}\left(\frac{1}{(n-mi)^2}+\frac{1}{(n+mi)^2}\right)\tag{4}$$ the connection with the Weierstrass elliptic function $\wp$ should now be clear.

Alhough when I came up with the sum above I couldn't find a proper way how to prove it, I have only flawn ideas which arised from "magic" manipulations of some "magic" formula, now, however, working on some different problem, I finally came up with proper solution - It is not my style to answer my own questions, but a friend of mine persuaded me to write it here anyway, so it goes like that :

This aproach requires simple techniques from complex analysis - let us define a meromorphic function $f(z)$ :

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

This function has poles $z_k$ at $z=n$ and $z=ni$, where $n\in\mathbb{Z}$

Consider a square contour in the picture below $(m\in\mathbb{N})$ By residue theorem :

$$\oint _{\gamma} f(z)\, \mathrm{d}z=2\pi i\sum\mathrm{Res}_{z=z_k}{f(z)}\tag{1}$$

For the residues we have :

\begin{align} & \mathrm{Res}_{z=n}{\frac{\cot\pi z}{\sinh^2\pi z}} = \frac{1}{\pi\sinh^2\pi n} \\ \\ & \mathrm{Res}_{z=ni}{\frac{\cot\pi z}{\sinh^2\pi z}} = \frac{1}{\pi\sinh^2\pi n} \\ \\ & \mathrm{Res}_{z=0}{\frac{\cot\pi z}{\sinh^2\pi z}} = -\frac{2}{3\pi} \end{align}

When $m\rightarrow\infty$ we have

$$\frac{\cot\pi (x\pm (mi+\frac12))}{\sinh^2\pi (x\pm (mi+\frac12))}\rightarrow \frac{\mp i}{\cosh^2\pi x}$$

Since integrals along the sides vanishes as $m\rightarrow\infty$ we rewrite $(1)$ using residues taking the limit as $m\rightarrow\infty$:

$$-2i\int_{-\infty}^\infty\frac{\mathrm{d}x}{\cosh^2\pi x}=2\pi i\left(-\frac{2}{3\pi}+\frac{4}{\pi}\sum_{n=1}^\infty\frac{1}{\sinh^2\pi n}\right)$$

Immediately since $\int_{-\infty}^\infty\frac{\mathrm{d}x}{\cosh^2\pi x}=\frac{1}{\pi}\tanh\pi x\bigg{|}_{-\infty}^\infty=\frac{2}{\pi}$ after simple manipulations we get the desired result :

$$\sum_{n=1}^{\infty}\frac{1}{\sinh^2\!\pi n}=\frac{1}{6}-\frac{1}{2\pi}$$

DECLARATION: I am not first and nor last who discovered exact value of the sum, and it is not in my competence to even name it like the Sophomore's dream has a name, however, I have decided to make an exeption because of its breathtaking beauty and after the tradition of "dreams" we shall refer to it as the Nike's dream, after greek goddess of victory - Nike.

• I wouldn't say it's particularly breathtaking. It's a garden-variety sum of the sort one usually encounters in a complex analysis course. Sep 7, 2016 at 21:15
• What makes it more than just a garden-variety sum that one encounters in a complex analysis course is the fact that the integral doesn't vanish along the top and bottom of the square. Sep 8, 2016 at 17:56