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This question and this question are about $$\sum_{n=1}^\infty \frac{1}{\cosh n\pi}=\frac{1}{2}\left(\frac{\sqrt{\pi}}{\Gamma ^2(3/4)}-1\right)$$ and $$\sum_{n=1}^\infty \frac{1}{\sinh ^2n\pi}=\frac{1}{6}-\frac{1}{2\pi}$$ But I couldn't find the closed form (in terms of $\Gamma (a)$ for rational $a$) of $$S=\sum_{n=1}^\infty \frac{1}{\sinh n\pi}.$$

The series $S$ can be rewritten as Lambert series $$2\sum_{n=1}^\infty \frac{q^n}{1-q^{2n}}$$ where $q=e^{-\pi}$, but unlike $$\sum_{n=1}^\infty \frac{1}{\cosh n\pi}=2\sum_{n=1}^\infty \frac{q^n}{1+q^{2n}}=2\sum_{n=0}^\infty (-1)^n \frac{q^{2n+1}}{1-q^{2n+1}},$$ $S$ doesn't seem to be expressible in terms of $\vartheta_3(x)$ (Jacobi theta function).

The other series $\sum_{n=1}^\infty \frac{1}{\sinh ^2n\pi}$ is suggestive of the Weierstrass elliptic function $\wp$: $$\sum_{n=1}^\infty \frac{1}{\sinh ^2n\pi}=\frac{1}{2\pi ^2}\sum_{n=1}^\infty \sum_{m\in\mathbb{Z}}\left(\frac{1}{(n-mi)^2}+\frac{1}{(n+mi)^2}\right)$$ but this approach is exploiting the square of $\sinh$ in the denominator which isn't of any use for $S$.

Poder Rac
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    I am pessimistic to a closed form (at least on same simplicity level of $\Gamma(1/4)$): the natural range of summation for all such series are from $-\infty$ to $\infty$, your cited example of $\cosh (n\pi), \sinh^2(n\pi)$ can be reduced into this summation range becase they're even function, on other hand, $\sinh(n\pi)$ on a half infinite interval isn't anything. Comparing to $\sum 1/n^2 = \pi^2/6$ while we can do nothing about $\sum 1/n^3$ – pisco Jul 07 '21 at 22:15
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    To illustrate more, $$\sum_{n=1}^\infty \frac{1}{\sinh(n\pi + z)}$$ isn't an elliptic function of $z$ (it only has $2i\pi$ as period) , but $$\sum_{n=-\infty}^\infty \frac{1}{\sinh(n\pi + z)}$$ is (now π is also a period), therefore this bilateral sum can be converted to Weierstrass elliptic, and then we have many machineries available to grind for special values. All this does not work if you're working on $\sum_1^\infty$. – pisco Jul 07 '21 at 22:30
  • @pisco So it's just unfortunate that we can't write $\sum_{n=1}^\infty$ as $\sum_{n=-\infty}^\infty$ because $\sinh n\pi$ is $0$ at $n=0$? – Poder Rac Jul 07 '21 at 22:50
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    $1/\sinh(n\pi)$ being undefined at $0$ isn't a fatal point, what fatal is, the function is odd. – pisco Jul 07 '21 at 22:52

1 Answers1

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Mathematica gives the following result where $\psi _{q}^{(0)}(z)$ is the q-Polygamma Function.

$$\sum _{n=1}^{\infty } \frac{1}{\sinh (\pi n)}=\frac{\psi _{e^{\pi }}^{(0)}(1-i)-\psi _{e^{\pi }}^{(0)}(1)+i \pi }{\pi }$$

Steven Clark
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  • I see the "derivative" definition of q-polygamma but is there a way to simplify it to $\Gamma$ only, at least the case $n=0$, $z=1$, $q=e^\pi$? – Poder Rac Jul 07 '21 at 23:45