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I'm aware that it is possible to compute $$\sum_{m=1}^{\infty} \frac{\coth(m\pi)}{m^{4k-1}}$$ in closed form for $k = 1, 2, 3\ldots$ as demonstrated in the links below:

Cauchy-Ramanujan Formula $ \displaystyle \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} $

Method of proof of $\sum\limits_{n=1}^{\infty}\tfrac{\coth n\pi}{n^7}=\tfrac{19}{56700}\pi^7$

But is it possible to find similar a simlar formula for $$\sum_{m=1}^{\infty} \frac{\coth(m\pi)}{m^{9}}$$ ?

lcv
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succubus
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  • At what point does the method of Marko Riedel fail? – K.defaoite Oct 27 '24 at 22:50
  • $9$ is a $4k+1$ I suspect OP wants a plus where we have a minus in the series above? – Mason Oct 28 '24 at 01:27
  • I might look into this some more later but the point of failure is in the establishment of the functional equation for the sum with its fixed point which gives for $x=\pi$ a Bernoulli number identity that does not involve $S(\pi)$ because it appears once with the same sign on each side of the equation. – Marko Riedel Oct 28 '24 at 02:04
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    Using the pole expansion of the $coth$ works for powers of the form $4p-1$, but fails to yield any meaningful result for powers of the form $4p+1$ (since partial fractions only lead to a trivial equality). – Stefan Lafon Oct 28 '24 at 16:08

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