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I quite like this question from here that was answered by Mr @Jack D'Aurizio

If he or anyone else could verify the following conjecture

$$-{1\over 2}(2\pi)^{4m-1}\sum_{j=0}^{2m}(-1)^j{B_{2j}\over (2j)!}\cdot{B_{4m-2j}\over (4m-2j)!}=\sum_{n=1}^{\infty}{\coth(n\pi)\over n^{4m-1}}$$

where $B_{m}$ are the Bernoulli numbers.

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We can change a bit the fantastic proof of robjohn to find the answer. Using $$\pi\coth\left(\pi n\right)=\frac{1}{n}+2n\sum_{k\geq1}\frac{1}{n^{2}+k^{2}}$$ we get $$\begin{align} \sum_{n\geq1}\frac{\coth\left(\pi n\right)}{n^{4m-1}}= & \frac{1}{\pi}\sum_{n\geq1}\frac{1}{n^{4m}}+\frac{2}{\pi}\sum_{n\geq1}\sum_{k\geq1}\frac{1}{n^{4m-2}\left(n^{2}+k^{2}\right)} \\ = & \frac{\zeta\left(4m\right)}{\pi}+\frac{2}{\pi}\sum_{n\geq1}\sum_{k\geq1}\frac{1}{k^{2}}\left(\frac{1}{n^{4m-2}}-\frac{1}{n^{4m-4}\left(n^{2}+k^{2}\right)}\right) \\ = & \frac{\zeta\left(4m\right)}{\pi}+\frac{2\zeta\left(4m-2\right)\zeta\left(2\right)}{\pi}-\sum_{n\geq1}\sum_{k\geq1}\frac{1}{k^{2}n^{4m-4}\left(n^{2}+k^{2}\right)} \\ = & \frac{\zeta\left(4m\right)}{\pi}+\frac{2\zeta\left(4m-2\right)\zeta\left(2\right)}{\pi}-\sum_{n\geq1}\sum_{k\geq1}\frac{1}{k^{4}}\left(\frac{1}{n^{4m-4}}-\frac{1}{n^{4m-6}\left(n^{2}+k^{2}\right)}\right) \\ = & \frac{\zeta\left(4m\right)}{\pi}+\frac{2\zeta\left(4m-2\right)\zeta\left(2\right)}{\pi}-\frac{2\zeta\left(4m-4\right)\zeta\left(4\right)}{\pi}+\frac{2}{\pi}\sum_{n\geq1}\sum_{k\geq1}\frac{1}{k^{4}n^{4m-6}\left(n^{2}+k^{2}\right)} \end{align}$$ and so on. So it is quite simple to note the pattern. And we also note that, by symmetry, we have $$\sum_{n\geq1}\sum_{k\geq1}\frac{1}{k^{a}n^{b}\left(n^{2}+k^{2}\right)}=\sum_{n\geq1}\sum_{k\geq1}\frac{1}{k^{b}n^{a}\left(n^{2}+k^{2}\right)} $$ hence

$$\sum_{n\geq1}\frac{\coth\left(\pi n\right)}{n^{4m-1}}=\color{red}{\frac{\zeta\left(4m\right)}{\pi}+2\sum_{r=1}^{m-1}\frac{\left(-1\right)^{r+1}\zeta\left(4m-2r\right)\zeta\left(2r\right)}{\pi}+\frac{\left(-1\right)^{m+1}\zeta\left(2m\right)^{2}}{\pi}} $$

and now since $$\zeta\left(2n\right)=\frac{2^{2n-1}B_{2n}\pi^{2n}}{\left(2n\right)!}$$ we can find the relation.

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