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In a recent problem I was attempting to solve, I hit a road block when I reduced the problem to that of finding a closed form for the following sum $$ \mathcal{S}_n(x)\equiv\sum_{k=1}^n{\rm csch}^2\left(\frac xk\right) $$ It looks deceptively simple, like there's some trick involved in evaluating it that I'm just not seeing. I've attempted a few things such as rewriting the ${\rm csch}$'s in terms of $\exp$, multiplying the top and the bottom of the summand by $\exp(2x/k)$ to get $\displaystyle\frac{4\exp(2x/k)}{(\exp(2x/k)-1)^2}$, then expanding as a power series in $\exp(2x/k)$, but I didn't end up with anything nicer than the original. On a whim, I think if I continued this process I would've ended up with a sum of zeta functions, which is certainly not what I was looking for.

Specifically, I'm looking for a closed form for $\mathcal{S}_n(x)$ in terms of $n$ and $x$. I feel like Fourier analysis or complex analysis would be necessary to get something nice. Any ideas?

By the way, I have no reason to believe there is a closed form for this sum, other than the fact that I've seen far more unlikely things that possess closed forms.

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  • $\begingroup$ The only reason I would expect anything clean is if there were a geometric series you could use to collapse things, and since there seems to be an ineffable term of the form $a+be^{cx}$ in your denominators, no such term is likely to come around... $\endgroup$ – Steven Stadnicki Oct 14 '15 at 0:21
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It seems that

$$ \text{csch}^2(x) = \sum_{j=0}^\infty \dfrac{2^jB_{j} \; (1-j)}{j!} x^{j-2} = \dfrac{1}{x^2} - \dfrac{1}{3} + \sum_{j=4}^\infty \dfrac{2^jB_{j} \; (1-j)}{j!} x^{j-2}$$ where $B_{j}$ are Bernoulli numbers. Thus

$$ \eqalign{\sum_{k=1}^\infty \left(\text{csch}^2\left(\dfrac{x}{k}\right) - \dfrac{k^2}{x^2} + \dfrac{1}{3}\right) &= \sum_{j=4}^\infty \dfrac{ 2^jB_{j} \; (1-j)}{j!} x^{j-2} \zeta(j-2) \cr &= \sum_{i=2}^\infty \dfrac{(-1)^i 2^{4i-1} B_{2i}^2 (2i-1) \pi^{2i}}{(2i)!^2} x^{2i-2}}$$

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  • $\begingroup$ Interesting result! How might this be extended to the partial sums $\mathcal{S}_n$? $\endgroup$ – user3002473 Oct 25 '15 at 21:00
  • $\begingroup$ To sum up to $N$, replace $\zeta(j-2)$ by $\zeta(j-2) - \zeta(j-2,N+1)$ where $\zeta(s,q)$ is the Hurwitz zeta function. However, I don't think there's a nice formula for $\zeta(s,q)$ when $s$ is an even positive integer. $\endgroup$ – Robert Israel Oct 25 '15 at 23:20

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