Series of Squared Hyperbolic Cosecant

Question

I would be interested in finding a closed form for the following series:

$$f(t):=\sum_{n=1}^\infty \text{csch}^2\left (\frac{2n\pi^2}{t} \right )$$

For come complex $t \ne 0$.

I have tried to use the main definition of Hyperbolic Cosecant, and then making the change of variables $k=e^{\frac{2\pi^2}{t}}$:

$$f(t)=4\sum_{n=1}^\infty \frac{1}{\left (k^{n}-k^{-n}\right )^2}$$

Using this definition, it is easier to test the convergence of the series. However, I do not know how to find a closed form nor how to continue working with the function in this way.

By another change of variables, we could get something similar to:

$$f(t) = -\sum_{n=1}^\infty \frac{1}{\sin^2(nz)}$$

Which seems to look like a more common/famous series.

Any help/hint/happy idea would be welcome.

Thank you.

Answer 1

Let's consider the function $$F(x) =\sum_{n=1}^{\infty}\frac{1}{\sinh^{2}nx}=4\sum_{n=1}^{\infty} \frac{e^{-2nx}}{(1-e^{-2nx})^{2}}=4\sum_{n=1}^{\infty}\frac{q^{2n}}{(1-q^{2n})^{2}}=4\sum_{n=1}^{\infty}\frac{nq^{2n}}{1-q^{2n}}$$ where $q=e^{-x} $. Thus the function $F(x) $ is essentially the same as Ramanujan function $P(q) $ defined by $$P(q) =1-24\sum_{n=1}^{\infty}\frac{nq^{n}}{1-q^{n}}$$ and we have $$F(x) =\frac{1-P(q^{2})}{6}$$ It is well known that the values of $P(q) $ where $q=e^{-\pi\sqrt{r}}, r\in\mathbb{Q} ^{+} $ can be expressed in terms of $\pi$ and values of Gamma function at rational points. And therefore so is the case with $F(x)$ for $x =\pi\sqrt{r} $. Your function $f(t) = F(2\pi^{2}/t)$ has the same story for $t=\pi\sqrt{r} $. In particular we have $P(e^{-2\pi})=3/\pi$ and hence $$f(2\pi)=F(\pi)=(1-P(e^{-2\pi}))/6=1/6-1/2\pi$$

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Such sums belong more properly to the theory of elliptic functions, elliptic integrals and theta functions and it is difficult to appreciate their beauty in isolation. In this answer I have given some references which deal with this beautiful theory.