# Series of Squared Hyperbolic Cosecant

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.

• $f(t)$ can be written in a alternative way by invoking the Poisson summation formula. It is related to a QPolyGamma function and has very few special values, like $f(2\pi)=\frac{1}{6}-\frac{1}{2\pi}$. See Zucker, THE SUMMATION OF SERIES OF HYPERBOLIC FUNCTIONS. – Jack D'Aurizio Nov 16 '17 at 17:02
• @JackD'Aurizio Thank you for your comment. However, I do not have access to that paper. Is there any other good bibliography related to this subject? – user3141592 Nov 17 '17 at 9:28
• You should be able to access this article of Simon Plouffe dealing with similar series through special functions and the residue theorem. – Jack D'Aurizio Nov 17 '17 at 12:44
• Related: math.stackexchange.com/q/346713 – Random Variable Nov 18 '17 at 22:33

Let's consider the function $$F(x) =\sum_{n=1}^{\infty}\frac{1}{\sinh^{2}x}=4\sum_{n=1}^{\infty} \frac{e^{-2nx}}{(1-e^{-2nx})^{2}}=4\sum_{n=1}^{\infty}\frac{q^{2n}}{(1-q^{2n})^{2}}=4\sum_{n=1}^{\infty}\frac{nq^{2n}}{1-q^{2n}}$$ 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 $$f(2\pi)=F(\pi)=(1-P(e^{-2\pi}))/6=1/6-1/2\pi$$
• Thank you! Could you please provide some bibliography/links about the Ramanujan Function $P(x)$ you mentioned? I have not been able to find information about that precise function around the internet, and I would like to learn about that type os series. – user3141592 Dec 13 '17 at 12:25
• In spite that I do not have access to all of the bibliography you mentioned in the other answer, I noticed that there are very few series involving a denominator of the type $\frac{f(q)}{1-q^n}$, and that's why I would like to get more information about the precise function you mentioned ($P(x)$) – user3141592 Dec 13 '17 at 13:06
• @user3141592: such series are called Lambert series. The Ramanujan function $P(q)$ belongs to elliptic function theory and in modern parlance it goes by the name Eisenstein series. You should have a look at the blog posts paramanands.blogspot.com/2013/05/… and paramanands.blogspot.com/2013/05/… – Paramanand Singh Dec 13 '17 at 14:42