$ \frac{\pi^2\csc^2(\pi/x)}{x^2}= \sum_{n=-\infty}^\infty\frac{1+(xn)^2}{(1-(xn)^2)^2}$. Where can I find some more series of this class?

Question

After reading Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ and noodling around on wolfram alpha, I discovered

$$ \begin{align} &\coth(x \pi)=\frac{x}{\pi}\sum_{n=-\infty}^\infty\frac{1}{x^2+n^2} & \cot(x \pi)= \frac{x}{\pi}\sum_{n=-\infty}^\infty\frac{1}{x^2-n^2} \\ & \text{csch}(x \pi )=\frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{(-1)^n }{x^2+n^2}} &\csc(x \pi)= \frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{(-1)^n }{x^2-n^2}} \\ & \tanh(x \pi)=\frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2+4x^2}} &\tan(x \pi) = \frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2-4x^2}} \end{align} $$

I suspect (but don't know for sure) that these can all be justified by milking the techniques from the link above.

Go Go Gadget Calculus

We should be able to derive a few more identities. For example,

$$ \Big(\cot(x \pi) \Big)'= \Big(\frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{1}{x^2-n^2}} \Big )'$$

$$ \Big(\pi csc(\pi x)\Big)^2= \sum_{n=-\infty}^\infty\frac{x^2+n^2}{(x^2-n^2)^2}$$

And after some manipulations we can find

$$ \frac{\pi^2\csc^2(\pi/x)}{x^2}= \sum_{n=-\infty}^\infty\frac{1+(xn)^2}{(1-(xn)^2)^2} $$ Lovely. We can write then $$\frac{\pi^2}{9}=\sum_{n=-\infty}^\infty {\frac{1+(6n)^2}{(1-(6n)^2)^2}}$$ I have a feeling at this point that this must be a well-studied subject and I wonder where I can find some more identities of this class. Does anyone have a link/resource where I can read more on these. I don't really need their derivations if they are just the techniques of the link above + elementary calculus techniques. I am just looking for a well organized list that I can refer to.

Answer 1

I don’t have a list, but I can present the method to make a list systematically.

Let $P(x,n),Q(x,n)$ be two polynomials of $x,n$.

Suppose $$Q(x,r_1(x))=Q(x,r_2(x))=\cdots=Q(x,r_k(x))=0$$ for all $x$.

Suppose the sum $$S(x)=\sum^\infty_{n=-\infty}\frac{P(x,n)}{Q(x,n)}$$ converges whenever $r_{(\cdot)}(x)\not\in\mathbb Z$.

Then, by residue theorem, $$S(x)=-\pi\sum^k_{n=1}\operatorname*{Res}_{z=r_n(x)}\frac{\cot(\pi z)P(x,z)}{Q(x,z)}$$

(It is an issue that for some $x$, we might have $r_p(x)=r_q(x)$. When two or more $r$ functions take the same value, the residue should only be evaluated once.)

Answer 2

There is some nice things to see here:

$$\tanh(x \pi)=\frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2+4x^2}}$$

$$\Big(\tanh(x \pi) \Big)'= \Big( \frac{1}{\pi}\sum_{n=-\infty}^\infty{\frac{4x}{(2n+1)^2+4x^2}} \Big)'$$

$$ \pi\operatorname{sech}^2(\pi x)=\frac{1}{\pi}\sum_{n=-\infty}^{\infty}\frac{4\left(4n^2+4n-4x^2+1\right)}{\left(4n^2+4n+4x^2+1\right)^2}$$

Which is nice: taking $x=0$ we have

$$\pi^2=\sum_{n=-\infty}^{\infty}\frac{4}{4n^2+4n+1}=8\Big(1+\sum_{n=1}^\infty{\frac{1}{4n^2+4n+1}}\Big)$$

Interesting. I wonder if we can use this to demonstrate Takebe Kenko's somewhat similar looking (this can be found- with a tiny typo in denominator- on the last page of this):

$$\pi^2=4\Big(1+\sum_{n=1}^{\infty} \frac{2^{2n+1}(n!)^2}{(2n+2) !} \Big) $$