Evaluate $\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(z-n)^2}=\frac{\pi^2}{(\sin \pi z)(\tan \pi z)}$

Question

I am trying to prove $\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(z-n)^2}=\frac{\pi^2}{(\sin \pi z)(\tan \pi z)}$ using the identity $\sum_{n=-\infty}^{\infty}\frac{1}{(z-n)^2}=\frac{\pi^2}{\sin^2 \pi z}$.

Ideas?

Answer 1

I know this is too complicated, but I'm on a roll with hypergeometric series today, so let's consider two functions:

$$f_1(x,z) = \sum_{n=0}^{\infty}\frac{x^n}{(z-n)^2}$$

$$f_2(x,z) = \sum_{n=0}^{\infty}\frac{\frac{1}{x^n}}{(z+n)^2}$$

Obviously:

$$f_1+f_2-\frac{1}{z^2}=\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(z-n)^2}$$

Considering the ratio of subsequent terms for $f_1$ we have:

$$\frac{t_{n+1}}{t_n}=\frac{(n-z)^2}{(n+1-z)^2} x=\frac{(n-z)^2(n+1)}{(n+1-z)^2} \frac{x}{n+1}$$

$$t_0=\frac{1}{z^2}$$

Which means:

$$f_1(x,z)=\frac{1}{z^2} {_3 F_2} \left(1,-z,-z;1-z,1-z;x \right)$$

And similarly:

$$f_2(x,z)=\frac{1}{z^2} {_3 F_2} \left(1,z,z;1+z,1+z; \frac{1}{x} \right)$$

These functions for $x=-1$ have the following 'closed forms' (actually, just series in disguise):

$$f_1(-1,z)=\frac{1}{4} \left( \psi^{(1)} \left(- \frac{z}{2} \right)- \psi^{(1)} \left( - \frac{z}{2}+\frac{1}{2} \right)\right)$$

$$f_2(-1,z)=\frac{1}{4} \left( \psi^{(1)} \left( \frac{z}{2} \right)- \psi^{(1)} \left( \frac{z}{2}+\frac{1}{2} \right)\right)$$

Where we have the derivative of digamma function, or the polygamma function of the 1st order.

And finally:

$$\psi^{(1)} \left(- \frac{z}{2} \right)- \psi^{(1)} \left( - \frac{z}{2}+\frac{1}{2} \right)+ \psi^{(1)} \left( \frac{z}{2} \right)- \psi^{(1)} \left( \frac{z}{2}+\frac{1}{2} \right) = 4 \left( \frac{1}{z^2}+\frac{\pi^2}{\tan \pi z \sin \pi z} \right)$$

Which confirms the proposed closed form for the series.

I know this answer is not very useful, so I'm prepared to delete it if the community doesn't want it here.