Let $z \in \mathbb{C}$. I want to prove that
$$\sum_{n\in\mathbb{Z}} \frac{(-1)^n}{z-n} \sum_{n\in\mathbb{Z}} \frac{(-1)^n}{z-n} = \sum_{n\in\mathbb{Z}} \frac{1}{(z-n)^2}.$$
Using the Cauchy Product I get $$\sum_{n \in\mathbb{Z}} (-1)^n \sum_{k=0}^{n}\frac{1}{(z-k)}\frac{1}{(z-n-k)}$$ Is this correct?
