Show that:
$$ \dfrac{\pi}{\sin \pi z} = \dfrac{1}{z} + \sum_{n \in \mathbb N} (-1)^n \left(\dfrac{1}{z-n} + \dfrac{1}{z+n}\right) $$
where $z\in \mathbb C \backslash \mathbb Z$
Let's note $\phi(z) = \dfrac{1}{z} + \sum_{n \in \mathbb N} (-1)^n \left(\dfrac{1}{z-n} + \dfrac{1}{z+n}\right)$.
I can "feel" it: both meromorphic functions have the same poles and have the same equivalent near each of them (I mean $\dfrac{\pi}{\sin \pi z} \underset{z \to p}{\sim} \dfrac{1}{(-1)^p(z-p)}$).
But what to conclude with the difference $\dfrac{\pi}{\sin(\pi z)} - \phi(z)$?