How to use Liouville's theorem to prove the formula of $\frac{\pi}{\sin \pi z}$?

Question

In my complex analysis textbook the author claims that the identity $$\frac{\pi }{ \sin \pi z} = \frac{1}{z} + \sum_{n \geq 1} (-1)^n \left[ \frac{1}{z-n} + \frac{1}{z+n} \right]$$ with $z \in \mathbb{C} - \mathbb{N}$ can be directly proved using Liouvilles's theorem, but I find it hard to verify that LHS-RHS is bounded. Can anyone give some ideas?

Answer 1

The RHS $$h(z) = \lim_{N \to \infty} \sum_{n=-N}^N \frac{(-1)^n}{z-n}=\frac1z+\sum_{n \ge 1} (-1)^n \frac{2z}{z^2-n^2}$$ converges locally uniformly away from $\Bbb{Z}$ so it is meromorphic with simple poles,

it is $2$-periodic because $h(z+2)-h(z)= \lim_{N \to \infty} \frac{(-1)^{N+1}}{z-N-1}+\frac{(-1)^{N+2}}{z-N-2}- \frac{(-1)^{-N}}{z-N}-\frac{(-1)^{-N+1}}{z-N+1}$

$$f(z)=\frac{\pi }{ \sin \pi z}-h(z)$$ is entire and $2$-periodic.

Also the second expression for $h$ means $h(z) = o(|z|)$ as $\Im(z) \to \pm \infty, \Re(z) \in [0,2]$.

Together with the decay of $\frac{\pi}{\sin \pi z}$ it means $f(z) = o(|z|)$ which implies from Liouville's theorem that $f$ is constant.

Since $f$ is odd then $f(0) = 0$.