I considered the following function: $$f:\, \mathbb{C}\mapsto\mathbb{C}_{\infty},\, z\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}\right)^{-1}.$$

It can be seen that $f(z)=0$ at every integer (division of a non-zero complex number by zero is assumed to be complex infinity, so its reciprocal is $0$). It also *seems* that the function is periodic on the real axis with period $2$, i.e.
$$\forall x\in\mathbb{R}:\, f(x+2)=f(x),$$
though I was not able to prove that rigorously. I tried to use the $\sinh$ addition formula, but that's probably not useful.

On the whole real axis, $f$ is very close to $$g:\, \mathbb{C}\mapsto \mathbb{C},\, z\mapsto 0.31837572\sin \pi z,$$ but on the imaginary axis, there's a big difference between $f$ and $g$, for instance: $$f(2i)\approx 5.3796i,\,\text{but}\, g(2i)\approx 85.2499i.$$

Why is that? Can a closed form of $f$ in terms of Weierstrass/Jacobi elliptic functions be found?

**Note:**
Curiously, any
$$h:\,\mathbb{R}\mapsto\mathbb{R},\, x\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{a^{x+n}-a^{-x-n}}\right)^{-1}$$
where $a\gt 0$ and $a\ne 1$ seems to be $2$-periodic...

**Edit:**
The function
$$f:\, \mathbb{C}\mapsto\mathbb{C}_{\infty},\, z\mapsto \left(\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}\right)^{-1}$$
is probably a doubly-periodic elliptic function with periods $2$ and $2\pi i$, so I added an appropriate tag.