Closed form of $\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}$? 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.
 A: Let $$g(z) = \sum_{n\in \mathbb{Z}} \frac{(-1)^n}{\sinh(z+n)}$$
$g$ has $2$ and $2\pi i$ as period, so an elliptic function. Poles of $g$, which are all simple, are exactly $0,1,\pi i,1+\pi i$ modulo periods, with residues $1,-1,-1,1$.
$\DeclareMathOperator{\sn}{sn}\DeclareMathOperator{\ns}{ns}$
Since $g$ has only simple poles, it is easier to express it as Jacobi elliptic function. Recall that $\ns(z,k)$ has periods $4K,2iK'$ and residues $1$ at $0$. Hence $g(z)$ must be of form
$$g(z) =a \ns(az,k) - a \ns(a(z-\pi i),k) + C$$
then it is easy to determine $K'/K = 2\pi, a = 2K, C=0$. This puts $g(z)$ in terms of Jacobi-elliptic function:
$$g(z) = 2K[\ns(2Kz, k) - k \sn(2Kz,k)] \qquad K' = 2\pi K$$

Mathematica code to calculate $g(z)$ numerically:
g[z_] := Module[{m = InverseEllipticNomeQ[Exp[-2 Pi^2]], K}, 
   K = EllipticK[m]; 
   2 K (JacobiNS[2 K*z, m] - Sqrt[m]*JacobiSN[2 K*z, m])];

This can give, within an eyeblink, value of any $g(z)$ to great precision. For example, try
N[g[1/3], 1000]

A: We can evaluate it in terms of Jacobi theta functions.
This identity is occasionally useful:
$$
\sum _{n=-\infty }^{\infty }{\frac {b{u}^{n}}{1-a{q}^{2\,n}}}=
\frac{b}{2i}\;
{\frac {
{\theta_1} \left( \frac{i}{2}\ln  \left( au \right) ,q \right) {
\theta_2} \left( 0,q \right) {\theta_3} \left( 0,q
 \right) {\theta_4} \left( 0,q \right) }{{\theta_1}
 \left( \frac{i}{2}\ln  \left( u \right) ,q \right) {\theta_1} \left( 
\frac{i}{2}\ln  \left( a \right) ,q \right) }}
$$
provided $1 > |u| > |q^2|$ so that the series converges.
For this problem, write $\sinh$ in terms of $\exp$ to get the left side of this,
with
$$
a=e^{-2 z},\quad b=2e^{-z},\quad q=e^{-1},\quad u=-e^{-1}
$$
to get the answer
$$
\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}
=
{\frac {i{{\rm e}^{-z}}{\theta_2} \left( iz+i/2,{{\rm e}^{-1}}
 \right) {\theta_2} \left( 0,{{\rm e}^{-1}} \right) { 
\theta_3} \left( 0,{{\rm e}^{-1}} \right) {\theta_4}
 \left( 0,{{\rm e}^{-1}} \right) }{{\theta_2} \left( i/2,{
{\rm e}^{-1}} \right) {\theta_1} \left( iz,{{\rm e}^{-1}}
 \right) }}
$$
Note, despite all those $i$s in there, this is real when $z$ is real.
