Analytical continuation of $F(p) = \sum_{n \neq 0, n \in \mathbb{Z}} \frac{e^{ipn}}{\sinh^2\kappa n}$ I am trying to find out the behaviour of the series 
$$
F(p) = \sum_{n \neq 0, n \in \mathbb{Z}} \frac{e^{ipn}}{\sinh^2\kappa n}
$$
under analytical continuation in the complex $p$-plane ($\kappa$ is a positive constant). This series converges for all $p$ with $|$Im$(p)|<2\kappa$ and I have found a representation for this series in terms of Weierstrass elliptic functions. This representation shows that $F$ has poles at all points with Re$(p) = 2\pi k$ for $k\in \mathbb{Z}\setminus\{0\}$ and Im$(p) = 2\kappa n$,  $n\in \mathbb{Z}\setminus\{0\}$. As far as I understand the issue, around these points, non-trivial monodromy can exist, i.e. continuation around such poles could lead to a multi-valued function. 
Performing explicit continuation using Taylor series seems to be unpractical, but I am not sure whether another approach exists. My question therefore is the following: how best to investigate the monodromy properties of this power series?
 A: Ok, I try to answer. As this is too long, I do not use "comment" but "answer". Note that your question is not easy, so you must verify.
We have a function defined in a nhood of the real line, $F(p)$, and we have for this function a first meromorphic continuation as a meromorphic function on all $\mathbb{C}$, say it is $M(z)$. So $M(z)$ is, on a disk $B(0,\rho)$, with $\rho$ small, analytic and equal to $F$. Suppose that there exists another meromorphic continuation, on $\mathbb{C}$ say $N(z)$, and also $N$ is analytic and equal to $F$ on $B(0,\rho)$ (Hence $M$=$N$ on $B(0,\rho)$). If $N$ is not equal to $M$, this imply that there exist $t\in \mathbb{C}$, that we can suppose not to be a pole of $N$ or $M$, such that $N(t) \not =M(t)$. 
Now we can find  $U=B(0,R)$ with $R$ large, containing $B(0,\rho)$ and $t$. The functions  $M$ and $N$ have a finite number of poles in $U$. Let $A$, $B$ be polynomials such that $A(t)B(t)$ is non zero, and $A(z)M(z)$, $B(z)N(z)$ analytic in $U$. Then $A(z)B(z)M(z)$ and $A(z)B(z)N(z)$ are analytic in $U$, and they are equal on $B(0,\rho)$. Hence they are equal on $U$. But then we can put $z=t$, and this gives $M(t)=N(t)$, a contradiction.
A: Inspired by the discussion with Kelenner, I believe I can answer my own question:
We have a function $F$, which is given by a convergent series for $|$Im$(p)| <2\kappa$ (call this open neighbourhood of the real line $\Omega$). By rewriting, one can show that $F(p) = G(\cos(p))$ where 
$$
G(z) = \sum_{m=1}^{\infty}e^{-2\kappa m} \frac{z-e^{-2\kappa m}}{1-z e^{-2\kappa m} +e^{-4\kappa m}}.
$$
the partial sums of this series converge uniformly to $G$ is we restrict the domain of $G$ to the set $\cos(\Omega)$, we find that $G$ is analytic on this set as the uniform limit of analytic functions. Moreover, $F$ is analytic as the composition of analytic functions. Consider now the subset $V \subset \mathbb{C}$ being the entire complex plane without the poles mentioned in the question. This set is open and there exists a function $F_e$ given by 
$$
F_e(p) = \wp\left( r_p\right) + \left( \zeta(r_p) - \frac{2r_p}{\omega}\zeta\left(\frac{\omega}{2}\right) \right) \frac{\wp''(r_p)}{\wp'(r_p)}+ 2\left( \zeta(r_p) - \frac{2r_p}{\omega}\zeta\left(\frac{\omega}{2}\right) \right)^2,
$$
where $r_p = \frac{ip}{4\kappa}$, $\wp$ is the Weierstrass elliptic function, $\zeta$ is the Weierstrass $\zeta$-function and both are defined on the lattice (1,$\omega$). This function is holomorphic on $V$ and equal to $F$ on $\Omega$. Therefore, it is the unique analytic continuation of $F$ to $V$ (see proof here). We cannot extend $F$ any further, since its analytic continuation shows that it is not finite at any of the unincluded points of $\mathbb{C}$. 
