The function $$S(z)=\sum_{n=1}^\infty \frac{1}{e^{1/z}+e^n}$$ is meromorphic in $\mathbb{C}\backslash \{z_{n,m}=\frac{1}{i\pi(2m+1) +n}:n\in \mathbb{N},m\in \mathbb{Z}\}$ with simple poles at every $z_{n,m}$.
I would like to prove that $S(z)$ can not be extended at $0$. This is my justification: Suppose $S(z)$ cab be extended analytically at $0$. Then exists an open neighborhood $U$ of $0$ such that $S$ is bounded in $U\backslash \{0\}$. But since $z_{n,0}\rightarrow 0$ are poles of $S(z)$ then exists $z_{n,0}\in U$ such that $S(z)\rightarrow \infty$ when $z\rightarrow z_{n,0}$. Therefore, $S(z)$ can not be extended analytically at $0$.
Is correct my justification?
I find it well. By definition a complex valued function is analytical at a point if it is analytical in a full neighborhood of such a point, but in this case this cannot be fulfilled as the origin is an accumulation point of poles of the function $S(z)$.