Trying to sum a series by writing it as the residue of a complex function, I was led to the following function
\begin{align*} f(z) = \frac{e^{-az-\frac{b}{z}}}{z}n(z)n\left(z^{-1}\right) \end{align*} where $a,b\in[0,1)$ are real constants and $n(z) = \left(e^{z}+1\right)^{-1}$. $f(z)$ has poles at $z=\pi(2n+1)$ due to $n(z)$ and at $z=\frac{1}{\pi(2n+1)}$ due to $n(1/z)$ for $n\in\mathbb{Z}$. Further, it has an essential singularity at $z=0$. I would like to compute the residue at $z=0$. My thought was to look at the Laurent series of each term around $z=0$, but I am struggling with the factor $n(1/z)$.
As a real function $n(1/x)$, it has a well defined limit as $x\to 0$. However at any complex neighbourhood of $z=0$ (say a punctured circle of radius $\epsilon>0$) there $\exists n\in\mathbb{Z}$ such that $\frac{1}{\pi(2n+1)}<\epsilon$, i.e. there are always infinitely many poles in the neighbourhood.
Does that mean the Laurent series around $z=0$ does not exist?