I have a problem, I know that the function $f(z)=\dfrac{1}{e^{z}-1}$ has a simple pole at $z=2i\pi k$ with $k\in \mathbb Z$, now for the behavior at infinity, we see if $f(\dfrac{1}{z})$ is holomorphic at $0$, in this case it is not analytic at zero. At the time of doing the Laurent series I get:
$\dfrac{1}{\sum_{k=1}^{\infty} \dfrac{1}{k!z^{k}}}$
Now, $Res_\infty f(z)=Res_0 \dfrac{1}{z^2}f(\dfrac{1}{z})$ and with what we would have to $Res_\infty f(z)=1$
Am I correct or is there an error? first of all, Thanks