I'm trying to find the Laurent expansion for
$$\frac{e^{1/z^2}}{z - 1}$$ about $z_0 = 0$.
Writing the series for $e^{1/z^2}$ and $1/(z-1)$ individually gives
$$\frac{e^{1/z^2}}{z - 1} = -\left(\sum_{n=0}^\infty z^n \right) \left(\sum_{n=0}^\infty \frac{1}{z^{2n}n!} \right)$$ How do I multiply these two series, and under what conditions is the convergence of the product guaranteed?