Does anyone know of a good way to calculate the residue at zero of the following function? I was able to calculate it with the higher order pole formula for residues and then used Mathematica to find the limit. Really just looking for a nice trick without having to get too dirty.
$$ f(z)=\frac{1}{z^2(e^{-z}-1)} $$
Thank you for any help.
By the way, the answer is $\frac{-1}{12}$
$$\begin{eqnarray} \frac{1}{z^2(e^{-z}-1)} &=& \frac{1}{z^2\left(1-z+\frac{z^2}{2}-\frac{z^3}{6} +\ldots - 1\right)} \\ &=& -\frac{1}{z^3\left(1-\left(\frac{z}{2}-\frac{z^2}{6}+\ldots\right)\right)} \\ &=& -\frac{1}{z^3}\left(1+\left(\frac{z}{2}-\frac{z^2}{6}+\ldots\right)+\left(\frac{z}{2}-\frac{z^2}{6}+\ldots\right)^2\right) \\ &=& -\frac{1}{z^3}\left(1+\frac{z}{2}+\frac{z^2}{12}+\ldots\right) \\ &=& -\frac{1}{z^3} - \frac{1}{2z^2} - \frac{1}{12 z} + \ldots \end{eqnarray}$$