How would one go about determining the Laurent series of such a series as
$$\frac{1}{z(e^z-1)}$$
This is tricky since I know its a pole of second order at z = 0 but just unsure what to do past that point. I tried expanding the exponential but it just gives me an infinite series that I do not know how to deal with.
On
By Taylor $$z(e^z-1)=z^2+\frac{z^3}{2}+\frac{z^4}{6}+\frac{z^5}{24}+\frac{z^6}{120}+\frac{z^7}{720}+\frac{z^8}{5040}+O\left(z^9\right)$$ Using the long division $$\frac 1{z(e^z-1)}=\frac{1}{z^2}-\frac{1}{2 z}+\frac{1}{12}-\frac{z^2}{720}+\frac{z^4}{30240}+O\left(z^5\right)$$
If you want the infinite series, just use the fact that $$\frac 1{e^z-1}=\sum_{m=0}^\infty \frac{B_m }{m!}z^{m-1}$$ and divide by $z$.
Apply the Taylor series for $e^z$
$$\frac{1}{z(e^z-1)} = \frac{1}{z^2\left(1+\frac{z}{2}+\frac{z^2}{6}+\cdots\right)} = \frac{1}{z^2}\left(1-\frac{z}{2}-\frac{z^2}{6}-\cdots+\left(\frac{z}{2}+\cdots\right)^2-\cdots\right)$$
$$= \frac{1}{z^2}\left(1-\frac{z}{2}-\frac{z^2}{6}+\frac{z^2}{4}+O(z^3)\right) = \frac{1}{z^2}-\frac{1}{2z}-\frac{1}{12}+O(z)$$
by the geometric series formula for $\frac{1}{1+z}$