Laurent series for $1/(e^z-1)$ Trying to compute the first five coefficients of the Laurent series for $$\frac{1}{e^z-1}$$ centered at the point $0$. I'm not seeing a way to use the geometric series due to the exponential. Any ideas?
 A: Consider that $e^z-1$ has a simple zero in $z=0$, since:
$$\lim_{z\to 0}\frac{e^z-1}{z}=1.$$
This gives that $z=0$ is a simple pole for $f(z)=\frac{1}{e^z-1}$ with residue $1$, hence
$$ g(z)=\frac{1}{e^z-1}-\frac{1}{z}$$
is a holomorphic function in a neighbourhood of zero. 
The coefficients of the Taylor series of $g(z)$ depend on the Bernoulli numbers, since we have:
$$\frac{z}{e^z-1}=\sum_{n\geq 0}\frac{B_n}{n!}z^n,$$
giving:

$$\frac{1}{e^z-1}=\frac{1}{z}+\sum_{n=0}^{+\infty}\frac{B_{n+1}}{(n+1)!}z^n.$$

A: You can do this formally. Since $\frac{1}{e^z-1}$ has a simple pole at $0$, it must have a Laurent series with no coefficient of $z^k$ for $k<-1$. If the coefficient of $z^k$ is $a_k$, we must then have
$$\left(\frac{a_{-1}}{z}+a_0+a_1z+a_2 z^{2} + a_3z^3+ O(z^4)\right)\left(z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\frac{z^5}{5!}+O(z^6)\right)=1$$
and you can multiply this out and equate coefficients to find the values of the $a_i$. For example, equating order-zero coefficients immediately gives $a_{-1}=1$. Equating order-one coefficients gives $\frac{a_{-1}}{2}+a_0=0$, so $a_0=-\frac{1}{2}$. Equating order-two coefficients gives $\frac{a_{-1}}{6}+\frac{a_0}{2}+a_1=0$, and so $a_1=-\frac{1}{6}+\frac{1}{4}=\frac{1}{12}$; and so on.
