Question: I want to find the first 4 terms of the Laurent series expansion of the expression $\frac{1}{e^z-1}$, around $z\_0=0$. I tried expanding first $e^z-1=(1+\frac{z}{1}+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+...)-1$ and then consider putting this under the numerator of the fraction. However, I really don't see how to get the terms of the Laurent series expansion after that, since I am not sure that $|\frac{z}{1}+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+...|<1$ and so cannot apply geometric series. Can anyone give me a hint to solve this problem?

Answer: 
Observe that 

$$e^z=\sum\_{n=0}^\infty\frac{z^n}{n!}\implies e^z-1=z\sum\_{n=1}^\infty\frac{z^{n-1}}{n!}$$

So

$$\frac1{e^z-1}=\frac1{z\left(1+\frac z2+\frac{z^2}6+\ldots\right)}$$

Now, doing long division:

$$\frac1{1+\frac z2+\frac{z^2}6+\ldots}=a\_0+a\_1z+a\_2z^2+\ldots\iff\left(1+\frac z2+\frac{z^2}6+\ldots\right)\left(a\_0+a\_1z+\ldots\right)=1$$

$$\iff a\_0+\left(a\_1+\frac12a\_0\right)z+\left(\frac16a\_0+\frac12a\_1+a\_2\right)z^2+\left(\frac1{24}a\_0+\frac16a\_1+\frac12a\_2+a\_3\right)z^3+\ldots=1$$

$$\implies a\_0=1\;,\;\;a\_1=-\frac12a\_0=-\frac12\;,\;\;a\_2=\frac14-\frac16=\frac1{12}\;,\;\;a\_3=-\frac1{24}+\frac1{12}-\frac1{24}=0$$

and thus the first four summands in the wanted Laurent series are:

$$\frac1{e^z-1}=\frac1z-\frac12+\frac z{12}+0$$
$$$$
Using polynomial long division to compute the reciprocal the power series you obtained gives the proper coefficients.

Alternatively, use the fact that the generating function for the alternated Bernoulli numbers is
$$\frac{t}{e^t-1}=\sum\_{n=0}^\infty(-1)^nB\_n\frac{t^n}{n!}$$