Exercise: find the Laurent series of $\frac{1}{1+z}$ where $|z|>1$
my solution: $\frac{1}{1+z}= \frac{1}{z(1+\frac{1}{z})}$, since $|z|>1 \Rightarrow \frac{1}{|z|}<1$
$$\frac{1}{z(1+\frac{1}{z})}= \frac{1}{z}\sum_{n=0}^{+\infty }\left (\frac{-1}{z} \right )^n=\sum_{n=0}^{+\infty } \frac{(-1)^n}{z^{n+1}} $$
Is it correct ?
Yes, you are right; when you have this kind of problems the tip is trying to factorize in an estrategic way to have a geometric sequence.