I'm working on the following exercise:
Prove that a Laurent series \begin{align*} \sum_{n = -\infty}^\infty a_n(z-z_0)^n = \sum_{n = 0}^\infty a_n(z-z_0)^n + \sum_{n = 1}^\infty a_{-n}(z-z_0)^{-n} \end{align*} converge for all $z$ such that $r < |z-z_0| < R$, where \begin{align*} r = \limsup |a_{-n}|^{1/n}, \quad R^{-1} = \limsup |a_n|^{1/n}, \end{align*} with $0 \le r$, $R \le \infty$.
I already tried the ratio test, but I didn't find the limit to be $<1$ and also with comparison test I didn't reach a result. Maybe there is another approach which I don't know. I wanted to prove the convergence for both parts of the series, because if both parts converge, then also the whole Laurent series.
Thanks.
This is a very good idea, indeed:
The power series $\sum_{n=0}^\infty a_n(z-z_0)^n$ converges in the disk $|z-z_0|<R$ where $R$ is given by the Cauchy-Hadamard formula $$R^{-1}= \limsup |a_n|^{1/n} $$
The series of negative terms can be transformed into a power series with positive powers using the substitution $\frac{1}{z-z_0}=w$. The series then becomes $$\sum_{n=1}^\infty a_{-n} w^n $$ and it converges in the disk $|w|<r$, where, again by Cauchy-Hadamard's formula $r^{-1}=\limsup |a_{-n}|^{1/n}$. Clearly the disk $|w|<r$ is equivalent to the annulus $\left| \frac{1}{z-z_0}\right|<r$.
Combining the above results, the problem is solved.