TITLE: Laurent series, radii of convergence.
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: I wanted to prove the convergence for both parts of the series, because if both parts converge, then also the whole Laurent series.

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.