Prove Laurent series prove uniformly convergence if there are two convergent series I have tried some methods and think long, but I sitll have no idea how to prove the following problem.

Given $a_n\in\mathbb{C}$ and $(n\in \mathbb{Z})$, 
if there are $z_1,z_2$ with $|z_2|<|z_1|$ such that both $$\sum_{n=-\infty}^{\infty}a_n(z_1)^n$$ and $$\sum_{n=-\infty}^{\infty}a_n(z_2)^n$$ converge,
then $$\sum_{n=-\infty}^{\infty}a_n(z)^n$$ converge uniformly and absolutly 
on the closed annulus $\overline{B_{R1}(0)}$ \ $B_{R2}(0)$
for any $R_1,R_2$ such that $$|z_2|<R_2<R_1<|z_1|$$

 A: Let $S(z)$ be the series given by
$$\begin{align}
S(z)&=\sum_{n=-\infty}^\infty  a_nz^n \\\\
&=\sum_{n=1}^\infty  a_{-n}z^{-n}+\sum_{n=1}^\infty  a_{n}z^{n} \tag 1 
\end{align}$$
If $S(z)$ converges, then both series on the right-hand side of $(1)$ converge.

Now suppose $z_1$ and $z_2$ are complex numbers such that $|z_1|>|z_2|$.  Furthermore, suppose that the series $S(z_1)$ and $S(z_2)$ converge.  Then, the Root Test guarantees that 
$$\limsup_{n\to \infty}|a_n|^{1/n}\le \frac{1}{|z_{1}|} \tag 2$$
and 
$$\limsup_{n\to \infty}|a_{-n}|^{1/n}\le |z_2| \tag 3$$
Let $z$ be any complex number such that $|z_2|<|z|<|z_1|$.  Using $(2)$ and $(3)$, we find respectively that 
$$\limsup_{n\to \infty} |a_n z^n|^{1/n}<1$$
and
$$\limsup_{n\to \infty} |a_{-n} z^{-n}|^{1/n}<1$$
Therefore, the series $S(z)$ converges for all $z$ such that $|z_2|<|z|<|z_1|$.

Next, suppose that $z$ is in the annular region such that $|z_2|<R_2<|z|<R_1<|z_2|$.  Then, we can write
$$\begin{align}
\left|\sum_{n=-N^-}^{N^+} a_nz^n\right|&\le \sum_{n=-N^-}^{N^+} |a_n||z|^n \\\\
&=\sum_{n=1}^{N^-} |a_{-n}z^{-n}| +\sum_{n=0}^{N^+} |a_{n}z^{n}| \\\\
&<\sum_{n=1}^{N^-} |a_{-n}|R_2^{-n} +\sum_{n=0}^{N^+} |a_{n}|R_1^n \tag 4
\end{align}$$
Since the partial sums on the right-hand side of $(4)$ converge (to see this, simply use $(2)$ and $(3)$), then $S(z)$ converges absolutely.  Moreover, the Weierstrass M-test guarantees that the series $S(z)$ converges uniformly.
And we are done!
A: *

*If $\displaystyle \sum_{n= -\infty}^\infty c_n (z_1)^n$ converges then $c_n z_1^n \to 0$ as $n \to \pm \infty$, in particular there is a $A$ such that $|c_n z_1^n| < A$ 

*If $\displaystyle \sum_{n= -\infty}^\infty c_n (z_2)^n$ converges then $c_n z_2^n \to 0$ as $n \to \pm \infty$ and $|c_n z_2^n| < B$ 

*If $|z_1| < |z| < |z_2| $ then : 


*

*$|c_n z^n| = |z/z_2|^n |c_n z_2^n| < |z/z_2|^n B$ so that $\sum_{n=0}^\infty c_n z^n$ converges absolutely

*and $|c_n z^n| = |z/z_1|^n |c_n z_1^n| < |z/z_1|^n A$ so that $\sum_{n=-\infty}^{-1} c_n z^n$ converges  absolutely
Hence $$f(z) = \sum_{n=-\infty}^\infty c_n z^n$$
converges uniformly on $|z_1|+\epsilon < |z| < |z_2|-\epsilon $, as all its derivatives, and it is analytic on $|z_1| < |z| < |z_2| $
