Show that $\lim\limits_n \frac{a_n}{a_{n+1}} = z_0$ I was wondering if anyone could give me a hint on the following problem:

Let $f$ be meromorphic on the unit disc with only a simple pole at $0 \neq z_0 \in D$.  Let $a_0 +a_1z + \cdots$ be a power series representation for $f$ about $0$.  Show that $z_0 = \lim\limits_n \frac{a_n}{a_{n+1}}$ 

I have no idea what to do.  Since $z_0$ is the first place where $f$ isn't holomorphic, if the limit of $\frac{|a_n|}{|a_{n+1}|}$ exists, then this limit should be $|z_0|$.
 A: The function $f(z)$ could be represented in the unit disc as $$f(z) = \frac c {z - z_0} + h(z)$$ where $h(z)$ is a holomorphic function and $c \neq 0 $ is some constant because $z_0$ is the only pole in the unit disc.
Because $h(z)$ is holomorphic it could be represented with a Taylor series in the unit disc: $$h(z) = \sum\limits_{n=0}^{\infty} b_n z^n,$$
and the function $\frac {c} {z - z_0}$ could be expanded using the formula for infinite geometric series:
 $$\frac c {z -z_0} = -\frac{c}{z_0}\frac{1}{1 - z/z_0} = - \frac{c}{z_0}\sum_{n=0}^{\infty}\left(\frac{z}{z_0}\right)^n.$$
So we could write $f(z)$ as a series like 
$$f(z) = \sum\limits_{n=0}^{\infty}
\left(b_n - \frac{c}{z_0^{n+1}}\right)z^n \equiv 
\sum\limits_{n=0}^{\infty}a_n z^n$$
Because the coefficients near the same powers of $z$ should be the same in each series
$$\frac{a_n}{a_{n+1}} = 
\frac{b_n - \frac{c}{z_0^{n+1}}} {b_{n+1} - \frac{c}{z_0^{n+2}}} = 
z_0 \frac{(b_n z_0^n) z_0 - c} {(b_{n+1}z_0^{n+1}) z_0 - c} =
z_0 \frac{z_0 h_n - c}{z_0 h_{n+1} - c}$$
where the $h_n$ is the $n$-th term in the Taylor expansion of $h(z)$ evaluated in the point $z=z_0$.
The Taylor series for $h(z)$ converges for $z=z_0$ because $h(z)$ is holomorphic in the unit disk, so $\lim_{n\to \infty}\limits h_n = 0$ and
$$\lim_{n\to \infty}\limits \frac{a_n}{a_{n+1}} = 
\lim_{n\to \infty}\limits \left(
z_0 \frac{z_0 h_n - c}{z_0 h_{n+1} - c} \right) = z_0.$$
