If $f$ is holomorphic except for $z_0$, then $\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0$ The question is from Stein & Shakarchi - Complex Analysis Chapter 2, Exercise 14.

Suppose that $f$ is holomorphic in an open set $\Omega$ containing the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $f$ is given by a power series expansion
  $$f(z)=\sum^\infty_{n=0}a_n z^n$$
  in the unit disc $D_1(0)$, then
  $$\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0.$$

I solved this problem by using the pole formula
$$f(z)=(z-z_0)^{-m}g_0(z)\implies f^{(n)}(z)=(z-z_0)^{-m-n}g_n(z)$$
where $g_0$ and $g_n$ are holomorphic and not zero at $z_0$, and $m$ is a positive integer. These are defined on $D_{1+\epsilon}(0)\subset\Omega$, which contains $z_0$ and $\epsilon$ is sufficiently small. I've got below :
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}}=\lim_{n\to\infty} \dfrac{\dfrac{f^{(n+1)}(0)}{(n+1)!}}{\dfrac{f^{(n)}(0)}{(n)!}}=\lim_{n\to\infty}\dfrac{1}{n+1}(\dfrac{-m-n}{0-z_0}+H(0))=\dfrac{1}{z_0}.$$
where $H(z)$ is holomorphic in the disc.
Actually, the pole formula I've used appears in Chapter 3 so I should not use this, but I think it is worth to try. Is there something wrong about this solution?
Very thanks.

I see you are trying a different approach. It's good.
However your argument does not work since you do not use the condition that $f$ is holomorphic in an open set $\Omega$ containing the closed unit disc, except for a pole at $z_0$ on the unit circle.   
Since your argument is a local one, if $g_0(z)$ is defined on $D_1(0)\cup \{|z-z_0|<\delta \}$, holomorphic and not zero at $z_0$, your argument goes and concludes $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\frac{1}{z_0}$.  
If your argument is correct, you can prove:  
Suppose that $f$ is holomorphic in an open set $\Omega$ containing the closed unit disc, except for two poles at $z_0$ and $z_1$ on the unit circle. Show that if $f$ is given by a power series expansion 
$$f(z)=\sum^\infty_{n=0}a_n z^n$$
in the unit disc $D_1(0)$, then $$
\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0\quad \text{and}\quad \lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_1.$$
Proof. Consider $f(z)=(z-z_0)^{-m}g_0(z)$ on $D_1(0)\cup \{|z-z_0|<\delta \}$. Then $$
\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_0.$$
Consider $f(z)=(z-z_1)^{-m}g_1(z)$ on $D_1(0)\cup \{|z-z_1|<\delta \}$. Then 
$$
\lim_{n\to\infty} \frac{a_n}{a_{n+1}}=z_1.$$
Do you recognize your argument is wrong ?
