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.
 A: First note that the proof you gave works perfectly well for $f(z)=\frac{1}{z+1} + \frac{1}{z-1}$. Your functions $g_n$ will not be defined in $D_{1+\epsilon}(0)$, but they are defined on an open set containing $0$, which is enough for your proof to go through. This however would yield a contradiction, since then you would have shown that $-1=\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=1$.
So what is the flaw in your proof? I see two. First, you assumed that none of the $a_n$ are $0$. Indeed if you look at the Taylor series for the function $f$ I defined, you will see that every even term is $0$. Thus the limit doesn't even exist.
Second, I'm not sure how you got that function $H$. Your proof is missing a lot of details, and when I try to reproduce it instead of $H$ I get
$$\frac{g_n'(0)}{g_n(0)}\frac{1}{n+1}$$
If you're looking for solutions, this has been asked before.
A: 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 ?
A: Hint: Suppose the pole of order $d\geq 1$. Then $(z-z_0)^d f(z)$ is holomorphic in a disc of radius $1+\epsilon$. Show that:
$$ f(z) = \sum_{j=1}^d \frac{p_j}{(z-z_0)^j} + g(z) $$
with $p_d\neq 0$ and $g$ holomorphic in a disc of radius $1+\epsilon$. Now, expand each term in a powerseries at $0$ and show that the series for $p_d(z-z_0)^{-d}$ (a negative binomial expansion) will dominate all other terms and give the provided limit ratio.
