Complex Analysis Question from Stein
The question is #$14$ from Chapter $2$ in Stein and Shakarchi's text Complex Analysis:
Suppose that $f$ is holomorphic in an open set containing the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $$\sum_{n=0}^\infty a_nz^n$$ denotes the power series expansion $f$ in the open unit disc, then $$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=z_0.$$
I've shown that we can take $z_0=1$ without a loss of generality, but I'm having trouble showing the proof otherwise. One of the problems I'm having is because we aren't told the definition of a pole except that it is a place where the function isn't holomorphic. Disregarding this fact, the other problem I'm running into is that I don't know the order of the pole.
Making some additional assumptions, including that the pole is simple so we can write $F(z)=(z-1)f(z)$ as a holomorphic function, we see that $$F(z)=-a_0+z(a_0-a_1)+z^2(a_1-a_2)+\cdots$$ This almost gets me to the end with these added assumptions, but I don't think it's quite enough (why do we know some of the $a_i$'s aren't $0$, for example).
On another note, if we know $\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ exists, then it is easy to see that $\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}=1=|z_0|$; I, however, do not see why the limit must exist.
Are there any hints that someone can provide? Even a solution would be nice, especially if one can avoid making any assumptions about what a pole is or is not.
EDIT: So there isn't any confusion, I know the definition of a pole and I'm inclined to believe that the problem, as stated, necessarily has a pole at $z_0$. The problem is that the exercise is in Chapter $2$, and poles are introduced in Chapter $3$.
I will assume we can write
$$f(z) = \frac{c}{z_0-z} + \sum_{n=0}^{\infty} b_n z^n$$
for some value of $c \ne 0$, and $\lim_{n \to \infty} b_n = 0$. Then
$$f(z) = \sum_{n=0}^{\infty} a_n z^n$$
where
$$a_n = b_n + \frac{c}{z_0^{n+1}}$$
Then
$$\begin{align}\lim_{n \to \infty} \frac{a_n}{a_{n+1}} &= \lim_{n \to \infty} \frac{\displaystyle b_n + \frac{c}{z_0^{n+1}}}{\displaystyle b_{n+1}+ \frac{c}{z_0^{n+2}}}\\ &= \lim_{n \to \infty} \frac{\displaystyle \frac{c}{z_0^{n+1}}}{\displaystyle \frac{c}{z_0^{n+2}}}\\ &= z_0\end{align}$$
as was to be shown. Note that the second step above is valid because $z_0$ is on the unit circle.
For a nonsimple pole, we may write
$$f(z) = \frac{c}{(z_0-z)^m} + \sum_{n=0}^{\infty} b_n z^n$$
for $m \in \mathbb{N}$. It might be known that
$$(1-w)^{-m} = \sum_{n=0}^{\infty} \binom{m-1+n}{m-1} w^n$$
Then
$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{n+1}{n+m} z_0$$
EDIT
@TCL observed that we can simply require that $b_n z_0^n$ goes to zero as $n \to \infty$. Then for a simple pole
$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{\displaystyle b_n z_0^n + \frac{c}{z_0}}{\displaystyle b_{n+1} z_0^n + \frac{c}{z_0^2}}$$
which you can see goes to $z_0$.
This is just an attempt to give a rigorous proof of Ron Gordon's answer.
Suppose the order of the pole at $z_0$ is $m$. Then for some constants $c_0,\ldots,c_m, c_m\neq 0,$ $$g(z)=f(z)-\left(\frac{c_m}{(z_0-z)^m}+\cdots +\frac{c_1}{(z_0-z)}\right)$$ is analytic at $z_0$. Since $f(z)$ is analytic on an open set containing the unit disk except at $z_0$, we see that $g(z)$ is analytic on an (possibly different) open set containing the unit disk. So $$g(z)=\sum_{n=0}^\infty b_nz^n$$ has radius of convergence greater than 1, in particular the series converges at $z_0$ and $\lim_n b_nz_0^n=0$. So for $|z|<1$, $$f(z)=\sum_{n=0}^\infty a_nz^n=\sum_{n=0}^\infty b_nz^n+\left(\frac{c_m}{(z_0-z)^m}+\cdots +\frac{c_1}{(z_0-z)}\right)$$ For $1\le k\le m$, we have $$\frac{c_k}{(z_0-z)^k}=\sum_{n=0}^\infty \frac{(k)_nc_k}{n!z_0^{n+k}} z^n$$ where $(k)_n=k(k+1)\cdots(k+n-1)$ and $(k)_0=1$. It follows that $$a_n=b_n+\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{n+k}}$$ and \begin{align*}\frac{a_n}{a_{n+1}}&=\frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{1+k}}+b_{n+1}z_0^n}\\ &= z_0 \frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{k}}+b_{n+1}z_0^{n+1}} \end{align*} Now divide the numerator and denominator of the last expression by $$\frac{(m)_{n+1}}{(n+1)!}$$, which is equal to 1 when $m=1$ and approaches infinity as $n\to\infty$ if $m>1$. Furthermore, for $1\le k\le m$, $$\frac{(k)_n}{n!}\cdot \frac{(n+1)!}{(m)_{n+1}}=\frac{(k)_n (n+1)}{(m)_n (n+m)}$$ approaches 0 as $n\to\infty$ if $k<m$ and approaches 1 if $k=m$. It follows that the fraction $$\frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{k}}+b_{n+1}z_0^{n+1}}$$ approaches 1 as $n\to\infty$, and the proof is complete.
EDIT. If $k<m$, $$\frac{(k)_n }{(m)_n }=\prod_{i=0}^{n-1} \frac{k+i}{m+i}=\prod_{i=0}^{n-1} \left(1-\frac{m-k}{m+i}\right)\to 0$$ because $\sum_{i=0}^\infty \frac{m-k}{m+i}=\infty$.