Complex Analysis Question from Stein
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 $\sum_{i=0}^\infty \frac{m-k}{m+i}=\infty$.
Suppose that $$ \begin{align} f(z)(z-z_0)^n &=(z-z_0)^n\sum_{k=0}^\infty a_kz^k\\ &=\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}z^jz_0^{n-j}\sum_{k=0}^\infty a_kz^k\\ &=\sum_{k=0}^\infty\left(\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right)z^k \end{align} $$ converges in a neighborhood of $z_0$. Thus, for some $r\lt1$, we have $$ \left|\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right|\le c\,r^k $$ That is, the the $n^{\text{th}}$ finite difference of the sequence $a_kz_0^k$ satisfies $$ \Delta^na_kz_0^k=O(r^k) $$ Inverting the finite difference operator yields that there is an $n-1$ degree polynomial $P$ so that $$ a_kz_0^k=P(k)+O(r^k) $$ Taking the limit of the ratio of terms yields $$ \lim_{k\to\infty}\frac{a_kz_0^k}{a_{k+1}z_0^{k+1}}=\lim_{k\to\infty}\frac{P(k)+O(r^k)}{P(k+1)+O(r^{k+1})}=1 $$ That is, $$ \lim_{k\to\infty}\frac{a_k}{a_{k+1}}=z_0 $$
Inverting Finite Difference Operators
If we define the finite difference operator as $$ \Delta a_k=a_k-a_{k-1} $$ then, as with indefinite integrals, when inverting $\Delta$, we need to include a constant: $$ \Delta^{-1}a_k=c+\sum_{j=1}^k a_j $$ Suppose that $\Delta a_k=P_n(k)+O(r^k)$, where $P_n$ is a degree $n$ polynomial and $0\le r\lt1$. Since the sum of a degree $n$ polynomial is a degree $n+1$ polynomial and the sum of $O(r^k)$ is $c+O(r^k)$, $a_k=P_{n+1}(k)+O(r^k)$.
Iterating, we get that if $\Delta^n a_k=O(r^k)$, then $a_k=P_{n-1}(k)+O(r^k)$.