A problem regarding meromorphic functions I was trying to solve problems from Gamelin's complex analysis book, and I came across the following question:
Suppose $f(z) = \sum_{k} a_k z^k$ is analytic for $|z| < R$, and suppose that $f$ extends to be meromorphic for $|z| < R+\epsilon,$ with only one pole $z_0$ on the circle $|z|=R$. Prove that $\frac {a_k}{a_{k+1}}\rightarrow z_0$ as $k\rightarrow \infty$.
Now it's clear that $\lim_{k\rightarrow \infty}|a_k/a_{k+1}| = |z_0| = R,$ as the limit is the radius of convergence for the power series of $f$, and $f$ extends to be analytic only upto $|z|< R$. But I can't figure out how to prove the actual statement. I was trying to work with the Laurent series of $f$ near $z_0$, but I got stuck. Any help would be appreciated. 
 A: Let $h$ be the principal part of $f$ in $z_0$. Then $f-h$ is holomorphic on the disk $\lvert z\rvert < R+\epsilon$ with Taylor series
$$(f-h)(z) = \sum_{n = 0}^\infty c_n z^n.$$
For the principal part we have
$$h(z) = \sum_{k = 1}^m \frac{b_k}{(z - z_0)^k},$$
where $m$ is the order of the pole. Now
\begin{align}
\frac{1}{(z-z_0)^k} &= \frac{(-1)^{k-1}}{(k-1)!}\biggl(\frac{d}{dz}\biggr)^{k-1}\frac{1}{z-z_0}\\
&= \frac{(-1)^{k-1}}{(k-1)!}\biggl(\frac{d}{dz}\biggr)^{k-1} \frac{-1}{z_0}\sum_{n = 0}^\infty \frac{z^n}{z_0^n}\\
&= \frac{(-1)^k}{(k-1)!} \sum_{n = 0}^\infty \frac{1}{z_0^{n+1}} \biggl(\frac{d}{dz}\biggr)^{k-1}z^n\\
&= \frac{(-1)^k}{(k-1)!} \sum_{n = k-1}^\infty \frac{1}{z_0^{n+1}} \frac{n!}{(n+1-k)!}z^{n+1-k}\\
&= (-1)^k \sum_{r = 0}^\infty \binom{r+k-1}{k-1}\frac{z^r}{z_0^{k+r}},
\end{align}
so
$$h(z) = \sum_{r = 0}^\infty \underbrace{\Biggl(\sum_{k = 1}^m (-1)^kb_k\binom{r+k-1}{k-1}\frac{1}{z_0^{k}}\Biggr)}_{d_r}\biggl(\frac{z}{z_0}\biggr)^r.$$
Then we have
$$a_n = c_n + \frac{d_n}{z_0^n}$$
and
$$\frac{a_n}{a_{n+1}} = \frac{c_nz_0^{n} + d_n}{c_{n+1}z_0^{n+1}+ d_{n+1}}\cdot z_0.$$
Now we know that $c_n z_0^n \to 0$ since $f-h$ converges in $z_0$, so the assertion follows if we can show that $\liminf\limits_{n\to\infty} \lvert d_n\rvert \geqslant \delta > 0$ and $\dfrac{d_n}{d_{n+1}} \to 1$. If $m = 1$, then we have
$$d_n = -\frac{b_1}{z_0}$$
independent of $n$, so the conditions hold. Let's now suppose $m > 1$.
For fixed $k$ we have
$$\binom{n+k}{k} \sim \frac{n^k}{k!},$$
so
$$d_n = (-1)^m b_m \binom{n+m-1}{m-1}z_0^{-m} + O(n^{m-2}) \sim (-1)^m \frac{b_m}{z_0^m (m-1)!} n^{m-1},$$
which shows a) $\lvert d_n\rvert \to +\infty$ if $m > 1$ and
$$\frac{d_n}{d_{n+1}} \sim \biggl(\frac{n}{n+1}\biggr)^{m-1} \to 1,$$
so the conditions also hold for $m > 1$.
