a series derived from a holomorphic function converges implies that the coefficients converge to $0$

Question

Let $D=\{z\in\mathbb{C}\mid |z|<2\}$. Let $f:D\setminus\{\frac{i}{2}\}\longrightarrow \mathbb{C}$ be holomorphic with $f(z)=\sum_{n=0}^\infty a_nz^n$ for any $|z|<\frac{1}{2}$. Suppose $a_n\neq 0$ for all $n$. If $\frac{i}{2}$ is a simple pole of $f$, prove that $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=-2i$.

How to prove?

my attempt: a simple pole means that $\lim_{z\to \frac{i}{2}}(z-\frac{i}{2})f(z)=C\neq 0$. That is, $\lim_{z\to \frac{i}{2},|z|<\frac{1}{2}}(z-\frac{i}{2})\sum_{n=0}^\infty a_nz^n=\lim_{z\to \frac{i}{2},|z|<\frac{1}{2}}\sum_{n=0}^\infty (a_{n+1}-\frac{i}{2}a_n)z^n=C$. But I cannot obtain $\lim_{n\to\infty}a_{n+1}-\frac{i}{2}a_n=0$. How to solve?

Answer 1

Also with the approach taken in the question, $g(z)=(z-\frac i2)f(z)$ is holomorphic in the full disk $D$, so its power series in $z=0$ has a radius of convergence of at least $2$, so that $$ (a_{n-1}-\tfrac i2a_n)(2-\varepsilon)^n $$ is bounded for any $ε>0$.

Answer 2

Hint: if you subtract $c/(z - i/2)$ for some $c$, you get a function that is holomorphic in $D$.