# Show the limit converges to $z_0$

The problem 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 $$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=z_0.$$

This is already answered here. But in the answer, this and this, it say that

$$\lim_{n \to \infty} \frac{\displaystyle b_n z_0^n + \frac{c}{z_0}}{\displaystyle b_{n+1} z_0^n + \frac{c}{z_0^2}}=z_0$$

But I can't see. Why this is true?

• Note: Limits don't converge. They either exist or don't and when they do, they are just numbers that sit there.
– zhw.
Jun 2, 2021 at 4:32

You have $$\frac{a_n}{a_{n+1}} = \frac{ b_n + \frac{c}{z_0^{n+1}}}{ b_{n+1}+ \frac{c}{z_0^{n+2}}} = z_0 \cdot \frac{\frac{z_0^{n+1}b_n}{c} + 1}{\frac{z_0^{n+2}b_{n+1}}{c} + 1}$$ where

• $$|z_0| = 1$$,
• $$b_n \to 0$$, and
• $$c \ne 0$$.

Then $$\frac{z_0^{n+1}b_n}{c} \to 0$$ and $$\frac{z_0^{n+2}b_{n+1}}{c} \to 0$$, so that $$\frac{a_n}{a_{n+1}} \to z_0$$ for $$n \to \infty$$.

It's because the series $$\sum_{n=0}^{+ \infty} b_nz_0^n$$ converges; therefore, we have $$b_nz_0^n \underset{n \rightarrow \infty}{\rightarrow} 0$$.