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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?

one potato two potato
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  • Note: Limits don't converge. They either exist or don't and when they do, they are just numbers that sit there. – zhw. Jun 02 '21 at 04:32

2 Answers2

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

Jujustum
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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 $n \to \infty$.

Martin R
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