28 - Radius of Convergence

Radius of Convergence

Let $\sum c_k{z}^k$ be a power series
Radius of Convergence $R$ is defined as

$$R=\{\begin{array}{ll}\sup \left\{|z|\in \mathbb{R}:\sum |c_k{z}^k|\text{ converges}\right\}& \text{ if the }\sup \text{ exists}\\ & \\ \infty & \text{otherwise}\end{array}$$

Note that the set is not empty as $0$ is always on it

Radius of Convergence with Convergence

Let $\sum c_k{z}^k$ be a power series with radius of convergence $R$

1. If $R>0$ and $|z|<R$ then

$$\sum c_k{z}^k\text{ converges absolutely and hence converges}$$

2. If $|z|>R$ then

$$\sum c_k{z}^k\text{ diverges}$$

Proof

1. Case $R\in \mathbb{R}$
   Assume that $R>0$ and let $z\in \mathbb{C}$ with $|z|<R$
   Then there is $S$ with $|z|<S<R$
   Let $ϵ=R-S>0$
   Since $R=\sup \left\{|w|\in \mathbb{R}:\sum |c_k{w}^k|\text{ converges}\right\}$
   By Approximation Property there is $\rho$ such that

$$S=R-ϵ<\rho \le R\text{ and }\sum |c_k{\rho }^k|\text{ converges}$$

Then $0\le |z|<\rho$ and $\sum |c_k{\rho }^k|$ converges so by Comparison Test

$$\sum |c_k{z}^k|\text{ converges}$$

Since absolute convergence implies convergence then

$$\sum c_k{z}^k\text{ converges}$$

Case $R=\infty$
Similar to **Case $R\in \mathbb{R}$

2. Let $z\in \mathbb{C}$ with $|z|>R$
   Suppose for a contradiction that $\sum c_k{z}^k$ converges
   Then $c_k{z}^k\to 0$ as $k\to \infty$, hence $(c_k{z}^k)$ is bounded
   So there is $M$ such that

$$|c_k{z}^k|\le M\text{ for all }k$$

For $R<\rho <|z|$ then

$$0\le |c_k{\rho }^k|\le |c_k{z}^k|{\left|\frac{\rho }{z}\right|}^k\le M{\left|\frac{\rho }{z}\right|}^k$$

As $\sum {\left|\frac{\rho }{z}\right|}^k$ converges then by Comparison Test $\sum |c_k{\rho }^k|$ converges
This contradicts the definition of $R$

Disc of Convergence for the Power Series

$$\{z\in \mathbb{C}:|z|<R\}$$

where $R$ is the Radius of Convergence