The Radius of Convergence for Complex Power Series

Consider the following complex power series $\displaystyle{S^{z_0}(z) = \sum_{n=0}^{\infty} a_n(z - z_0)^n}$. The center of this power series is $z_0$, and we know that this series converges at $z = z_0$ since:

(1)

\begin{align} \quad S^{z_0}(z_0) = \sum_{n=0}^{\infty} a_n(z_0 - z_0)^n = \sum_{n=0}^{\infty} a_n 0^n = a_0 \end{align}

We would like to determine whether a generic power series converges for any other points $z \in \mathbb{C}$.

Theorem 1: Consider the complex power series $\displaystyle{\sum_{n=0}^{\infty} a_n(z - z_0)^n}$ and let $\displaystyle{R = \frac{1}{\limsup_{n \to \infty} \mid a_n \mid^{1/n}}}$ if $\displaystyle{\limsup_{n \to \infty} \mid a_n \mid^{1/n} \neq 0}$. Then for all $R'$ with $0 \leq R' < R$, $\displaystyle{\sum_{n=0}^{\infty} a_n(z - z_0)^n}$ converges uniformly on $D(z_0, R')$.

In this proof we will show that for each $R'$ with $0 \leq R' < R$ that we can define a sequence of nonnegative real numbers $(M_n)_{n=0}^{\infty}$ for which we can invoke the Weierstrass M-Test on to show that $\displaystyle{\sum_{n=0}^{\infty} a_n(z - z_0)^n}$ converges uniformly on $D(z_0, R')$.

• Proof: Let $\displaystyle{R = \frac{1}{\limsup_{n \to \infty} \mid a_n \mid^{1/n}}}$ and let $R' < R$. For each $n \in \{ 0, 1, ... \}$ let:

(2)

\begin{align} \quad M_n = \mid a_n \mid R'^n \end{align}

• Note that each $M_n \geq 0$. This is because $\mid a_n \mid \geq 0$ and $R' \geq 0$. For any $z \in \overline{D(z_0, R')}$ we have that:

(3)

\begin{align} \quad \mid a_n(z - z_0)^n \mid \leq \mid a_n \mid \mid R' \mid^n = M_n \end{align}

• Therefore:

(4)

\begin{align} \quad \limsup_{n \to \infty} M_n^{1/n} = \limsup_{n \to \infty} (\mid a_n \mid R'^n)^{1/n} = \limsup_{n \to \infty} \mid a_n \mid^{1/n} R' = \frac{R'}{R} < 1 \end{align}

• So by the root test for nonnegative series of real numbers, $\displaystyle{\sum_{n=0}^{\infty} M_n}$ converges. So by the Weierstrass M-Test we have that for all $R'$ with $0 \leq R' < R$ that $\displaystyle{\sum_{n=0}^{\infty} a_n(z - z_0)^n}$ converges uniformly on $D(z_0, R')$. $\blacksquare$.