Existence of Radius of Convergence of Complex Power Series/Divergence

Theorem

Let $\xi \in \C$.

Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n $ be a (complex) power series about $\xi$.

Let $R$ be the radius of convergence of $\map S z$.

Let $\map { {B_R}^-} \xi$ denote the closed $R$-ball of $\xi$.

Let $z \notin \map { {B_R}^-} \xi$.

Then $\map S z$ is divergent.

Let $z \notin \map { {B_R}^-} \xi$.

Then by definition of the closed $R$-ball of $\xi$:

$\cmod {z - \xi} > R$

where $\cmod z$ denotes the complex modulus of $z$.

By definition of radius of convergence, there exists $w \in \C$ such that:

$\cmod {w - \xi} < \cmod {z - \xi}$

and $S \paren w$ is divergent.

Then:

$\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n}$

$>$

$\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {w - \xi}^n}^{1/n}$

$\ds $

$\ge$

$\ds 1$

From the $n$th Root Test, it follows that $\map S z$ is divergent.

$\blacksquare$