Existence of Radius of Convergence of Complex Power Series/Divergence

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Theorem

Let $\xi \in \C$.

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

Let $R$ be the radius of convergence of $S \left({z}\right)$.


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

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

Then $S \paren z$ is divergent.


Proof

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

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:

\(\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n }^{1/n}\) \(>\) \(\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n \paren {w - \xi}^n }^{1/n}\)
\(\displaystyle \) \(\ge\) \(\displaystyle 1\) $n$th Root Test: $S \paren w$ is divergent

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

$\blacksquare$


Also see