Existence of Radius of Convergence of Complex Power Series

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 power series about $\xi$.

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

Then:


 * $(1) \quad $ If $z \in B_R \left({\xi}\right)$, then $S \left({z}\right)$ converges absolutely.


 * $(2) \quad$ If $z \notin B_R^- \left({\xi}\right)$, then $S \left({z}\right)$ is divergent.

Here, $B_R \left({\xi}\right)$ denotes the open $R$-ball of $\xi$, and $B_R^- \left({\xi}\right)$ denotes the closed $R$-ball of $\xi$.

If $R = +\infty$, we define $B_R \left({\xi}\right) = \C$.

Proof of First Result
Suppose that $z \in B_R \left({\xi}\right)$, so $\left\vert{z - \xi}\right\vert < R$.

By definition of radius of convergence, it follows that $S \left({z}\right)$ converges.

If $R$ is finite, let $\epsilon = R - \left\vert{z - \xi}\right\vert > 0$. Now, let $w \in B_R \left({\xi}\right)$ be a complex number such that $R - \left\vert{w - \xi}\right\vert = \dfrac \epsilon 2$. Such a $w$ exists because, if at a loss, we can always let $w=\xi+R-\dfrac \epsilon 2$.

If $R = +\infty$, then let $w \in C$ be any complex number such that $\left\vert{z - \xi}\right\vert < \left\vert{w - \xi}\right\vert$.

Then:

From the $n$th Root Test, it follows that $S \left({z}\right)$ converges absolutely.

Proof of Second Result
Suppose that $z \notin B_R^- \left({\xi}\right)$, so $\left\vert{z - \xi}\right\vert > R$.

By definition of radius of convergence, there exists $w \in \C$ such that $\left\vert{w - \xi}\right\vert < \left\vert{z - \xi}\right\vert$, and $S \left({w}\right)$ is divergent.

Then:

From the $n$th Root Test, it follows that $S \left({z}\right)$ is divergent.

Also see

 * Existence of Interval of Convergence of Power Series for a proof of the same result in real numbers.