Existence of Radius of Convergence of Complex Power Series/Absolute Convergence
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 open $R$-ball of $\xi$.
Let $z \in \map {B_R} \xi$.
Then $\map S z$ converges absolutely.
If $R = +\infty$, we define $\map {B_R} \xi = \C$.
Proof
Let $z \in \map {B_R} \xi$.
By definition of the open $R$-ball of $\xi$:
- $\cmod {z - \xi} < R$
where $\cmod z$ denotes the complex modulus of $z$.
By definition of radius of convergence, it follows that $\map S z$ converges.
Suppose $R$ is finite.
Let $\epsilon = R - \cmod {z - \xi} > 0$.
Now, let $w \in \map {B_R} \xi$ be a complex number such that $R - \cmod {w - \xi} = \dfrac \epsilon 2$.
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 $\cmod {z - \xi} < \cmod {w - \xi}$.
Then:
\(\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n}\) | \(<\) | \(\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {w - \xi}^n}^{1/n}\) | as $\cmod {z - \xi} < \cmod {w - \xi}$, we have $\cmod {z - \xi}^n < \cmod {w - \xi}^n$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds 1\) | $n$th Root Test: $\map S w$ is not divergent |
From the $n$th Root Test, it follows that $\map S z$ converges absolutely.
$\blacksquare$
Also see
- Existence of Interval of Convergence of Power Series for a proof of the same result in real numbers.