Radius of Convergence of Geometric Progression in Complex Plane

Theorem

Consider the complex power series:

$S = \displaystyle \sum_{k \mathop = 0}^\infty z^n$

The radius of convergence $S$ is $1$.

Proof

 $\displaystyle \lim_{n \mathop \to \infty} \dfrac {\cmod {z^n} } {\cmod {z^{n - 1} } }$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {z^n} {z^{n - 1} } }$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \cmod z$ $\displaystyle$ $=$ $\displaystyle \cmod z$

By the Ratio Test, it follows that:

$S$ is convergent for $\cmod z < 1$
$S$ is divergent for $\cmod z > 1$.

Hence the result by definition of radius of convergence.

$\blacksquare$