# Absolutely Convergent Complex Series/Examples/(z over (1-z))^n

## Example of Absolutely Convergent Complex Series

The complex series defined as:

$\ds S = \sum_{n \mathop = 1}^\infty \paren {\dfrac z {1 - z} }^n$

is absolutely convergent, provided $\Re \paren z < \dfrac 1 2$.

## Proof

Suppose $S$ is absolutely convergent.

Then $\ds \sum_{n \mathop = 1}^\infty \cmod {\dfrac z {1 - z} }^n$ is convergent.

By Terms in Convergent Series Converge to Zero, this means that:

$\lim_{n \mathop \to \infty} \cmod {\dfrac z {1 - z} }^n \to 0$

which means in turn that:

 $\ds \cmod {\dfrac z {1 - z} }$ $<$ $\ds 1$ $\ds \leadsto \ \$ $\ds \cmod z^2$ $<$ $\ds \cmod {1 - z}^2$ $\ds \leadsto \ \$ $\ds x^2 + y^2$ $<$ $\ds \paren {1 - x}^2 + y^2$ Definition of Complex Modulus, where $z = x + i y$ $\ds \leadsto \ \$ $\ds x^2 + y^2$ $<$ $\ds 1 - 2 x + x^2 + y^2$ $\ds \leadsto \ \$ $\ds 0$ $<$ $\ds 1 - 2 x$ $\ds \leadsto \ \$ $\ds 2 x$ $<$ $\ds 1$ $\ds \leadsto \ \$ $\ds x$ $<$ $\ds \dfrac 1 2$

$\Box$

It remains to be shown that $S' := \ds \sum_{n \mathop = 1}^\infty \cmod {\dfrac z {1 - z} }^n$ is in fact a convergent series when $z < \dfrac 1 2$.

When $z < \dfrac 1 2$, we have that $\cmod {\dfrac z {1 - z} } < 1$, from above.

Let $w = \cmod {\dfrac z {1 - z} }$.

Then we have that:

 $\ds S'$ $=$ $\ds \sum_{n \mathop = 1}^\infty w^n$ $\ds$ $=$ $\ds \dfrac w {1 - w}$ Sum of Infinite Geometric Sequence: Corollary 1

The result follows.

$\blacksquare$