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

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Example of Absolutely Convergent Complex Series

The complex series defined as:

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

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


Proof

Suppose $S$ is absolutely convergent.

Then $\displaystyle \sum_{n \mathop = 1}^\infty \cmod {\dfrac z {1 - z} }^n$ 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:

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

$\Box$


It remains to be shown that $S' := \displaystyle \sum_{n \mathop = 1}^\infty \cmod {\dfrac z {1 - z} }^n$ 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:

\(\displaystyle S'\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty w^n\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac w {1 - w}\) Sum of Infinite Geometric Progression: Corollary 1

The result follows.

$\blacksquare$


Sources