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:
- $\ds S = \sum_{n \mathop = 1}^\infty \paren {\dfrac z {1 - z} }^n$
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.
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.3$. Series: Example $\text{(iii)}$