Convergent Complex Series/Examples/((-1)^n + i cos n theta) over n^2
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Example of Convergent Complex Series
The series $\ds \sum_{n \mathop = 1}^\infty a_n$, where:
- $a_n = \dfrac {\paren {-1}^n + i \cos n \theta} {n^2}$
is convergent.
Proof 1
\(\ds \dfrac {\paren {-1}^n + i \cos n \theta} {n^2}\) | \(=\) | \(\ds \dfrac {\paren {-1}^n} {n^2} + i \dfrac {\cos n \theta} {n^2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{n \mathop = 1}^\infty a_n\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \paren {\dfrac {\paren {-1}^n} {n^2} } + i \sum_{n \mathop = 1}^\infty \paren {\dfrac {\cos n \theta} {n^2} }\) |
Both of the terms on the right hand side are convergent real series.
Hence from Convergence of Series of Complex Numbers by Real and Imaginary Part, $\ds \sum_{n \mathop = 1}^\infty a_n$ is convergent.
$\blacksquare$
Proof 2
\(\ds \sum_{n \mathop = 1}^\infty \cmod {\dfrac {\paren {-1}^n + i \cos n \theta} {n^2} }\) | \(\le\) | \(\ds \sum_{n \mathop = 1}^\infty \dfrac {1 + \cmod {\cos n \theta} } {n^2}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{n \mathop = 1}^\infty \paren {\dfrac 2 {n^2} }\) |
Thus $\ds \sum_{n \mathop = 1}^\infty \paren {\dfrac {\paren {-1}^n + i \cos n \theta} {n^2} }$ is absolutely convergent.
The result follows from Absolutely Convergent Series is Convergent: Complex Numbers.
$\blacksquare$