Convergent Complex Series/Examples/((-1)^n + i cos n theta) over n^2

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.

$\blacksquare$