Let \(a\in\mathbb R\) with \(|a|<1\). Prove that \[ \sum_{k=1}^{\infty}a^k\cos(k\theta) =\frac{-a^2+a\cos\theta}{1+a^2-2a\cos\theta}. \]
Solution
We use the fact that for any complex number \(z = e ^ { i \theta } = \cos \theta + i \sin \theta \in \mathbb { C }\)
\[ { \frac { 1 } { 1 - a z } } = \sum _ { k = 0 } ^ { \infty } a ^ { k } z ^ { k } = \sum _ { k = 0 } ^ { \infty } a ^ { k } e ^ { i k \theta } = 1 + \sum _ { k = 1 } ^ { \infty } a ^ { k } ( \cos ( k \theta ) + i \sin ( k \theta ) ) . \]
Therefore
\[ \begin{array} { c } { { \displaystyle \sum _ { k = 1 } ^ { \infty } a ^ { k } \cos ( k \theta ) = \Re \left( \displaystyle \frac { 1 } { 1 - a z } - 1 \right) = \Re \left( \displaystyle \frac { a z } { 1 - a z } \right) = \Re \left( \displaystyle \frac { a z ( 1 - a \bar { z } ) } { | 1 - a z | ^ { 2 } } \right) } } \\ { { = \Re \left( \displaystyle \frac { a \bar { z } - a ^ { 2 } } { ( 1 - a \cos \theta ) ^ { 2 } + ( a \sin \theta ) ^ { 2 } } \right) = \displaystyle \frac { a \cos \theta - a ^ { 2 } } { 1 + a ^ { 2 } - 2 a \cos \theta } . } } \end{array} \]