How can I compute this infinite series with cosines?

Question

$\sum\limits_{h=1}^\infty (\frac{1}{2})^h\cos(\lambda h)$ where $\lambda$ is a real number.

According to WolframAlpha the answer is $\frac{1-2\cos(\lambda)}{4\cos(\lambda)-5}$

Thanks!

Answer 1

Recall that $\cos(t) = \dfrac{e^{it} + e^{-it}}2$. Hence, we have $$\sum_{h=1}^{\infty} \dfrac{\cos(\lambda h)}{2^h} = \dfrac12\sum_{h=1}^{\infty} \dfrac{e^{i \lambda h} + e^{-i \lambda h}}{2^{h}} = \dfrac12\left(\dfrac{e^{i \lambda}/2}{1-e^{i \lambda}/2} + \dfrac{e^{-i \lambda}/2}{1-e^{-i \lambda}/2}\right)=\dfrac{1-2\cos(\lambda)}{4 \cos(\lambda)-5}$$