Convergence of a convergence series with $e^{in}$

Question

I am facing difficulties with this question: It says show by using the comparison test that the folowing complex series converges:

$$\sum_{n=1}^\infty \frac{\Re(e^{in\phi})}{2^n}$$

The $\Re $ refers to the real part of the complex number. Thank you for your help, I don't know where to start from.

Answer 1

Note that $$ |\Re (e^{in\phi})| = |\cos(n\phi)| \le 1 $$ Hence \begin{align*} \sum_{n=1}^\infty \left|\frac{\Re (e^{in\phi})}{2^n}\right| &\le \sum_{n=1}^\infty\frac{1}{2^n}\\ &= 1 < \infty \end{align*}