Proving $\sum_{n=1}^\infty \frac{\cos \pi n}{2^n}$ converges by the comparison test

Question

$$\sum_{n=1}^\infty \frac{\cos \pi n}{2^n}$$

How can I show whether or not the following series converges using the comparison test?

Answer 1

For any $n$, $$ \bigg| \frac{\cos \pi n}{2^n} \bigg| \leq \frac{1}{2^n}. $$

So $$ 0\leq \bigg|\sum_{n=1}^\infty \frac{\cos \pi n}{2^n}\bigg| \leq \sum_{n=1}^\infty \bigg|\frac{\cos \pi n}{2^n}\bigg| \leq \sum_{n=1}^\infty \left( \frac{1}{2}\right)^n = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1 < \infty. $$ So the series converges.

Answer 2

Hint 1: $$|\cos(\pi n)| \le 1$$

Hint 2:

$$\sum_{n \ge 1} \frac{1}{2^n} < \infty$$