Answer
Converges to $0$
Work Step by Step
Given: $a_n=({2})^{-n}cos \pi n$
It can be re-written as
$a_n=\frac{cos n \pi}{{2}^{n}}$
$cos n \pi$ is a bounded function.
$-1 \leq cos n\pi\leq 1$
Place upper and lower bounds into inequality.
$\lim\limits_{n \to \infty}\frac{-1}{2^{n}} \leq \lim\limits_{n \to \infty}\frac{cos n \pi}{{2}^{n}}\leq \lim\limits_{n \to \infty}\frac{1}{2^{n}}$
$0 \leq \lim\limits_{n \to \infty}\frac{cos n \pi}{{2}^{n}}\leq 0$
Therefore, by Squeeze Theorem the given sequence converges to $0$.
