Answer
converges to $0$
Work Step by Step
Given: $a_n=\frac{cos^{2}n}{2^{n}}$
Note that we can write
$\frac{0}{2^{n}} \leq \frac{cos^{2}n}{2^{n}}\leq \frac{1}{2^{n}}$
Also,
$0 \leq \frac{cos^{2}n}{2^{n}}\leq \frac{1}{2^{n}}$
Now, $\lim\limits_{n \to \infty}\frac{1}{2^{n}}=\frac{1}{\infty}=0$
Therefore, by Sandwich Theorem $\lim\limits_{n \to \infty}\frac{cos^{2}n}{2^{n}}=0$
Hence, the given sequence converges to $0$.
