Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{\sin ^2 n}{2^n}$
We know that $0 \leq \sin^2 n \leq 1$
Now, $a_n \leq \dfrac{1}{2^n}$
and $\Sigma_{n=1}^\infty \dfrac{1}{2^n}$ is a geometric convergent series with $r=\dfrac{1}{2}$
Hence, the series converges due to the direct comparison test.
