Answer
Converges
Work Step by Step
Let $u_n=\dfrac{\sin ^2 n}{2^n}$
But $0 \leq \sin^2 n \leq 1$
This implies that $u_n \leq \dfrac{1}{2^n}$
Here, the series $\Sigma_{n=1}^\infty \dfrac{1}{2^n}$ is a geometric convergent series and has common ratio $r=\dfrac{1}{2}$
Thus, the series converges due to the limit comparison test.
