Answer
Converges
Work Step by Step
We have: $a_k=\dfrac{\sin^2 (k)}{k^2}$ and $b_k=\dfrac{1}{k^2}$
Now, we can see that
$\dfrac{\sin^2 (k)}{k^2} \leq \dfrac{1}{k^2}$ for $0 \lt \sin^2 k \leq 1$
But the series $\Sigma \dfrac{1}{k^2}$ is a convergent p-series with $p=2$.
Therefore, the series converges by the comparison test.
