Answer
Converges
Work Step by Step
Given
$$ \sum_{k=1}^{\infty}\frac{\sin^2k }{k^2} $$
Compare with $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^{2}}$, which is a convergent series ( $p-$series with $p>1$) and for $k\geq 1$
\begin{align*}
\frac{\sin^2k }{k^2} &\leq \frac{1 }{k^2}
\end{align*}
Then $\displaystyle\sum_{k=1}^{\infty}\frac{\sin^2k }{k^2} $ also converges.
