Answer
Converges
Work Step by Step
In fact, for all $n\geq 1$, we have $\sin^2(\frac{1}{n})\leq \frac{1}{n^2}$.
The series $\sum_{n=1}^\infty\frac{1}{n^2}$ converges since it is a $p-$series with $p=2$.
It follows by the Direct Comparison Test with $a_n=\sin^2(\frac{1}{n})$ and $b_n=\frac{1}{n^2}$ that the series $\sum_{n=1}^\infty \sin^2(\frac{1}{n})$ converges.
