Answer
The series $\mathop \sum \limits_{n = 1}^\infty {\sin ^2}\dfrac{\pi }{n}$ converges.
Work Step by Step
For large $n$, we have $\sin \dfrac{\pi }{n} \le \dfrac{\pi }{n}$. So,
${\sin ^2}\dfrac{\pi }{n} \le \dfrac{{{\pi ^2}}}{{{n^2}}}$
The larger series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{{\pi ^2}}}{{{n^2}}} = {\pi ^2}\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{{n^2}}}$ converges (because $p = 2 \gt 1$ by the Convergence of $p$-Series).
Therefore, by the Direct Comparison Test, the smaller series $\mathop \sum \limits_{n = 1}^\infty {\sin ^2}\dfrac{\pi }{n}$ also converges.
