Convergence of $\sum_{n=1}^\infty\sin^2\frac\pi n$ through comparison

Question

$$\sum_{n=1}^\infty\sin^2\frac\pi n$$ How can you show that this series converges using the limit comparison test?

Answer 1

The inequality that $|\sin u|\leq |u|$, so $\left|\sin\left(\dfrac{\pi}{n}\right)\right|^{2}\leq\dfrac{\pi^{2}}{n^{2}}$.

Other method: $\lim_{n\rightarrow\infty}\dfrac{\sin^{2}(\pi/n)}{(\pi/n)^{2}}=1$.