sequences and series - converges or diverges? $sum_{n=1}^infty sin^2(frac{pi}{n})$

$$\sum_{n=1}^\infty \sin^2(\frac{\pi}{n})$$

First, I tried Divergence test, but the limit is $0$, so that's inconclusive. I don't think I can do integral test, since it's not a clean integration.

Did I do this right ? If it's wrong, can you steer me in the right direction? I tried subbing in $1-\cos(x)$ for the $\sin(x)$. This would get:

$$\sum_{n=1}^\infty \sin^2(\frac{\pi}{n})$$
$$= \sum_{n=1}^\infty 1-\cos^2(\frac{\pi}{n})$$
$$= \sum_{n=1}^\infty 1 - \sum_{n=1}^\infty \sin^2(\frac{\pi}{n})$$
But that first sigma is divergent to infinity, so does the original series also diverge to infinity?

The tests I know are: Geometric, p-series, Divergence (nth term) test, Integral test, Direct Comparison test, Alternating Series Test, Absolute convergence, and Ratio test.

Answer

Using $\sin(x)\le x$ for all $x\ge0$ so by comparison with the convergent series $\sum\frac1{n^2}$ we conclude the convergence of the given series.