## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty} \sin\frac{1}{n^2}$$ Since for $n\geq1$ $$\sin\frac{1}{n^2}\leq \frac{1}{n^2}$$ Compare with the convergent series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ ($p-$series, $p>1$), then: $$\sum_{n=1}^{\infty} \sin\frac{1}{n^2}$$ also converges.