$\sum_1^\infty\frac{sin(\frac{1}{n})}{\sqrt{n}}$
2. Let $b_n=\frac{1}{n^{3/2}}$. We know $\sum_{n=0}^\infty b_n$ converges, and we have $\lim_{n\rightarrow \infty} \left(\frac{\sin(1/n)}{\sqrt{n}}\right)/b_n =\lim_{n\rightarrow \infty} {n\sin(1/n)} = 1$ and therefore, by the limit comparison test your series converges.