1. convergent/divergent series

Hi,

How to check whether $\sum_{n=1}^{\infty}{n\sin}\frac{1}{n^3}$ is convergent/divergent? I could show that $\sum_{n=1}^{\infty}{n\sin}\frac{1}{n^2}$ is divergent, but I don't know how to start with the first one... Won't someone drop a hint, please?

Input appreciated.

2. Since $\sin(x)\approx{x}$ when $x\rightarrow{0}$

we have $n\cdot{\sin\left(\frac{1}{n^3}\right)}\approx{\fra c{1}{n^2}}$ when $n\rightarrow{+\infty}$ ( $\lim_{n\rightarrow{+\infty}}\frac{n\cdot{\sin\left (\frac{1}{n^3}\right)}}{\frac{1}{n^2}}=1$ )

But: $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ does converge, so, by the Limit Comparison Test, $\sum_{n=1}^{\infty}{n\cdot{\sin\left(\frac{1}{n^3} \right)}}$ converges

Again: $\lim_{n\rightarrow{+\infty}}\frac{n\cdot{\sin\left (\frac{1}{n^2}\right)}}{\frac{1}{n}}=1$

and $\sum_{n=1}^{\infty}{\frac{1}{n}}$ diverges, thus: $\sum_{n=1}^{\infty}{n\cdot{\sin\left(\frac{1}{n^2} \right)}}$ diverges

3. Great! The way you showed seems to be valid for any positive $\alpha$.

$
\sum_{n=1}^{\infty}{n\sin}\frac{1}{n^\alpha}
$