# This series converges or diverges? $\sum\limits_{n=1}^\infty n^{3/2}\sin\left(\frac{3}{n^3}\right)$

I am studying the convergence of the series: $$\sum\limits_{n=1}^\infty n^{3/2}\sin\left(\frac{3}{n^3}\right)$$

I have tried the following: As, $|\sin x|\le 1$ and term of the series are positves we have, $$\sum\limits_{n=1}^\infty n^{3/2}\sin\left(\frac{3}{n^3}\right) \le \sum\limits_{n=1}^\infty n^{3/2} =\infty$$

but this estimate leads neither to convergence nor to the divergence.

What is an elegant to prove the convergence or the divergence of this series? thanks in advance.

• use that $|\sin(x)|\leq |x|$. – Kelenner Nov 21 '17 at 16:51

As $|\sin(x)|\le |x|$ and as in your series you are feeding the $\sin$ positive numbers you get
$$\sum_n n^{3/2}\sin\left(\frac{3}{n^3} \right)\le 3\sum_n \frac{1}{n^{3/2}}\,,$$
$$\frac{n^{3/2}\sin\frac3{n^3}}{\frac{n^{3/2}}{n^3}}=3\,\frac{\sin\frac3{n^3}}{\frac3{n^3}}\xrightarrow[n\to\infty]{}3$$
and thus, by the limit comparison test, our series converges since $\;\sum\limits_{n=1}^\infty\frac{n^{3/2}}{n^3}=\sum\limits_{n=1}^\infty\frac1{n^{3/2}}\;$ converges.