Finding convergence of Series using Comparison Test

Question

Determine the Convergence of $$ \sum_{n=1}^\infty n\sin \left (\frac{1}{n^3}\right) $$ Can i compare it to $$ \left (\frac{1}{n^3}\right) \geq n\sin \left (\frac{1}{n^3}\right) $$ and since $\left (\frac{1}{n^3}\right) $ it is a p-series with p>1 then it is convergent and the smaller series also converges I'm not sure though how to prove that the original is smaller then $\left (\frac{1}{n^3}\right) $

Answer 1

When $x\in(0,\pi/2)$, $$\sin x<x.$$ So $\sin \frac{1}{n^3}<\frac{1}{n^3}$, and hence $$0<n\sin \frac{1}{n^3}<\frac{1}{n^2}.$$ Take $p=2$ of $p-$series will work.

And also you can use Limit Comparison Test: $$\lim_{n\to\infty}\frac{n\sin \left (\frac{1}{n^3}\right)}{\frac{1}{n^2}} =\lim_{n\to\infty}\frac{\sin \left (\frac{1}{n^3}\right)}{\frac{1}{n^3}}=1>0.$$