Answer
Divergent
Work Step by Step
Consider $\lim\limits_{n \to \infty} n \sin (\dfrac{1}{n})=\lim\limits_{n \to \infty} \dfrac{\sin (\dfrac{1}{n})}{\dfrac{1}{n}}$
and $\lim\limits_{x \to 0} \dfrac{\sin x}{x}=1 \ne 0$
As per n-th term Test of Divergence, the series is divergent.