Answer
Divergent
Work Step by Step
$a_n=\Sigma sin(\frac{1}{n})$
$b_n=\Sigma \frac{1}{n}$
$\lim\limits_{n \to \infty}\frac{sin(\frac{1}{n})}{\frac{1}{n}}$
$t=\frac{1}{n}$
$\lim\limits_{n \to \infty}\frac{sin(\frac{1}{n})}{\frac{1}{n}}=\lim\limits_{n \to \infty}\frac{sin(t)}{t}=\lim\limits_{t \to 0}\frac{sin(t)}{t}= \lim\limits_{t \to 0}\frac{cos(t)}{1}=1$
So the series is divergent.