Answer
converges to $0$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} n (1-\dfrac{1}{n})$
But $\lim\limits_{n \to \infty} n (1-\dfrac{1}{n})=\dfrac{0}{0}$
This shows that the limit has an Indeterminate form, so we will use L-Hospital's rule.
This implies that $ \lim\limits_{n \to \infty} \dfrac{\sin (1/n) \cdot (1/n^2)}{(1/n^2)}=\lim\limits_{n \to \infty} \sin (1/n)$
and $\lim\limits_{n \to \infty} \sin (\dfrac{1}{n})=0$
Thus, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} converges to $0$.
