Answer
Converges to $0$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{1}{\sqrt n} tan^{-1} n$
Since, $ \lim\limits_{n \to \dfrac{\pi}{2}} \tan x=\dfrac{\pi}{2}$ and $\lim\limits_{n \to \infty} \dfrac{1}{\sqrt n}=0$
So, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{1}{\sqrt n} tan^{-1} n=(0) (\dfrac{\pi}{2})$
Hence, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is Convergent and converges to $0$
