Answer
Converges to $1$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \sqrt[n] {n^2}$
Since, $\lim\limits_{n \to \infty} \sqrt[n] {n}=1$ and $\lim\limits_{n \to \infty} x^{1/n}=1$ when $x \gt 0$
So, $\lim\limits_{n \to \infty} a_n=\sqrt[n] {n^2}=(\lim\limits_{n \to \infty} \sqrt[n] {n})^2=(1)^2=1$
Hence, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} is convergent and converges to $1$
