Answer
converges to $1$
Work Step by Step
As we know that when $x \gt 0$ so, $\lim\limits_{n \to \infty} \sqrt[n] {n}=1$ and $\lim\limits_{n \to \infty} x^{1/n}=1$
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \sqrt[n] {n^2}$
This implies that $\lim\limits_{n \to \infty} a_n=\sqrt[n] {n^2}=(\lim\limits_{n \to \infty} \sqrt[n] {n})^2=1^2=1$
Thus, $\lim\limits_{n \to \infty} a_n=1$ and {$a_n$} converges to $1$.
