Answer
converges to $9$
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 $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \sqrt [n] {3^{2n+1}}$
This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \sqrt [n] {3^{2n+1}}$
and $a_n= (3^2) \lim\limits_{n \to \infty}(3^{1/n}) = 9 \cdot 1=9$
Thus, $\lim\limits_{n \to \infty} a_n=9$ and {$a_n$} converges to $9$
