Answer
Converges to $9$
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \sqrt [n] {3^{2n+1}}$
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, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \sqrt [n] {3^{2n+1}}= \lim\limits_{n \to \infty} (3^2)(3^{1/n}) =(9)(1)=9$
Hence, $\lim\limits_{n \to \infty} a_n=9$ and {$a_n$} is Convergent and converges to $9$
