Answer
$\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}$
But $\lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}=\dfrac{\infty}{\infty}$
The limit shows an Indeterminate form.
So, we will use L-Hospital's rule.
, $\lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}=\lim\limits_{n \to \infty} \dfrac{3^n \ln 3}{3n^2}$
or, $\lim\limits_{n \to \infty} \dfrac{3^n \ln 3}{3n^2}=\dfrac{\infty}{\infty}$
Apply L-Hospital's rule again.
$\lim\limits_{n \to \infty} \dfrac{3^n (\ln 3)^3}{(6)}=\infty$
Thus, $\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
