Answer
$\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}$
But $\lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}=\dfrac{\infty}{\infty}$
We need to apply L-Hospital's rule:
So, $\lim\limits_{n \to \infty} \dfrac{ 3^n}{n^3}=\lim\limits_{n \to \infty} \dfrac{3^n \ln 3}{3n^2}=\dfrac{\infty}{\infty}$
We need to apply L-Hospital's rule again:
$\lim\limits_{n \to \infty} \dfrac{3^n (\ln 3)^3}{6}=\infty$
Hence, $\lim\limits_{n \to \infty} a_n=\infty $ and {$a_n$} is divergent.
