## Thomas' Calculus 13th Edition

$\lim\limits_{n \to \infty} a_n=\infty$ and {$a_n$} is divergent.
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.