Answer
Divergent
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 $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{\ln n}{n^{1/n}})$
This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{\ln n}{n^{(1/n)}})$
and $a_n= \dfrac{\lim\limits_{n \to \infty}\ln n}{\lim\limits_{n \to \infty} n^{(1/n)}}=\infty$
Hence, $\lim\limits_{n \to \infty} a_n=\infty$ and {$a_n$} is Divergent.
