Answer
Divergent
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{\ln n}{n^{1/n}})$
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, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (\dfrac{\ln n}{n^{1/n}})= (\dfrac{\lim\limits_{n \to \infty}\ln n}{\lim\limits_{n \to \infty} n^{1/n}})=\dfrac{\infty}{n}=\infty$
Hence, $\lim\limits_{n \to \infty} a_n=\infty$ and {$a_n$} is Divergent
