Answer
$\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent.
Work Step by Step
Consider $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{\ln (n+1)}{\sqrt n}$
But $\lim\limits_{n \to \infty} \dfrac{\ln (n+1)}{\sqrt n}=\dfrac{\infty}{\infty}$
We need to apply L-Hospital's rule:
So, $\lim\limits_{n \to \infty} \dfrac{\ln (n+1)}{\sqrt n}=\lim\limits_{n \to \infty} \dfrac{ 2\sqrt n}{n+1}$
or, $\lim\limits_{n \to \infty} \dfrac{\dfrac{2}{\sqrt n}}{1+\dfrac{1}{n}}=0$
Hence, $\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent.
