Answer
$\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent.
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} \dfrac{\ln (n+1)}{\sqrt n}$
and $\lim\limits_{n \to \infty} \dfrac{\ln (n+1)}{\sqrt n}=\dfrac{\infty}{\infty}$
The limit shows an Indeterminate form so will use L-Hospital's rule.
This implies that, $\lim\limits_{n \to \infty} \dfrac{\ln (n+1)}{\sqrt n}=\lim\limits_{n \to \infty} \dfrac{ 2\sqrt n}{n+1}$
and, $\lim\limits_{n \to \infty} \dfrac{ 2\sqrt n}{n+1}=\lim\limits_{n \to \infty} \dfrac{\dfrac{2}{\sqrt n}}{1+\dfrac{1}{n}}=0$
Thus, $\lim\limits_{n \to \infty} a_n=0 $ and {$a_n$} is convergent.
