University Calculus: Early Transcendentals (3rd Edition)

$\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is convergent.
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.