Calculus 8th Edition

Converges to $0$
Given:$a_n=\frac{(lnn)^{2}}{n}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{(lnn)^{2}}{n}$ Apply L-Hospital's Rule. $=\lim\limits_{n \to \infty}\frac{2ln (n).\frac{1}{n}}{1}$ $=\lim\limits_{n \to \infty}\frac{2ln (n)}{n}$ Again apply L-Hospital's Rule. $=\lim\limits_{n \to \infty}\frac{2}{n}$ $=0$ Hence, the sequence converges to $0$.