Answer
Converges to $0$
Work Step by Step
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 $0$.
