## Thomas' Calculus 13th Edition

converges to $0$.
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists. Consider $a_n=\dfrac{\ln (n^2)}{n}$ Re-write the given sequence as:$a_n=\dfrac{2 \ln n}{n}$ Apply limits to both sides. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{2 \ln n}{n}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{\infty}{\infty}$ Since, we can see that the limit has the form of $\frac{\infty}{\infty}$, use L-Hospital's rule. $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{\frac{2}{n}}{1}$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\dfrac{2}{n}$ $\lim\limits_{n \to \infty}a_n=\dfrac{2}{\infty}$ $\lim\limits_{n \to \infty}a_n=0$ Therefore, the sequence converges to $0$.