## Thomas' Calculus 13th Edition

converges to $0$
Here, $\int_1^1 \dfrac{1}{n} dx$ and $[\ln x]_1^n=\ln n -\ln 1=\ln n-0=\ln n$ Let $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{\ln n}{n}$ But $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{\ln n}{n}=\dfrac{\infty}{\infty}$ This shows that the limit has an Indeterminate form so, we will use L-Hospital's rule. This implies that $\lim\limits_{n \to \infty} \dfrac{1}{n}=0$ Thus, {$a_n$} is Convergent and converges to $0$.