University Calculus: Early Transcendentals (3rd Edition)

Converges to $0$
Since, we have $\int_1^1 \dfrac{1}{n} dx=[\ln x]_1^n=\ln n -\ln 1=\ln n$ Consider $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{\ln n}{n}$ But $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{\ln n}{n}=\dfrac{\infty}{\infty}$ Need to apply L-Hospital's rule. $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{\ln n}{n}=\lim\limits_{n \to \infty} \dfrac{1}{n}$ or, $=0$ Hence, $\lim\limits_{n \to \infty} a_n=0$ and {$a_n$} is Convergent and converges to $0$