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