## University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=\dfrac{(\ln n)^2}{n^3}$ and $b_n=\dfrac{1}{ n^2}$ Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{(\ln n)^2}{n^3}}{\dfrac{1}{ n^2}}=\lim\limits_{n \to \infty} \dfrac{(\ln n)^2 }{n}$ and $\lim\limits_{n \to \infty} \dfrac{(\ln n)^2 }{n}=\dfrac{\infty}{\infty}$ Thus, we need to apply L-Hospital's rule. so, we have $\lim\limits_{n \to \infty} \dfrac{2 \ln n(1/n)}{1}=0$ Hence, the series converges by the comparison test.