## Calculus 8th Edition

Let $a_{n}=\sqrt[n] 2-1$ and $b_{n}=\frac{1}{n}$ $\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty} \frac{\sqrt[n] 2-1}{1/n}=\lim\limits_{n \to \infty} \frac{\sqrt[n] 2}{1/n}$ [ Use L-Hospital's rule] $=\lim\limits_{n \to \infty}2^{1/n}ln2$ $=ln2$ The series is divergent.