## Calculus: Early Transcendentals 8th Edition

converges to $1$
Given: $a_n=\sqrt[n] n$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\sqrt[n] n$ Let $x=\lim\limits_{n \to \infty }\sqrt[n] n=n^{1/n}$ Take natural logarithmic to both sides. $lnx=\lim\limits_{n \to \infty }n^{1/n}$ $lnx=\lim\limits_{n \to \infty}\frac{lnn}{n}$ Use L-Hospital's Rule. $lnx=\lim\limits_{n \to \infty}\frac{1/n}{1}$ $lnx=0$ Raise the power to base $e$ $e^{lnx}=e^{0}$ $x=1$ Hence, the sequence converges to $1$.