Answer
converges to $1$
Work Step by Step
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 $1$.
