Answer
$a_1=1^{\frac{1}{1}}=1$
$a_2=2^{\frac{1}{2}}=1.141$
$a_3=3^{\frac{1}{3}}=1.442$
$a_4=4^{\frac{1}{4}}=1.414$
$a_5=5^{\frac{1}{5}}=1.38$
$a_6=6^{\frac{1}{6}}=1.348$
The sequence converges to 1
Work Step by Step
Whether or not the sequence converges can be judged by calculating $\lim\limits_{n \to \infty}S_n$.
$\lim\limits_{n \to \infty}S_n = \lim\limits_{n \to \infty}n^{1/n} = e^{\lim\limits_{n \to \infty}\frac{\ln n}{n}}$.
Let's first calculate this $\lim\limits_{n \to \infty}\frac{\ln n}{n}$
By L'Hospital's Rule,
$\lim\limits_{n \to \infty}\frac{\ln n}{n}=\lim\limits_{n \to \infty}\frac{(\ln n)^{'}}{n^{'}}=\lim\limits_{n \to \infty}\frac{1/n}{1}=0$
Therefore,
$\lim\limits_{n \to \infty}S_n = e^{\lim\limits_{n \to \infty}\frac{\ln n}{n}}= e^0 = 1$
Thus, the series converges and it converges to one
