Answer
$1$
Work Step by Step
We have
$$ \lim\limits_{n \to \infty}{b_n}=\lim\limits_{n \to \infty}n^{1/n} .$$
By using the proerties of $\ln$ and L'Hopital's rule, we have
$$\ln b_n=\frac{\ln n}{n} \Longrightarrow \lim_{n\to \infty} \ln b_n=\lim_{n\to \infty} \frac{\ln n}{n} =0.$$
Hence, $ \lim\limits_{n \to \infty}{b_n}=e^0=1.$ Using Theorem 1, we get that the sequence $b_n$ converges to $1$.
