Answer
converges to $1$.
Work Step by Step
Let $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (8)^{\frac{1}{n}}$
as we know that $\lim\limits_{n \to \infty} x^{\frac{1}{n}}=1$ when $x \gt 0$
This shows that the limit has an indeterminate form so we will use L-Hospital's rule.
This implies that $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty} (8)^{\frac{1}{n}}=1$
Thus, $\lim\limits_{n \to \infty} a_n=1 $ converges to $1$.