Answer
Converges to $8$
Work Step by Step
Given: $a_n=\sqrt[n] {2^{1+3n}}$
It can be re-written as
$a_n= ({2^{1+3n}})^{1/n}=2^(\frac{{1+3n}}{n})$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}2^(\frac{{1+3n}}{n})$
$=\lim\limits_{n \to \infty}2^{(\frac{1}{n}+\frac{{3n}}{n})}$
$=\lim\limits_{n \to \infty}2^{(\frac{1}{n}+3)}$
$=2^{(0+3)}$
$=2^{3}$
$=8$
Therefore, the given sequence converges to $8$.
