Answer
The series is divergent.
Work Step by Step
$a_{n} = \sqrt[n] 2$
$a_{n} = (2)^{\frac{1}{n}}$
$\lim\limits_{n \to \infty}a_{n} = \lim\limits_{n \to \infty} (2)^{\frac{1}{n}}$
$=(2)^{\lim\limits_{n \to \infty} \frac{1}{n}}$
$=2^{0}$
$\lim\limits_{n \to \infty}a_{n} = 1 \ne 0$
The series is divergent.