Answer
Diverges
Work Step by Step
Let $u_n=\dfrac{1}{2 \sqrt n+\sqrt [3] n}$ and $v_n=\dfrac{1}{\sqrt n}$
$\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{1}{2 \sqrt n+\sqrt [3] n}}{1/\sqrt n}$
$\implies \lim\limits_{n \to \infty} \dfrac{\sqrt n}{2 \sqrt n+\sqrt [3] n}=\lim\limits_{n \to \infty} \dfrac{1}{2 +\sqrt [3] n/\sqrt n}=\dfrac{1}{2}$
Thus, the series diverges due to the limit comparison test.