Answer
Diverges.
Work Step by Step
Let us consider that $I_n=\int_2^n \dfrac{4}{\sqrt [4] x} \ dx$
After integrating, we have: $I_n=16[\dfrac{x^{3/4}}{3}]_2^n =16[\dfrac{n^{3/4}}{3}-\dfrac{2^{3/4}}{3}]$
Now, $I=\lim\limits_{n \to \infty} I_n=16 \lim\limits_{n \to \infty} 16[\dfrac{n^{3/4}}{3}-\dfrac{2^{3/4}}{3}]$
Since, $n^k$ is unbounded and positive for any value of $k$.
Therefore, $I=\infty$
This means that the limit does not exist. So, the given integral diverges.