Answer
Divergent
Work Step by Step
Let $f(x)=\frac{1}{\sqrt[5] x}=x^{-1/5}$. Since $f$ is continuous, positive, and decreasing on $[1,\infty)$, we can use the Integral Test to determine whether the series is convergent or divergent.
We have $\int_1^\infty f(x)dx=\int_1^\infty x^{-1/5}dx=\lim\limits_{t \to \infty}\int_1^tx^{-1/5}dx=\lim\limits_{t \to \infty}\left[\frac{x^{4/5}}{4/5}\right]_1^t=\lim\limits_{t \to \infty}\frac{5}{4}\left(t^{4/5}-1\right)=\infty.$
Since the integral $\int_1^\infty f(x)dx$ diverges, the series $\displaystyle{\sum_{n=1}^\infty\frac{1}{\sqrt[5] n}}$ also diverges.