Answer
Divergent
Using the Limit Comparison Test
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{\sqrt[3]{n}}{n}$$
Use the Limit Comparison Test with $a_n = \dfrac{\sqrt[3]{n}}{n}$ and $b_n=\dfrac{1}{n^{2/3}}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{n^{2/3}n^{1/3}}{n }\\
&=\lim _{n \rightarrow \infty} \frac{n }{n }\\
&=1
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2/3}}$ is divergent $( p-\text{series} \ p<1)$ , then $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt[3]{n}}{n}$ is also divergent
