Answer
Diverges
Work Step by Step
Since
\begin{aligned} S_{N} &=\frac{1}{\sqrt[3]{1}}+\frac{1}{\sqrt[3]{2}}+\cdots+\frac{1}{\sqrt[3]{N}} \\ & \geq \frac{1}{\sqrt[3]{N}}+\frac{1}{\sqrt[3]{N}}+\cdots+\frac{1}{\sqrt[3]{N}} \\ &=N\left(\frac{1}{\sqrt[3]{N}}\right)=N^{1 / 3} \end{aligned}
Then \begin{aligned} S &=\lim _{N \rightarrow \infty} S_{N} \\ &=\lim _{N \rightarrow \infty} N^{2 / 3} \\ &=\infty \end{aligned}
Hence $\sum_{k=1}^{\infty}\frac{1}{k^{1/3}}$ diverges
