Answer
\[\text{The series diverges}\]
Work Step by Step
\[\begin{align}
& \sum\limits_{k=1}^{\infty }{\frac{1}{\sqrt[3]{27{{k}^{2}}}}} \\
& \text{Rewrite the series} \\
& \sum\limits_{k=1}^{\infty }{\frac{1}{\sqrt[3]{27{{k}^{2}}}}}=\sum\limits_{k=1}^{\infty }{\frac{1}{3\sqrt[3]{{{k}^{2}}}}} \\
& \text{ }=\frac{1}{3}\sum\limits_{k=1}^{\infty }{\frac{1}{{{\left( {{k}^{2}} \right)}^{1/3}}}} \\
& \text{ }=\frac{1}{3}\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{2/3}}}} \\
& \text{Using the p-series test }\left( \text{Theorem 8}\text{.11}\text{, page 632} \right) \\
& \text{The }p\text{-series }\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{p}}}\text{ converges for }p>1\text{ and diverges for }p\le 1} \\
& \text{Comparing} \\
& \underbrace{\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{2/3}}}}}_{\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{p}}}}}\Rightarrow p=\frac{2}{3}<1 \\
& p<1,\text{ then the series diverges} \\
\end{align}\]
