infinite series - 3

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \sum_{k=1}^{\infty}\frac{4}{\sqrt[3]{k}}\)

 

[logistic_guy]

\(\displaystyle k \geq \sqrt[3]{k}\)


\(\displaystyle \frac{1}{k} \leq \frac{1}{\sqrt[3]{k}}\)


\(\displaystyle \sum_{k=1}^{\infty}\frac{4}{k} \leq \sum_{k=1}^{\infty}\frac{4}{\sqrt[3]{k}}\)


\(\displaystyle 4\sum_{k=1}^{\infty}\frac{1}{k} \leq \sum_{k=1}^{\infty}\frac{4}{\sqrt[3]{k}}\)


Since \(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k}\) is a divergent series, then \(\displaystyle \sum_{k=1}^{\infty}\frac{4}{\sqrt[3]{k}}\) \(\displaystyle \textcolor{blue}{\text{diverges}}\) as well.