## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 11 - Infinite Series - 11.2 Summing and Infinite Series - Exercises - Page 548: 44

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

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.