Chapter 10.5, Problem 2E

To determine

To Explain: that is it convergence or divergence of the series.

Answer to Problem 2E

Converges

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }{n}^{-3/2}}$

Calculation:

Determine the improper integral

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{n}}={\displaystyle \sum _{n=1}^{\infty }{n}^{-3/2}}}\\ \underset{a\to \infty }{\lim }{{\displaystyle {\int }_1^a{x}^{-3/2}dx=\underset{a\to \infty }{\lim }\left[-2x^{-1/2}\right]}}_1^a\\ =\underset{a\to \infty }{\lim }-2\left[\frac{1}{\sqrt{a}}-1\right]\\ =2\end{array}$

The integral converges, therefore the addition converges also.