Chapter 9.5, Problem 24E

To determine

To calculate: The series converges absolutely, converges conditionally, or diverges.

Answer to Problem 24E

The series converges absolutely, and the truncation error is less than ${0.1}^{100}$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n+1}{(0.1)}^n}$

Calculation:

Let examine whether the series converges absolutely. If it does not, let use other tests in order to determine conditional convergence. The bound for the truncation error can be determined by using the Alternating Series Bound Theorem.

First let examine the series ${\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}$ . If this series converges, then the series converges absolutely.

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n+1}{(0.1)}^n}\\ {\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=1}^{\infty }{(0.1)}^n}\end{array}$

Since ${\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}$ is a geometric series with $r=0.1$ , and $\left|r\right|<1$ , it converges, from which the original series converges absolutely.

The truncation error can be determined by using the Alternating Series Bound Theorem. Based on this, the error after 99 terms is less than $u_{100}$ ,

Thus

  $\begin{array}{c}\text{ error }<u_{100}\\ $ ={(-1)}^{100}\cdot {(0.1)}^{100}$={0.1}^{100}\end{array}$

Therefore, the series converges absolutely, and the truncation error is less than ${0.1}^{100}$ .