Chapter 8.4, Problem 12E

To determine

To calculate: the first 10 partial sums of the given series and estimate the error in using the 10^th partial sum to approximate the total.

Answer to Problem 12E

  $s\in (0.900365,0.901867)$ .

Explanation of Solution

Given information: Given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^3}}$

Calculation:

Let $f(x)={\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{n^3}}$ .

Then, The first ten partial sums.

  $\begin{array}{l}f(1)=1\\ f(2)=0.875\\ f(3)=0.912037\\ f(4)=0.896412\\ f(5)=0.904412\\ f(6)=0.899782\\ f(7)=0.902697\\ f(8)=0.900744\\ f(9)=0.902116\\ f(10)=0.901116\end{array}$

Checking conditions for the Alternating Series Estimation Theorem.

  $\begin{array}{l}{b}_{n+1}\le b_n\\ \frac{1}{{(n+1)}^3}\le \frac{1}{n^3}.\\ $ {(n+1)}^3\ge n^3$n^3+3n^2+3n+1\ge n^3\\ 3n^2+3n+1\ge 0\\ \text{Because}\text{n}\text{is}\text{always}\text{positive}\text{,}\text{this}\text{condition}\text{is}\text{met}.\\ \underset{n\to \infty }{\lim }{b}_n=0\\ \underset{n\to \infty }{\lim }\frac{1}{n^3}=0\\ \text{Thus}\text{,}\text{this}\text{condition}\text{is}\text{met}\text{.}\end{array}$

Applying Alternating Series Estimation Theorem.

  $\begin{array}{c}\left|R_{10}\right|=\left|s-s_{10}\right|\le b_{11}\\ \left|s-s_{10}\right|\le b_{11}\\ \left|s-s_{10}\right|\le \frac{1}{{11}^3}\\ \left|s-s_{10}\right|\le \frac{1}{1331}\\ \left|s-0.901116\right|\le 0.000751\\ s\in (0.900365,0.901867).\end{array}$