Chapter 8.4, Problem 17E

To determine

To graph: both the sequence of the terms and the sequence of partial sums on the same screen and use the graph to make a rough estimate of the sum of the series then use the Alternating Series Estimation Theorem to estimate the sum correct to four decimal place.

Answer to Problem 17E

  $S_{10}=-0.5507$ .

Explanation of Solution

Given information: The given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{{(-0.8)}^n}{n!}}$ .

Calculation:

  $\begin{array}{c}S={\displaystyle \sum _{n=1}^{\infty }\frac{{(0.8)}^n}{n!}}.\\ S={\displaystyle \sum _{n=1}^{\infty }\frac{{(0.8)}^n{(-1)}^n}{n!}}\\ b_n=\frac{{(0.8)}^n}{n!}\\ \text{Let's}\text{see}\text{if}\text{it}\text{is}\text{true}\text{that}{b}_{n+1}\le b_n.\\ \frac{{(0.8)}^{n+1}}{(n+1)!}\le \frac{{(0.8)}^n}{n!}\\ \frac{0.8}{n+1}\le 1\end{array}$

Since n >0 the above statement is always true, so condition (i) is true next condition (ii):

  $\begin{array}{l}\underset{n\to \infty }{\lim }{b}_n=0\\ \underset{n\to \infty }{\lim }\frac{{0.8}^n}{n!}=\frac{{0.8}^{\infty }}{\infty }=0\end{array}$

Because 0.8 is less than 1 it means the above limit will always be 0 so condition (ii) is also true:

The graphs are shown below using graphing utility.

  

Red graph shows the partial sum up to that value of n , blue graph shows sequence.

Looking at the graphs when x =10 the error is very small and it will not affect the first 4 decimal values, from the red graph when n =10, $S_{10}=-0.5507$ .

Conclusion then from the alternative series estimation is $S_{10}=-0.5507$ .