Chapter 8.4, Problem 18E

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 18E

  $S_{10}=0.0988$ .

Explanation of Solution

Given information: The given series is ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{n}{8^n}}$ .

Calculation:

  $\begin{array}{c}S={\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{n}{8^n}}.\\ b_n=\frac{n}{8^n}\\ \text{Let's}\text{see}\text{if}\text{it}\text{is}\text{true}\text{that}{b}_{n+1}\le b_n.\\ \frac{n+1}{8^{n+1}}\le \frac{n}{8^n}\\ \frac{n+1}{8^n.8}\le \frac{n}{8^n}\\ \frac{n+1}{8}\le n\\ n+1\le 8n\\ 1\le 7n\end{array}$

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

  $\begin{array}{l}\underset{n\to \infty }{\lim }{b}_n=0\\ y=8^n\\ \text{then}\\ \ln y=n\ln 8.\\ \text{Differentiating}\text{with}\text{respect}\text{to}n.\\ y\text{'}\frac{1}{y}=\ln 8\\ y\text{'}=y\ln 8\\ y\text{'}=8^n\ln 8\\ \underset{n\to \infty }{\lim }\frac{n}{8^n}=\underset{n\to \infty }{\lim }\frac{1}{8^n\ln 8}=0\end{array}$

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, $S_{10}=0.0988$ .

Conclusion then from the alternative series estimation is $S_{10}=0.0988$ .