Chapter 8.3, Problem 54E

To determine

To create: a table showing the sequence of the first 10 partial sums $S_1,S_2,S_3,\mathrm{...},S_{10}$ for the series using graphing utility.

Explanation of Solution

Given information:

Given sequence

  ${\displaystyle \sum _{n=1}^{\infty }4{\left(0.2\right)}^{n-1}}$

Calculation:

Now determine the partial sums $S_1,S_2,S_3,\mathrm{...},S_{10}$ by using a graphing utility as shown below:

$S_1={\displaystyle \sum _{n=1}^{1}4{\left(0.2\right)}^{n-1}}=4$

$S_2={\displaystyle \sum _{n=1}^{2}4{\left(0.2\right)}^{n-1}}=4.8$

$S_3={\displaystyle \sum _{n=1}^{3}4{\left(0.2\right)}^{n-1}}=4.96$

$S_4={\displaystyle \sum _{n=1}^{4}4{\left(0.2\right)}^{n-1}}=4.992$

$S_5={\displaystyle \sum _{n=1}^{5}4{\left(0.2\right)}^{n-1}}=4.9984$

Continue tabulating the partial sums

$S_6={\displaystyle \sum _{n=1}^{6}4{\left(0.2\right)}^{n-1}}=4.99968$

$S_7={\displaystyle \sum _{n=1}^{7}4{\left(0.2\right)}^{n-1}}=4.999936$

$S_8={\displaystyle \sum _{n=1}^{8}4{\left(0.2\right)}^{n-1}}4.9999872$

$S_9={\displaystyle \sum _{n=1}^{9}4{\left(0.2\right)}^{n-1}}=4.9999974$

$S_{10}={\displaystyle \sum _{n=1}^{10}4{\left(0.2\right)}^{n-1}}=4.99999648$