Chapter 8.1, Problem 118E

a)

To determine

To evaluate: the fourth partial sum of the given infinite series.

Answer to Problem 118E

Fourth partial sum of ${\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}$ is ${\displaystyle \sum _{k=1}^4\frac{4}{{10}^k}}=\frac{2222}{5000}$ .

Explanation of Solution

Given information:

A summation expression is given as

  ${\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}$

Formula used:

The nth partial sum of ${\displaystyle \sum _{i=1}^{\infty }{a}_k}$ is ${\displaystyle \sum _{i=1}^n{a}_k}$ .

Calculation:

Consider the given infinite series.

  ${\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}$

Now, the fourth partial sum of ${\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}$ will be ${\displaystyle \sum _{k=1}^4\frac{4}{{10}^k}}$ .

Now, substitute 1, 2, 3, and 4 for $i$ in the expression $\frac{4}{{10}^k}$ and add those.

  $\begin{array}{c}{\displaystyle \sum _{k=1}^4\frac{4}{{10}^k}}=\frac{4}{{10}^1}+\frac{4}{{10}^2}+\frac{4}{{10}^3}+\frac{4}{{10}^4}\\ =\frac{4}{10}+\frac{4}{100}+\frac{4}{1000}+\frac{4}{10000}\\ =\frac{4\left(1000+100+10+1\right)}{10000}\\ =\frac{4\cdot 1111}{10000}\\ =\frac{2222}{5000}\end{array}$

Hence, the fourth partial sum of ${\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}$ is ${\displaystyle \sum _{k=1}^4\frac{4}{{10}^k}}=\frac{2222}{5000}$ .

b)

To determine

To evaluate: the sum of the given infinite series.

Answer to Problem 118E

The sum of the infinite series ${\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}$ is $\frac{4}{9}$ .

Explanation of Solution

Given information:

A summation expression is given as

  ${\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}$

Formula used:

Sum of an infinite G.P series will be ${\displaystyle \sum _{i=1}^{\infty }a{r}^{i-1}}$ is $\frac{a}{1-r}$ .

Here, a is the first term and $r$ is the common ration, could be found by

  $r=\frac{ar}{a}$

Calculation:

Consider the given infinite series.

  ${\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}$

Now, terms of the series will be

  $\frac{4}{{10}^1},\frac{4}{{10}^2},\frac{4}{{10}^3},\frac{4}{{10}^4},\mathrm{...}$

Now, first term of the series is $\frac{4}{{10}^1}$ .

Common ratio can be found by

  $\begin{array}{c}r=\frac{\frac{4}{{10}^2}}{\frac{4}{{10}^1}}\\ =\frac{1}{10}\end{array}$

So, sum of the infinite series is calculated as shown:

  $\begin{array}{c}{\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}=\frac{\frac{4}{10}}{1-\frac{1}{10}}\\ =\frac{\frac{2}{5}}{\frac{9}{10}}\\ =\frac{2}{5}\times \frac{10}{9}\\ =\frac{4}{9}\end{array}$

Hence, the sum of the infinite series ${\displaystyle \sum _{k=1}^{\infty }\frac{4}{{10}^k}}$ is $\frac{4}{9}$ .