Chapter 8, Problem 24RE

a)

To determine

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

Answer to Problem 24RE

Fourth partial sum of ${\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}$ is ${\displaystyle \sum _{k=1}^{4}2{\left(0.1\right)}^k}=\frac{1111}{5000}$ .

Explanation of Solution

Given information:

A summation expression is given as

  ${\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}$

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.

  ${\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}$

Now, the fourth partial sum of ${\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}$ will be ${\displaystyle \sum _{k=1}^{4}2{\left(0.1\right)}^k}$ .

Now, substitute 1, 2, 3, and 4 for $k$ in the expression $2{\left(0.1\right)}^k$ and add those.

  $\begin{array}{c}{\displaystyle \sum _{k=1}^{4}2{\left(0.1\right)}^k}=\frac{2}{{10}^1}+\frac{2}{{10}^2}+\frac{2}{{10}^3}+\frac{2}{{10}^4}\\ =\frac{2}{10}+\frac{2}{100}+\frac{2}{1000}+\frac{2}{10000}\\ =\frac{2\left(1000+100+10+1\right)}{10000}\\ =\frac{2\cdot 1111}{10000}\\ =\frac{1111}{5000}\end{array}$

Hence, the fourth partial sum of ${\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}$ is ${\displaystyle \sum _{k=1}^{4}2{\left(0.1\right)}^k}=\frac{1111}{5000}$ .

b)

To determine

To evaluate: the sum of the given infinite series.

Answer to Problem 24RE

The sum of the infinite series ${\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}$ is $\frac{2}{3}$ .

Explanation of Solution

Given information:

A summation expression is given as

  ${\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}$

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.

  ${\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}$

Now, terms of the series will be

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

Now, first term of the series is $\frac{2}{{10}^1}$ .

Common ratio can be found by

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

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

  $\begin{array}{c}{\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}=\frac{\frac{2}{10}}{1-\frac{1}{10}}\\ =\frac{\frac{1}{5}}{\frac{9}{10}}\\ =\frac{1}{5}\times \frac{10}{9}\\ =\frac{2}{9}\end{array}$

Hence, the sum of the infinite series ${\displaystyle \sum _{k=1}^{\infty }2{\left(0.1\right)}^k}$ is $\frac{2}{9}$ .