Chapter 8, Problem 22RE

a)

To determine

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

Answer to Problem 22RE

Fourth partial sum of ${\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}$ is ${\displaystyle \sum _{j=1}^4\frac{3}{{100}^j}}=\frac{33333333}{100000000}$ .

Explanation of Solution

Given information:

A summation expression is given as

  ${\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}$

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.

  ${\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}$

Now, the fourth partial sum of ${\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}$ will be ${\displaystyle \sum _{j=1}^4\frac{3}{{100}^j}}$ .

Now, substitute 1, 2, 3, and 4 for $j$ in the expression $\frac{3}{{100}^j}$ and add those.

  $\begin{array}{c}{\displaystyle \sum _{j=1}^4\frac{3}{{100}^j}}=\frac{3}{{100}^1}+\frac{3}{{100}^2}+\frac{3}{{100}^3}+\frac{3}{{100}^4}\\ =\frac{3}{100}+\frac{3}{10000}+\frac{3}{1000000}+\frac{3}{1000000000}\\ =\frac{3\left(1000000+10000+100+1\right)}{100000000}\\ =\frac{3\cdot 1111111}{100000000}\\ =\frac{33333333}{100000000}\end{array}$

Hence, the fourth partial sum of ${\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}$ is ${\displaystyle \sum _{j=1}^4\frac{3}{{100}^j}}=\frac{33333333}{100000000}$ .

b)

To determine

To evaluate: the sum of the given infinite series.

Answer to Problem 22RE

The sum of the infinite series ${\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}$ is $\frac{1}{33}$ .

Explanation of Solution

Given information:

A summation expression is given as

  ${\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}$

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.

  ${\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}$

Now, terms of the series will be

  $\frac{3}{{100}^1},\frac{3}{{100}^2},\frac{3}{{100}^3},\frac{3}{{100}^4},\mathrm{...}$

Now, first term of the series is $\frac{3}{{100}^1}$ .

Common ratio can be found by

  $\begin{array}{c}r=\frac{\frac{3}{{100}^2}}{\frac{3}{{100}^1}}\\ =\frac{1}{100}\end{array}$

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

  $\begin{array}{c}{\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}=\frac{\frac{3}{100}}{1-\frac{1}{100}}\\ =\frac{\frac{3}{100}}{\frac{99}{100}}\\ =\frac{3}{100}\times \frac{100}{99}\\ =\frac{1}{33}\end{array}$

Hence, the sum of the infinite series ${\displaystyle \sum _{j=1}^{\infty }\frac{3}{{100}^j}}$ is $\frac{1}{33}$ .