Chapter 8, Problem 15P

To determine

To show: The given series is convergent and the sum is less than 90.

Answer to Problem 15P

The series ${\displaystyle \sum _{n=1}^{\infty }{s}_n}$ is convergent which less than $90$ .

Explanation of Solution

Given:

The given series is as:

Consider the series whose terms are the reciprocals of the positive integers that can be written in base $10$ notation without using the digit $0$ .

Calculation:

Consider the series:

  $\begin{array}{l}s=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\\ +\frac{1}{11}+\frac{1}{12}\cdots +\frac{1}{19}\cdots +\frac{1}{99}+\cdots \end{array}$

This is the series whose terms of the reciprocals of the positive integers and written in base $10$ notation without using the digit $0$ .

Group the terms according to their number of digits in the denominators of the series.

  $s=s_1+s_2+s_3+\cdots +s_n$ .

Here,

  $\begin{array}{l}{s}_1=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\\ s_2=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\cdots +\frac{1}{19}\cdots +\frac{1}{99}\\ s_3=\frac{1}{111}+\frac{1}{112}+\frac{1}{113}+\cdots +\frac{1}{119}\cdots +\frac{1}{999}\end{array}$

Each term in $s_1$ is less than or equals to $\frac{1}{{10}^{1-1}}=1$ .

Therefore $9^1=9$ terms for $s_1$ and each term which are less than or equal to $\frac{1}{{10}^{1-1}}=1$ .

Each term in $s_2$ is less than or equals to $\frac{1}{{10}^{2-1}}=\frac{1}{10}$ .

Therefore $9^2=81$ terms for $s_2$ and each term which are less than $\frac{1}{10}$ .

Each term in $s_3$ is less than or equals to $\frac{1}{{10}^{3-1}}=\frac{1}{100}$ .

Therefore $9^2=729$ terms for $s_3$ and each term which are less than $\frac{1}{100}$ .

Similarly in $s_n$ , $9^n$ terms for $s_n$ and each term which are less than $\frac{1}{{10}^{n-1}}$ .

Therefore,

  $\begin{array}{l}{s}_n<9^n\left(\frac{1}{{10}^{n-1}}\right)\\ s_n<9\cdot 9^{n-1}\left(\frac{1}{{10}^{n-1}}\right)\\ s_n<9\cdot {\left(\frac{9}{10}\right)}^{n-1}\\ {\displaystyle \sum _{n=1}^{\infty }{s}_n}<{\displaystyle \sum _{n=1}^{\infty }9\cdot {\left(\frac{9}{10}\right)}^{n-1}}\end{array}$

This series represents a geometric series ${\displaystyle \sum _{n=1}^{\infty }a{r}^{n-1}}$ , series is converges if $\left|r\right|<1$ .

And the sum of the series is $\frac{a}{1-r}$ .

From the series ${\displaystyle \sum _{n=1}^{\infty }9\cdot {\left(\frac{9}{10}\right)}^{n-1}}$

Here,

  $\begin{array}{l}a=9\\ r=\frac{9}{10},\left|r\right|<1\end{array}$

Therefore the series ${\displaystyle \sum _{n=1}^{\infty }9\cdot {\left(\frac{9}{10}\right)}^{n-1}}$ is convergent.

Sum of the series is

  $\begin{array}{c}\frac{a}{1-r}=\frac{9}{1-\frac{9}{10}}\\ =\frac{90}{10-9}\\ =90\end{array}$ .

Therefore ${\displaystyle \sum _{n=1}^{\infty }{s}_n}<{\displaystyle \sum _{n=1}^{\infty }9\cdot {\left(\frac{9}{10}\right)}^{n-1}}$

hence ${\displaystyle \sum _{n=1}^{\infty }{s}_n}<90$ .

Therefore the series ${\displaystyle \sum _{n=1}^{\infty }{s}_n}$ is convergent which less than $90$ .