Chapter 9.1, Problem 43E

To determine

To find: To find the geometric series and its sum.

Answer to Problem 43E

The Geometric series is $\frac{d}{10}+\frac{d}{{10}^2}+\frac{d}{{10}^3}+\frac{d}{{10}^4}+\frac{d}{{10}^5}+\mathrm{............}$ and the sum is $\frac{d}{9}$ .

Explanation of Solution

Given:

The given number is $\overline{0.d}$ .

Formula used:

If the sequence of partial sum has a limit as $n\to \infty$ , the series converges otherwise the sequence is divergence. For series to be convergence the term should approach to zero. That is common ratio should be less than $1$ . That is $\left|r\right|<1$ , where $r$ is the common ratio.

Calculation:

The given number is $\overline{0.d}$ .

It can be written as,

  $\begin{array}{l}\overline{0.d}=0.dddddddddddddd\mathrm{.....}\\ \overline{0.d}=0.d+0.0d+0.00d+0.000d+\mathrm{....}\\ \overline{0.d}=\frac{d}{10}+\frac{d}{{10}^2}+\frac{d}{{10}^3}+\frac{d}{{10}^4}+\frac{d}{{10}^5}+\mathrm{............}\end{array}$

It is a geometric series.

The sum of the infinite geometric series whose ratio is less than $1$ . $S=\frac{a}{1-r}$ where a is the initial term and r is the common ratio.

Put $a=0.d$ and $r=\frac{1}{10}$ ,

  $\begin{array}{l}S=\frac{0.d}{\left(1-\frac{1}{10}\right)}\\ S=\frac{d}{9}\end{array}$