Chapter 9.1, Problem 41E

To determine

To find the geometric series and its sum.

Answer to Problem 41E

The Geometric series is $\frac{234}{{10}^3}+\frac{234}{{10}^6}+\frac{234}{{10}^9}+\frac{234}{{10}^{12}}+\mathrm{............}$ and the sum is $\frac{26}{111}$ .

Explanation of Solution

Given:

The given number is $\overline{0.234}$ .

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 $\overline{0.234}$ .

It can be written as

  $\begin{array}{l}\overline{0.234}=\mathrm{0.234234234..........234234234}\\ \overline{0.234}=0.234+0.000234+0.0000000234+0.0000000000234+\mathrm{...........}\\ \overline{0.234}=\frac{234}{{10}^3}+\frac{234}{{10}^6}+\frac{234}{{10}^9}+\frac{234}{{10}^{12}}+\frac{234}{{10}^{15}}+\mathrm{....}\end{array}$

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.

  $a=0.234$ and $r=\frac{1}{100}$

  $\begin{array}{l}S=\frac{0.234}{\left(1-\frac{1}{1000}\right)}\\ S=\frac{234}{999}\\ S=\frac{26}{111}\end{array}$