Q. 77 - Calculus

Question

Express each of the repeating decimals in Exercises 71–78 as a geometric series and as the quotient of two integers reduced to lowest terms.

$0.199999 ...$

Step-by-Step Solution

Answer

The given repeating decimal as a geometric series is $\frac{1+\sum _{k=0}^{\infty }0.9{\left(0.1\right)}^k}{10},$ and as the quotient of two integers reduced to lowest terms is $\frac{1}{5}.$

Step 1. Given Information.

The given repeating decimal is $0.199999 ...$

Step 2. Express the repeating decimal as a geometric series.

The given repeating decimal starts repeating after the tenths place so, to express it as a geometric series, let $y=0.199999 ...$

Now, multiply both the sides by 10

$$\begin{gathered} \left(10\right)y=\left(10\right)0.199999 ... \\ 10y=1.99999 ... \\ y=\frac{1.99999 ...}{10} \\ y=\frac{1+0.99999 ...}{10} \\ y=\frac{1+0.9\left(1+0.1+{\left(0.1\right)}^2+{\left(0.1\right)}^3+...\right)}{10} \\ y=\frac{1+\sum _{k=0}^{\infty }0.9{\left(0.1\right)}^k}{10}..........\left(i\right) \end{gathered}$$

Step 3. Express the repeating decimal as the quotient of two integers reduced to the lowest terms.

The given repeating decimal as the quotient of two integers reduced to the lowest terms can be expressed as 

$$\begin{gathered} 0.199999 ... \\ =\frac{1+\sum _{k=0}^{\infty }0.9{\left(0.1\right)}^k}{10} \mathrm{Use}\left(i\right) \\ \mathrm{Now}, \mathrm{use} S_{\infty }=\frac{a}{1-r} \\ =\frac{1+{\displaystyle \frac{0.9}{1-0.1}}}{10} \\ =\frac{1+{\displaystyle \frac{0.9}{0.9}}}{10} \\ =\frac{2}{10} \\ =\frac{1}{5} \end{gathered}$$