# Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.05 Converting an Infinite Decimal Expansion to a Rational Number

## Question

Express

{\begin{aligned}1.7979797979\ldots =1.{\overline {79}}\end{aligned}} as a ratio of integers.

## Complete Solution

{\begin{aligned}1.{\overline {79}}&=1+0.79+0.0079+0.000079+\ldots &&(1)\\&=1+79\cdot 10^{-2}+79\cdot 10^{-4}+79\cdot 10^{-6}+\ldots &&(2)\\&=1+\sum _{k=1}^{\infty }(79\cdot 10^{-2})(10^{-2})^{k-1}&&(3)\\&=1+{\frac {(79\cdot 10^{-2})}{1-10^{-2}}}&&(4)\\&=1+{\frac {79}{99}}\\&={\frac {178}{99}}\end{aligned}} ## Explanation of Each Step

### Step (1)

Although not necessary, writing the repeating decimal expansion into a few terms of an infinite sum allows us to see more clearly what we need to do: relate each term to each other in some way to write the given number using sigma notation.

### Step (2)

Each term in the sum is equal to 79 times 10 to a power. Explicitly writing out what these powers are helps us look for a pattern in the individual terms of our sum.

### Step (3)

Suppose we allow our infinite series to start with the term $a=79\cdot 10^{-2}$ . Then each term in the infinite series (after the second term) is related to its previous term by a factor of $r=10^{-2}$ . Using the form of the geometric series:

{\begin{aligned}\sum _{k=1}^{\infty }ar^{k-1}\end{aligned}} and substituting our identified values for $r,a$ into this formula yields Step (3).

### Step (4)

Here, we apply our formula for the sum of an infinite series,

{\begin{aligned}\sum _{k=1}^{\infty }ar^{k-1}={\frac {a}{1-r}}.\end{aligned}} The rest of the problem is algebraic manipulation of fractions to find a simplified ratio of two integers.

## Possible Areas of Confusion

### Getting Started

Any problem of this type could be started in the same way: by writing out the first few terms of an infinite series. This way, it is easier to see a pattern in the terms of the infinite series.

Essentially, we solved the given problem by writing $1.{\overline {79}}$ as $1+0.{\overline {79}}$ , which isolated the repeating digits, which can be written as a geometric series.
{\begin{aligned}\sum _{k=1}^{\infty }ar^{k-1}\end{aligned}} In Step (2), we can identify $a=79\cdot 10^{-2}$ , and $r=10^{-2}$ . We may substitute these values into the above general form to obtain Step (3).