# How do I write a repeating decimal as an infinite geometric series?

Nov 14, 2015

$0. {b}_{1} {b}_{2.} . . {b}_{j} \overline{{a}_{1} {a}_{2.} . . {a}_{k}} =$
$0. {b}_{1} {b}_{2.} . . {b}_{j} + {\sum}_{n = 0}^{\infty} \left({10}^{-} j \cdot 0. {a}_{1} {a}_{2.} . . {a}_{k}\right) {\left({10}^{-} k\right)}^{n}$

#### Explanation:

It helps to think about what digits in a base 10 system represent. The digits to the left of a decimal point represents an increasing power of 10 (starting from 0), whereas those to the right of a decimal point represent a decreasing power of 10 (starting from -1).

For example:
$2 = 2 \cdot {10}^{0}$
$24 = 2 \cdot {10}^{1} + 4 \cdot {10}^{0}$
$0.1 = 1 \cdot {10}^{-} 1$
$0.023 = 2 \cdot {10}^{-} 2 + 3 \cdot {10}^{-} 3$
$0. {a}_{1} {a}_{2} {a}_{3.} . . = {a}_{1} \cdot {10}^{-} 1 + {a}_{2} + {10}^{-} 2 + {a}_{3} \cdot {10}^{-} 3 + \ldots$

Now we can figure out how to write a repeating decimal as an infinite sum. An infinite geometric series is a series of the form
${\sum}_{n = 0}^{\infty} a {r}^{n}$
where $a$ is the first term in the series and $r$ is the common ratio between terms.

From the properties of decimal digits noted above, we can see that the common ratio will be a negative power of 10. What the power is will be determined by how many digits are repeating, and what $a$ is will be determined by which decimals are repeating.

$0. \overline{3} = {\sum}_{n = 0}^{\infty} \left(0.3\right) {\left({10}^{-} 1\right)}^{n}$
$0. \overline{13} = {\sum}_{n = 0}^{\infty} \left(0.13\right) {\left({10}^{-} 2\right)}^{n}$

or, in general,

$0. \overline{{a}_{1} {a}_{2} {a}_{3.} . . {a}_{k}} = {\sum}_{n = 0}^{\infty} \left(0. {a}_{1} {a}_{2} {a}_{3.} . . {a}_{k}\right) {\left({10}^{-} k\right)}^{n}$

Finally, note that a repeating decimal may only start repeating after a finite number of initial decimal digits. In that case, we simply add the finite digits to the series separately. For example:

$0.1 \overline{3} = 0.1 + {\sum}_{n = 0}^{\infty} \left(0.03\right) {\left({10}^{-} 1\right)}^{n}$

(Note that it is the digits that are repeating which determine $a$ and the power of 10, and not the number of digits after the decimal point.)

$0.101 \overline{23} = 0.101 + {\sum}_{n = 0}^{\infty} \left(0.00023\right) {\left({10}^{-} 2\right)}^{n} = 0.101 + {\sum}_{n = 0}^{\infty} \left({10}^{-} 3 \cdot 0.23\right) {\left({10}^{-} 2\right)}^{n}$

Thus, we get the most general form:

$0. {b}_{1} {b}_{2.} . . {b}_{j} \overline{{a}_{1} {a}_{2.} . . {a}_{k}} =$
$0. {b}_{1} {b}_{2.} . . {b}_{j} + {\sum}_{n = 0}^{\infty} \left({10}^{-} j \cdot 0. {a}_{1} {a}_{2.} . . {a}_{k}\right) {\left({10}^{-} k\right)}^{n}$