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

1 Answer
Nov 14, 2015

Answer:

#0.b_1b_2...b_jbar(a_1a_2...a_k)=#
#0.b_1b_2...b_j + sum_(n=0)^oo(10^-j*0.a_1a_2...a_k)(10^-k)^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*10^0#
#24 = 2*10^1 + 4*10^0#
#0.1 = 1*10^-1#
#0.023 = 2*10^-2 + 3*10^-3#
#0.a_1a_2a_3... = a_1*10^-1 + a_2+10^-2+a_3*10^-3+...#

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)^ooar^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.bar3 = sum_(n=0)^oo(0.3)(10^-1)^n#
#0.bar13 = sum_(n=0)^oo(0.13)(10^-2)^n#

or, in general,

#0.bar(a_1a_2a_3...a_k) = sum_(n=0)^oo(0.a_1a_2a_3...a_k)(10^-k)^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:

#0.1bar3 = 0.1 + sum_(n=0)^oo(0.03)(10^-1)^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.)

#0.101bar23=0.101 + sum_(n=0)^oo(0.00023)(10^-2)^n = 0.101 + sum_(n=0)^oo(10^-3*0.23)(10^-2)^n#

Thus, we get the most general form:

#0.b_1b_2...b_jbar(a_1a_2...a_k)=#
#0.b_1b_2...b_j + sum_(n=0)^oo(10^-j*0.a_1a_2...a_k)(10^-k)^n#