The Journal of Online Mathematics and Its Applications, Volume 7 (2007)

Bouncing Balls and Geometric Series, Robert Styer and Morgan Besson

We can also find the sum of an infinite geometric series using classical high school algebra.

Given an infinite repeating decimal such as 0.33333..., there is a standard technique to convert it to a fraction:

`A` = 0.333333...

10 `A` = 3.33333...

Subtracting the first equation from the second, we get 9 `A` = 3 and from this we see that `A` = 3 / 9 = 1 / 3.

A repeating decimal is a disguised form of an infinite geometric series, so it is no accident that we can use this same idea to calculate infinite geometric series!

`A` = 1 + `r` + `r`^{2} + `r`^{3} + ...

`r` `A` = `r` + `r`^{2} + `r`^{3} + `r`^{4} + ...

Subtracting, `A` − `r` `A` = 1 and so `A` (1− `r` ) = 1 and so `A` = 1 / (1 − `r` ).

In other words,

1 + `r` + `r`^{2} + `r`^{3} + `r`^{4} + ··· = 1 / (1 − `r` ).

For instance, if `r` = 1 / 10, we get

1 + 0.1 + 0.01 + 0.001 + ··· = 1 / (1 − 1 / 10 ) = 10 / 9

If `r` = 1 / 2, we get

1 + 1 / 2 + 1 / 4 + 1 / 8 + ··· = 1 / ( 1 − 1 / 2 ) = 2

Can you see why our derivation fails if `r` ≥ 1 or if `r` ≤ − 1? If you are interested in more proofs of the geometric series sum, please see the annotated bibliography.