- vCommons Home
- Infinite Geometric Series

# Infinite Geometric Series

CalebSvobodny.Infinite Geometric Series

The **infinite geometric series** is any series of the form

`sum_(n=0)^infty ar^n`.

The series converges for any `r` where `|r|<1` and otherwise diverges. When `|r|<1`, the series converges to `a/(1-r)`.

#### Derivation of formula

First, consider the sum of the first `n` terms of the geometric series. If we write out the sum, it looks like

`a+ar+ar^2 + ... + ar^(n-1)`

Call this sum `S`. Then

`S=a+ar+ar^2 + ... + ar^(n-1)`

`rS=ar+ar^2 + ... + ar^(n-1)+ar^n`

`S-rS=a-ar^n`

`S(1-r)=a(1-r^n)`

`S=a((1-r^n)/(1-r))` (Assuming `r!=1`.)

Now we can see that as `n->infty`, the series can't converge unless `|r|<1`, since otherwise `|r^n| ->infty`. But when `|r|<1`, `r^n ->0`, so

`a+ar+ar^2 + ... = lim_(n->infty) a((1-r^n)/(1-r))`

`a+ar+ar^2 + ... = a(1/(1-r))`

`sum_(n=0)^infty ar^n = a/(1-r)`.

(Note: This formula also holds for `r in CC`, provided that the modulus of `r` is strictly less than one.)

#### Example

Geometric series can be used to convert repeating decimals into fractions. For example, we can use geometric series to show that `0.999... =1`.

Notice that

`0.999... = 9/10 + 9/100 + 9/1000 + 9/10000 + ...`

is the same as the geometric series with `a=9/10` and `r=1/10`. Since `|1/10|<1`, we know the series converges, so we can find the value

`sum_(n=0)^infty (9/10)(1/10)^n = (9/10)/(1-1/10)`

`sum_(n=0)^infty (9/10)(1/10)^n = (9/10)/(9/10)`

`sum_(n=0)^infty (9/10)(1/10)^n = 1`.

So

`0.999... = 1`.

**Infinite Geometric Series**, is listed in 1 Collection.