# Infinite Geometric Series

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Equation / Last modified by CalebSvobodny on 2016/07/07 14:11
sum_(n=0)^infty ar^n =
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CalebSvobodny.Infinite Geometric Series
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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.

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