Infinite Geometric Series

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


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`.

     `0.999... = 1`.


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