Question 419137
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Hi
"The sum of an infinite geometric series is twice its first term.
 Find the common ratio of the series
{{{sum( a[i], i=1, n )}}} = {{{a[1]/(1-r)}}} when |r| < 1
a/(1-r) = 2a
 1/(1-r) = 2
   1 = 2 - 2r
   2r = 1
    r = 1/2

In general: the sum of a geometric series is:
  {{{sum( a[i], i=1, n ) = a[1]((1-r^n)/(1-r))}}}
As the 'infinite' sum being just twice it's first term, was safe, in my mind,
to assume that r was a fraction and r^n would become so...insignificant that (1-r^n)= 1  
For ex:  (1/2)^20 = .0000001 demonstrates that processs