Question 817179
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Hi,
The sum of  1 1/3 1/3^2 1/3^3 ..... 1/3^n-1 is:
r = {{{1/3 }}} < 1
{{{sum((1/3)^(n-1), i=1 infinity )}}} = {{{a[1]/(1-r) =  (1/(2/3)) = 3/2}}}