Question 42555
The sum of the first 'n'-terms of a geometric series: a, ar, {{{ar^2}}},{{{ar^3}}},.........,{{{ar^(n-1)}}} is given by {{{S = a*(1-r^n)/(1-r)}}} when {{{red(0 < r < 1)}}}.


When 'n' is very large, {{{r^n}}} << 1 [means {{{r^n}}} is very very less than 1].
So, obviously when 'n' is infinity, {{{r^n}}} << 1 so it can be neglected (means taken as 0) in comparison to 1.
Thus the summation formula for n = {{{infinity}}} becomes
{{{S = a*(1-0)/(1-r)}}}
or {{{S = a/(1-r)}}} 


Here, a = 1 and {{{red(r=3/5)}}} so {{{red(0 < r < 1)}}}. 
Hence the formula of summation of geometric series for infinite number of terms (n -> {{{infinity}}}) is applicable.


Thus the summation of the given infinite geometric series is
{{{S = 1/(1-3/5)}}}
or {{{S = 1/(2/5)}}}
or {{{S = 5/2 = 2.5}}}


Hence, the summation of the given series exists and its value is 2.5.