sum_(n=0)^oo 3/5^n Determine the convergence or divergence of the series.

Recall that infinite series converges to single finite value S   if the limit if the partial sum S_n as n approaches oo converges to S . We follow it in a formula:
lim_(n-gtoo) S_n=sum_(n=1)^oo a_n = S .
To evaluate the sum_(n=0)^oo 3/5^n , we may express it in a form:
sum_(n=0)^oo 3/5^n =sum_(n=0)^oo 3* (1/5^n)
               =sum_(n=0)^oo 3 *(1/5)^n
 This resembles form of geometric series with an index shift: sum_(n=0)^oo a*r^n .
By comparing "3 *(1/5)^n " with  "a*r^n ", we determine the corresponding values: a = 3 and r =1/5 or 0.2 .
 The convergence test for the geometric series follows the conditions:
 a) If |r|lt1  or -1 ltrlt1 then the geometric series converges to sum_(n=0)^oo a*r^n = a/(1-r) .
 b) If |r|gt=1 then the geometric series diverges.
The r=1/5 or 0.2 from the given infinite series falls within the condition |r|lt1 since |1/5|lt1 or |0.2|lt1 . Therefore, we may conclude that sum_(n=0)^oo 3/5^n is a convergent series.
By applying the formula: sum_(n=0)^oo a*r^n= a/(1-r) , we determine that the given geometric series will converge to a value:
sum_(n=0)^oo 3/5^n =sum_(n=0)^oo 3 *(1/5)^n
               = 3/(1-1/5)
                =3/(5/5-1/5)
                =3/(4/5)
                =3*(5/4)
                = 15/4 or 3.75