sum_(n=0)^oo (2/3)^n Determine the convergence or divergence of the series.

The given series sum_(n=0)^oo (2/3)^n is in a form of the geometric series.
 Recall that the sum of geometric series follows the formula: sum_(n=1)^oo a*r^(n-1) .
or with an index shift: sum_(n=0)^oo a*r^n = a+a*r + a*r^2 +...
The convergence test for the geometric series follows the conditions:
a) If |r|lt1  or -1 ltrlt 1 then the geometric series converges to sum_(n=0)^oo a*r^n =sum_(n=1)^oo a*r^(n-1)= a/(1-r) .
b) If |r|gt=1 then the geometric series diverges.
By comparing   sum_(n=0)^o(2/3)^n or sum_(n=0)^oo1*(2/3)^n with the geometric series form sum_(n=0)^oo a*r^n , we determine the corresponding values as:
a=1 and r= 2/3 .
The r= 2/3 falls within the condition |r|lt1 since |2/3|lt1 or |0.67| lt1 .
Note: 2/3 ~~0.67 .
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=1)^oo(2/3)*(2/3)^(n -1) =1/(1-2/3)
                                  =1/(3/3-2/3)
                                  =1/(1/3)
                                  =1*(3/1)
                                  = 3