sum_(n=1)^oo 2(-1/2)^n Verify that the infinite series converges

To verify if the given infinite series: sum_(n=1)^oo 2(-1/2)^n converges, recall that infinite series converge to a single finite value S  if the limit of 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=1)^oo 2(-1/2)^n , we apply the Law of exponent : x^(n+m) = x^n*x^m .
Then, (-1/2)^n =(-1/2)^(n -1+1)
                        =(-1/2)^(n -1)*(-1/2)^1
                        = (-1/2)^(n -1)*(-1/2)
Plug-in (-1/2)^n =(-1/2)^(n -1)*(-1/2) , we get:
sum_(n=1)^oo 2(-1/2)^n =sum_(n=1)^oo 2*(-1/2)^(n -1)*(-1/2)
                          =sum_(n=1)^oo -1*(-1/2)^(n -1)
By comparing given infinite series  sum_(n=1)^oo -1*(-1/2)^(n -1) with the geometric series form sum_(n=1)^oo a*r^(n-1) , we determine the corresponding values as: 
a=-1 and r= -1/2 .
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.
The r=-1/2 falls within the condition |r|lt1 since |-1/2|lt1 or |-0.5| lt1 .
Therefore, the series converges.
 
By applying the formula: sum_(n=1)^oo a*r^(n-1)= a/(1-r) , we determine that the given geometric series will converge to a value:
 
sum_(n=1)^oo2(-1/2)^n=sum_(n=1)^oo -1*(-1/2)^(n -1)
                            =(-1)/(1-(-1/2))
                             =(-1)/(1+1/2)
                            =(-1)/(2/2+1/2)
                             =(-1)/(3/2)
                             =(-1)*(2/3)
                             = -2/3 or -0.67 (approximated value)