sum_(n=0)^oo (2x)^n Find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.)

Recall the Root test determines the limit as:
lim_(n-gtoo) root(n)(|a_n|)= L
or 
lim_(n-gtoo) |(a_n)|^(1/n)= L
Then, we follow the conditions:
a) Llt1 then the series is absolutely convergent
b) Lgt1 then the series is divergent.
c) L=1 or does not exist  then the test is inconclusive. The series may be divergent, conditionally convergent, or absolutely convergent.
For the given series sum_(n=0)^oo (2x)^n , we have a_n = (2x)^n .
Applying the Root test, we set-up the limit as:
lim_(n-gtoo) |((2x)^n )^(1/n)| =lim_(n-gtoo) |(2x)^(n*1/n)|
                                  =lim_(n-gtoo) |(2x)^(n/n)|
                                  =lim_(n-gtoo) |(2x)^1|
                                  =lim_(n-gtoo) |(2x)|
                                  =|2x|
Applying  Llt1 as the condition for an absolutely convergent series, we let L=|2x| and set-up the interval of convergence as:
|2x|lt1
-1 lt2xlt1
Divide each part by 2:
(-1)/2 lt(2x)/2lt1/2
-1/2ltxlt1/2
The series may converges when L =1 or |2x|=1 . To check on this, we test for convergence at the endpoints: x=-1/2 and x=1/2 by using geometric series test.
The convergence test for the geometric series sum_(n=0)^oo a*r^n  follows the conditions:
a) If |r|lt1  or -1 ltrlt 1 then the geometric series converges to a/(1-r) .
b) If |r|gt=1 then the geometric series diverges.
When we let x=-1/2 on sum_(n=0)^oo (2*(1/2))^n , we get a series:
sum_(n=0)^oo 1*(-2/2)^n =sum_(n=0)^oo 1*(-1)^n
 
It shows that r=-1 and |r|= |-1|=1 which satisfies |r|>=1. The series diverges at the left endpoint.
When we let x=1/2 on sum_(n=0)^oo (2*1/2)^n , we get a series:
sum_(n=0)^oo 1*(2/2)^n =sum_(n=0)^oo 1*(1)^n
It shows that r=1 and |r|= |-1|=1 which satisfies |r|gt=1 . The series diverges at the right endpoint.
Conclusion:
The interval of convergence of the power series sum_(n=0)^oo (2x)^n is -1/2ltxlt1/2 .