sum_(n=0)^oo (x/4)^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) |(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 (x/4)^n , we have a_n = (x/4)^n .
Applying the Root test, we set-up the limit as:
lim_(n-gtoo) |((x/4)^n )^(1/n)| =lim_(n-gtoo) |(x/4)^(n*1/n)|
                                  =lim_(n-gtoo) |(x/4)^(n/n)|
                                  =lim_(n-gtoo) |(x/4)^1|
                                 =lim_(n-gtoo) |(x/4)|
                                 =|x/4|
Applying Llt1 as the condition for absolutely convergent series, we plug-in L = |x/4| on Llt1 . The interval of convergence will be:
|x/4|lt1
-1 ltx/4lt1
Multiply each part by 4 :
(-1)*4 ltx/4*4lt1*4
-4ltxlt4
The series may converges when L =1 or |x/4|=1 . To check on this, we test for convergence at the endpoints: x=-4 and x=4 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=-4 on sum_(n=0)^oo (x/4)^n , we get a series:
sum_(n=0)^oo 1*(-4/4)^n =sum_(n=0)^oo 1*(-1)^n
It shows that r=-1 and |r|= |-1|=1 which satisfies |r|gt=1 . Thus, the series diverges at the left endpoint.
When we let x=4 on sum_(n=0)^oo (x/4)^n , we get a series:
sum_(n=0)^oo 1*(4/4)^n =sum_(n=0)^oo 1*(1)^n
It shows that r=1 and |r|= |-1|=1 which satisfies |r|gt=1 . Thus, the series diverges at the right endpoint.
Conclusion:
The interval of convergence of sum_(n=0)^oo (x/4)^n is -4ltxlt4.