sum_(n=0)^oo e^(-3n) Use the Root Test to determine the convergence or divergence of the series.

To apply the Root test on a series  sum a_n , we determine 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.
To apply the Root Test to determine the convergence or divergence of the series sum_(n=0)^oo e^(-3n) , we  let a_n = e^(-3n) .
Apply Law of Exponent: x^(-n) = 1/x^n . 
a_ n = 1/e^(3n).
Applying the Root test, we set-up the limit as:
lim_(n-gtoo) |1/e^(3n)|^(1/n) =lim_(n-gtoo) (1/e^(3n))^(1/n)
 Apply the Law of Exponents: (x/y)^n = x^n/y^n and (x^n)^m= x^(n*m) .
lim_(n-gtoo) (1/e^(3n))^(1/n) =lim_(n-gtoo) 1^(1/n)/(e^(3n))^(1/n)
                            =lim_(n-gtoo) 1^(1/n)/e^(3n*1/n)
                          =lim_(n-gtoo) 1^(1/n)/e^((3n)/n)     
                          =lim_(n-gtoo) 1^(1/n)/e^3        
Apply the limit property: lim_(x-gta)[(f(x))/(g(x))] =(lim_(x-gta) f(x))/(lim_(x-gta) g(x)) .
lim_(n-gtoo) 1^(1/n)/e^3 =(lim_(n-gtoo) 1^(1/n))/(lim_(n-gtoo)e^3 )
                  = 1^(1/oo) /e^3
                  =1^0/e^3
                     =1/e^3 or 0.0498 (approximated value)
The limit value L = 1/e^3 or 0.0498  satisfies the condition: Llt1 since 0.0498lt1.
Thus, the series sum_(n=0)^oo e^(-3n)  is absolutely convergent.