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.