sum_(n=1)^oo n^3/3^n Use the Root Test to determine the convergence or divergence of the series.

When using Root test on a series sum a_n , we determine a 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.
We may apply the Root Test to determine the convergence or divergence of the series sum_(n=1)^oo n^3/3^n . when we let: a_n =n^3/3^n .
Applying the Root test, we set-up the limit as: 
lim_(n-gtoo) |n^3/3^n|^(1/n) =lim_(n-gtoo) (n^3/3^n)^(1/n)
Apply Law of Exponents: (x/y)^n = x^n/y^n and (x^n)^m= x^(n*m) .
lim_(n-gtoo) (n^3/3^n)^(1/n) =lim_(n-gtoo) (n^3)^(1/n)/(3^n)^(1/n)
                        =lim_(n-gtoo) n^(3*1/n)/3^(n*1/n)
                        =lim_(n-gtoo) n^(3/n)/3^(n/n)
                        =lim_(n-gtoo) n^(3/n)/3^1
                        =lim_(n-gtoo) n^(3/n)/3
Evaluate the limit.
lim_(n-gtoo)n^(3/n)/ 3 =1/3 lim_(n-gtoo) n^(3/n)   
                  =1/3 *1
                  =1/3
The limit value L =1/3 satisfies the condition: Llt1 since 1/3lt1 .
Therefore, the series sum_(n=1)^oo n^3/3^n is absolutely convergent.