# `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`

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.

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