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* ...

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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**.