Recall the **Root test** determines 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.

We may apply the **Root Test** to determine the convergence or divergence of the **series** `sum_(n=0)^oo 6^n/(n+1)^n` .

For the given series `sum_(n=0)^oo 6^n/(n+1)^n` , we have `a_n =6^n/(n+1)^n`

Applying the Root test, we set-up the limit as:

`lim_(n-gtoo) |6^n/(n+1)^n|^(1/n)=lim_(n-gtoo) (6^n/(n+1)^n)^(1/n)`

Apply the Law of Exponents: `(x^n/y^n) = (x/y)^n` and `(x^n)^(m)= x^(n*m)` .

`lim_(n-gtoo) (6^n/(n+1)^n)^(1/n) =lim_(n-gtoo) ((6/(n+1))^n)^(1/n)`

`=lim_(n-gtoo) (6/(n+1))^(n*(1/n))`

`=lim_(n-gtoo) (6/(n+1))^(n/n)`

`=lim_(n-gtoo) (6/(n+1))^1`

`=lim_(n-gtoo) (6/(n+1))`

Evaluate the limit.

`lim_(n-gtoo) (6/(n+1)) =6lim_(n-gtoo) 1/(n+1)`

`=6 *(lim_(n-gtoo) 1)/(lim_(n-gtoo)(n+1))`

`=6* (1/(oo+1))`

`=6* (1/oo)`

`=6*0`

`=0`

The limit value `L =0 ` satisfies the condition: `Llt1` .

Thus, the series `sum_(n=0)^oo 6^n/(n+1)^n` is **absolutely convergent**.

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