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

Expert Answers
marizi eNotes educator| Certified Educator

Recall the Root test determines the limit as:

`lim_(n-gtoo) root(n)(|a_n|)= L`


`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 1/n^n` .

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

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

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

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

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

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

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

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

Evaluate the limit.

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

                ` = 1^(1/oo)/oo`

                ` = 1^0/oo`

                ` = 1/oo`

                ` =0`

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

Therefore, the series `sum_(n=1)^oo 1/n^n`   is absolutely convergent.