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=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**.

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.