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

To determine the convergence or divergence of a series `sum a_n` using Root test, we evaluate 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.

...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

To determine the convergence or divergence of a series `sum a_n` using Root test, we evaluate 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.

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

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

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

Apply Law of Exponent: `(x^n)^m = x^(n*m)` .

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

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

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

`=lim_(n-gtoo) (ln(n)/n)`

Evaluate the limit using direct substitution: `n = oo` .

`lim_(n-gtoo) (ln(n)/n) = oo/oo `

When the limit value is indeterminate `(oo/oo)` , we may apply L'Hospital's Rule:

`lim_(x-gta) (f(x))/(g(x)) =lim_(x-gta) (f'(x))/(g'(x))` .

Let: `f(n) = ln(n)` then

`g(n) = n` then `g'(n) =1` .

Then, the limit becomes:

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

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

` = 1/oo`

` =0`

The limit value   `L=0` satisfies the condition: `L lt1` since `0lt1.`

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

Approved by eNotes Editorial Team