`sum_(n=2)^oo n/(nlnn)^n` Use the Root Test to determine the convergence or divergence of the series.

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marizi eNotes educator| Certified Educator

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`

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.

For the given series `sum_(n=2)^oo n/(nln(n))^n` , we have `a_n =n/(nln(n))^n` .

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

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

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

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

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

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

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

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

Apply the limit property: `lim_(x-gta)[(f(x))/(g(x))] =(lim_(x-gta) f(x))/(lim_(x-gta) g(x)).`

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

                      `=1/ oo`

                      `= 0 `                                                   

Note: `lim_(n-gtoo) n^(1/n) = 1 ` and 

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

                                ` = oo*oo`

                                ` =oo`

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

Conclusion: The series `sum_(n=2)^oo n/(nln(n))^n` is absolutely convergent.

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