`sum_(n=1)^oo n/3^n` Use the Root Test to determine the convergence or divergence of the series.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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) `L lt1` 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 n/3^n` .

For the given series `sum_(n=1)^oo n/3^n` , we have `a_n =n/3^n` .

 We set up the limit as:

`lim_(n-gtoo) |n/3^n|^(1/n)`

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


        ` =n^(1/n)/(3^n)^(1/n)`

        ` = n^(1/n)/3^(n/n)`

         `= n^(1/n)/3^1`

         ` = 1/3 n^(1/n)`

Applying `|(n/3^n)|^(1/n)=1/3 n^(1/n)` , we get:

`lim_(n-gtoo) |(n/3^n)|^(1/n)`

`=lim_(n-gtoo)1/3 n^(1/n)`

`= 1/3lim_(n-gtoo)n^(1/n)`

`= 1/3[1]`


 Note: lim_(n->oo) n^(1/n) = 1

The limit value  `L=1/3 ` satisfies the condition: `L lt1` .

 Therefore, the series `sum_(n=1)^oo n/3^n` is absolutely convergent.

Approved by eNotes Editorial Team