`sum_(n=1)^oo n(6/5)^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

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`


`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 Root test on the given series `sum_(n=1)^oo n(6/5)^n` when we let: `a_n=n(6/5)^n` .

Then, set-up the limit as:

`lim_(n-gtoo) |n(6/5)^n|^(1/n) =lim_(n-gtoo) (n(6/5)^n)^(1/n)`

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

`lim_(n-gtoo) (n(6/5)^n)^(1/n)=lim_(n-gtoo) n^(1/n) ((6/5)^n)^(1/n)`

                               `=lim_(n-gtoo) n^(1/n) (6/5)^(n*1/n)`

                               ` =lim_(n-gtoo) n^(1/n) (6/5)^(n/n)`

                               `=lim_(n-gtoo) n^(1/n) (6/5)^1`

                               `=lim_(n-gtoo) 6/5n^(1/n)`

Evaluate the limit.

`lim_(n-gtoo) 6/5n^(1/n) =6/5lim_(n-gtoo) n^(1/n) `         

                   ` =6/5 *1`  

                  ` =6/5 or 1.2`

The limit value `L =6/5 or 1.2` satisfies the condition: `Lgt1` since `6/5gt1 or 1.2gt1` .

Therefore, the series `sum_(n=1)^oo n(6/5)^n` is divergent.

Approved by eNotes Editorial Team