# `sum_(n=1)^oo (n/(2n+1))^n` Use the Root Test to determine the convergence or divergence of the series.

## Expert Answers

To apply the Root test on a series `sum a_n` , we determine 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 (n/(2n+1))^n` .

For the given series `sum_(n=1)^oo(n/(2n+1))^n` , we have `a_n =(n/(2n+1))^n.`

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

`lim_(n-gtoo) |(n/(2n+1))^n|^(1/n) =lim_(n-gtoo) ((n/(2n+1))^n)^(1/n)`

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

`lim_(n-gtoo) ((n/(2n+1))^n)^(1/n) =lim_(n-gtoo) (n/(2n+1))^(n*(1/n) )`

`=lim_(n-gtoo) (n/(2n+1))^(n/n )`

`=lim_(n-gtoo) (n/(2n+1))^1`

`=lim_(n-gtoo)n/(2n+1)`

Evaluate the limit.

`lim_(n-gtoo) n/(2n+1)=lim_(n-gtoo) (n/n)/((2n)/n+1/n)`

` =lim_(n-gtoo) 1/(2+1/n)`

` =(lim_(n-gtoo) 1)/(lim_(n-gtoo)(2+1/n))`

` = 1 /(2 +1/oo)`

` = 1 /(2 +0)`

` = 1/2`

The limit value `L =1/2` satisfies the condition: `Llt1` .

Conclusion: The series `sum_(n=1)^oo(n/(2n+1))^n` converges absolutely

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