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

a) `Llt1` then the series is absolutely convergent.

b) `Lgt1` then the series is divergent.

c) `L=1` or does not exist   then...

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Applying 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`

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(2root(n)(n)+1)^n` , we have `a_n =(2root(n)(n)+1)^n` .

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

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

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

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

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

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

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

Evaluate the limit.

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

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

`= 2 * 1 + 1`

`=2 +1`

`=3`

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

The limit value `L =3` satisfies the condition: `Lgt1` .

Conclusion: The series `sum_(n=1)^oo(2root(n)(n)+1)^n`   is divergent .

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