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
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=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 .