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 .
Tuesday, October 2, 2018
Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 43
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