Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 29

Recall the Root test determines 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=0)^oo 6^n/(n+1)^n .
For the given series sum_(n=0)^oo 6^n/(n+1)^n , we have a_n =6^n/(n+1)^n
Applying the Root test, we set-up the limit as:
lim_(n-gtoo) |6^n/(n+1)^n|^(1/n)=lim_(n-gtoo) (6^n/(n+1)^n)^(1/n)
Apply the Law of Exponents: (x^n/y^n) = (x/y)^n and (x^n)^(m)= x^(n*m) .
lim_(n-gtoo) (6^n/(n+1)^n)^(1/n) =lim_(n-gtoo) ((6/(n+1))^n)^(1/n)
=lim_(n-gtoo) (6/(n+1))^(n*(1/n))
=lim_(n-gtoo) (6/(n+1))^(n/n)
=lim_(n-gtoo) (6/(n+1))^1
=lim_(n-gtoo) (6/(n+1))
Evaluate the limit.
lim_(n-gtoo) (6/(n+1)) =6lim_(n-gtoo) 1/(n+1)
=6 *(lim_(n-gtoo) 1)/(lim_(n-gtoo)(n+1))
=6* (1/(oo+1))
=6* (1/oo)
=6*0
=0
The limit value L =0 satisfies the condition: Llt1 .
Thus, the series sum_(n=0)^oo 6^n/(n+1)^n is absolutely convergent.