Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 17

To determine the convergence or divergence of a series sum a_n using Root test, we evaluate a 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 Root test on the given series sum_(n=1)^oo n(6/5)^n when we let: a_n=n(6/5)^n .
Then, set-up the limit as:
lim_(n-gtoo) |n(6/5)^n|^(1/n) =lim_(n-gtoo) (n(6/5)^n)^(1/n)
Apply Law of Exponents: (x*y)^n = x^n*y^n and (x^n)^m = x^(n*m).
lim_(n-gtoo) (n(6/5)^n)^(1/n)=lim_(n-gtoo) n^(1/n) ((6/5)^n)^(1/n)
=lim_(n-gtoo) n^(1/n) (6/5)^(n*1/n)
=lim_(n-gtoo) n^(1/n) (6/5)^(n/n)
=lim_(n-gtoo) n^(1/n) (6/5)^1
=lim_(n-gtoo) 6/5n^(1/n)
Evaluate the limit.
lim_(n-gtoo) 6/5n^(1/n) =6/5lim_(n-gtoo) n^(1/n)
=6/5 *1
=6/5 or 1.2
The limit value L =6/5 or 1.2 satisfies the condition: Lgt1 since 6/5gt1 or 1.2gt1 .
Therefore, the series sum_(n=1)^oo n(6/5)^n is divergent.