Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 48

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.
For the given series sum_(n=1)^oo (ln(n)/n)^n , we have a_n =(ln(n)/n)^n .
Applying the Root test, we set-up the limit as:
lim_(n-gtoo) |(ln(n)/n)^n|^(1/n) =lim_(n-gtoo) ((ln(n)/n)^n)^(1/n)
Apply Law of Exponent: (x^n)^m = x^(n*m) .

lim_(n-gtoo) ((ln(n)/n)^n)^(1/n)=lim_(n-gtoo) (ln(n)/n)^(n*1/n)
=lim_(n-gtoo) (ln(n)/n)^(n/n)
=lim_(n-gtoo) (ln(n)/n)^1
=lim_(n-gtoo) (ln(n)/n)
Evaluate the limit using direct substitution: n = oo .
lim_(n-gtoo) (ln(n)/n) = oo/oo
When the limit value is indeterminate (oo/oo) , we may apply L'Hospital's Rule:
lim_(x-gta) (f(x))/(g(x)) =lim_(x-gta) (f'(x))/(g'(x)) .
Let: f(n) = ln(n) then
g(n) = n then g'(n) =1 .
Then, the limit becomes:
lim_(n-gtoo) (ln(n)/n)=lim_(n-gtoo) ((1/n))/1
=lim_(n-gtoo) 1/n
= 1/oo
=0
The limit value L=0 satisfies the condition: L lt1 since 0lt1.
Therefore, the series sum_(n=1)^oo (ln(n)/n)^n is absolutely convergent.