Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 49

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=2)^oo n/(nln(n))^n , we have a_n =n/(nln(n))^n .
Applying the Root test, we set-up the limit as:
lim_(n-gtoo) |n/(nln(n))^n|^(1/n) =lim_(n-gtoo) (n/(nln(n))^n)^(1/n)
Apply Law of Exponent: (x/y)^n = x^n/y^n and (x^n)^m = x^(n*m) .
lim_(n-gtoo) (n/(nln(n))^n)^(1/n) =lim_(n-gtoo) n^(1/n)/((nln(n))^n)^(1/n)
=lim_(n-gtoo) n^(1/n)/(nln(n))^(n*1/n)
=lim_(n-gtoo) n^(1/n)/(nln(n))^(n/n)
=lim_(n-gtoo) n^(1/n)/(nln(n))^1
=lim_(n-gtoo) n^(1/n)/(nln(n))
Apply the limit property: lim_(x-gta)[(f(x))/(g(x))] =(lim_(x-gta) f(x))/(lim_(x-gta) g(x)).
lim_(n-gtoo) n^(1/n)/(nln(n))=(lim_(n-gtoo) n^(1/n))/(lim_(n-gtoo) nln(n))
=1/ oo
= 0
Note: lim_(n-gtoo) n^(1/n) = 1 and
lim_(n-gtoo) nln(n) = oo ln(oo)
= oo*oo
=oo
The limit value L=0 satisfies the condition: L lt1 since 0lt1 .
Conclusion: The series sum_(n=2)^oo n/(nln(n))^n is absolutely convergent.