Calculus of a Single Variable, Chapter 9, 9.6, Section 9.6, Problem 20

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 5^n/n^4 when we let: a_n =5^n/n^4 .
Applying the Root test, we set-up the limit as:
lim_(n-gtoo) |5^n/n^4|^(1/n) =lim_(n-gtoo) (5^n/n^4)^(1/n)
Apply Law of Exponent: (x/y)^n = x^n/y^n and (x^n)^m= x^(n*m) .
lim_(n-gtoo) (5^n/n^4)^(1/n) =lim_(n-gtoo) (5^n)^(1/n)/(n^4)^(1/n)
=lim_(n-gtoo)5^(n*1/n)/n^(4*1/n)
=lim_(n-gtoo)5^(n/n)/n^(4/n)
=lim_(n-gtoo)5^1/n^(4/n)
=lim_(n-gtoo)5/n^(4/n)
Evaluate the limit.
lim_(n-gtoo) 5/n^(4/n)=5 lim_(n-gtoo) 1/n^(4/n)
=5 *1/oo^(4/oo)
=5 *1/oo^(0)
=5 *1/1
= 5*1
=5
The limit value L =5 satisfies the condition: Lgt1 since 5gt1 .
Conclusion: The series sum_(n=1)^oo 5^n/n^4 is divergent.