sum_(n=1)^oo (n/500)^n Use the Root Test to determine the convergence or divergence of the series.

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=1)^oo(n/500)^n .
For the given series sum_(n=1)^oo(n/500)^n , we have a_n = (n/500)^n.
Applying the Root test, we set-up the limit as: 
lim_(n-gtoo) |(n/500)^n|^(1/n)=lim_(n-gtoo) ((n/500)^n)^(1/n)
 Apply Law of Exponent: (x^n)^m = x^(n*m) .
lim_(n-gtoo) ((n/500)^n)^(1/n) =lim_(n-gtoo) (n/500)^(n*(1/n))
                               =lim_(n-gtoo) (n/500)^(n/n)
                               =lim_(n-gtoo) (n/500)^1
                               =lim_(n-gtoo) (n/500)
Evaluate the limit as n approaches oo .
lim_(n-gtoo) (n/500) =1/500lim_(n-gtoo) n
                      = 1/500 * oo
                      =oo
The limit value L =oo satisfies the condition: Lgt1 .
Thus, the series sum_(n=1)^oo(n/500)^n  is divergent.