sum_(n=1)^oo n/3^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
lim_(n-gtoo) |a_n|^(1/n)= L
Then, we follow the conditions:
a) L lt1 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/3^n .
For the given series sum_(n=1)^oo n/3^n , we have a_n =n/3^n .
 We set up the limit as:
lim_(n-gtoo) |n/3^n|^(1/n)
 Apply Law of Exponent: (x/y)^n = x^n/y^m and simplify.
|n/3^n|^(1/n)=(n/3^n)^(1/n)
        =n^(1/n)/(3^n)^(1/n)
        = n^(1/n)/3^(n/n)
         = n^(1/n)/3^1
          = 1/3 n^(1/n)
Applying |(n/3^n)|^(1/n)=1/3 n^(1/n) , we get:
lim_(n-gtoo) |(n/3^n)|^(1/n)
=lim_(n-gtoo)1/3 n^(1/n)
= 1/3lim_(n-gtoo)n^(1/n)
= 1/3[1]
=1/3
 Note: lim_(n->oo) n^(1/n) = 1
The limit value  L=1/3 satisfies the condition: L lt1 .
 Therefore, the series sum_(n=1)^oo n/3^n is absolutely convergent.