sum_(n=2)^oo lnn/n^3 Determine the convergence or divergence of the series.

To evaluate the series sum_(n=2)^oo ln(n)/n^3 , we may apply Direct Comparison test.
Direct Comparison test is applicable when sum a_n and sum b_n are both positive series for all n where a_n lt=b_n .
If sum b_n converges then sum a_n converges.
If sum a_n diverges so does the sum b_n diverges.
Let b_n=1/n^2 and a_n =ln(n)/n^3  
It follows that a_n < b_n
Graph: 
Note: f(x) =1/x^2 for red graph and g(x)=ln(x)/x^3  for green graph.
Apply the p-series test where kgt0 : the sum_(n=k)^oo 1/n^p is convergent if pgt1 and divergent if plt=1 .
For the sum_(n=2)^oo 1/n^2 , we have the corresponding value p=2 . It satisfies the condition pgt1 since 2gt1 .Therefore, the series sum_(n=2)^oo 1/n^2 converges.
Conclusion:
Because a_n < b_n and sum b_n converges, then sum a_n = sum_(n=2)^oo lnn/n^3 converges