You've written:
sum_(n=1)^(oo)(ln^3n)/n^2
Let's call that sum_(n=1)^(oo)a_n.
The Limit Comparison Test says (paraphrased):
For any two series that can be written as suma_n >= 0 and sumb_n > 0, define some limit c as:
c = lim_(n->oo)suma_n/(b_n)
If the value of c turns out to be positive (c > 0) and finite, then the series a_n converges. Otherwise, a_n diverges.
There's an easy way to choose these series a_n and b_n. We can choose a series b_n that behaves just like a_n by adding a +1.
Therefore, let:
a_n = (ln^3n)/n^2
b_n = (ln^3(n+1))/(n+1)^2
So, now we get:
c = lim_(n->oo)(ln^3n)/n^2*(n+1)^2/(ln^3(n+1))
= lim_(n->oo)(n+1)^2/n^2(ln^3n)/(ln^3(n+1))
(Nono, don't use L'Hopital's Rule here; there's an easy way to do this.)
This limit can be split into a product of limits, like so:
= lim_(n->oo)(n+1)^2/n^2 * lim_(n->oo)(ln^3n)/(ln^3(n+1))
Imagine, as n->oo, the +1 becomes really minor, so both limits approach the result of omitting the +1. The two limits are each 1 because everything cancels out as n->oo. Therefore, the limit of the total result is 1.
lim_(n->oo)a_n/(b_n) = 1 > 0.
Therefore, the sum sum_(n=1)^(oo)a_n converges.
