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.
