How do you use the comparison test for sum (((ln n)^3) / (n^2)) n=1 to n=oo?

1 Answer
Jun 8, 2015

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.