Question

Can anyone help me identify which comparison test to use for the sequence from n=1 to infinity of (ln(n)^2)/(n^3)?

Answer 1

I just need to know weather it converges or diverges

Answer 2

it definitely converges

Answer 3

\(\ln(x)\) grows much slower than \(x\)

Answer 4

i am trying to find a nice inequality that will prove that this converges since you are supposed to use comparison test

Answer 5

ok so does that mean i should compare it to 1/n^3?

Answer 6

well i can use any test to prove it it converges

Answer 7

no you cannot because log is not less than 1, and neither is the square

Answer 8

i have to run, but one idea is that you can show that for large values of n, ln(n)^2 < n
then you can use comparison test

Answer 9

ok cook ty

Answer 10

so here was my logic if anyone would be willing to check it. for large values of n, ln(n)^2 is <n, there for you can compare the original series to n/n^3. The lim as n->inf of 1/n^2 converges so you must also know the original converges. Is this correct?