# Homework Help: Divergence/Convergence of the series (ln(n))^3/n^3

1. Oct 19, 2009

### beaubulldog

1. The problem statement, all variables and given/known data
Does the following series converge or diverge? give reasons for your answers.
$$\Sigma$$(ln(n))^3/n^3

2. Relevant equations
Is the direct comparison test appropriate for this question? limit comparison? I have issues with the direct comparison when attempting to compare to the harmonic series, and limit comparison doesn't seem to yield a very simple solution either.

3. The attempt at a solution
I attempted to show that the series above is greater than the series ln(n)/n, which is greater than 1/n, the harmonic series. this is apparently not the case, however, as made evident by my graphing calculator on window -1<x<10 & -2<y<2.

2. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Compare ln(n) and n1/3

3. Oct 19, 2009

### beaubulldog

could you be a little more specific...? im not sure how comparing ((ln(n))^3)/(n^3) to ln(n) or n^1/3 would help...you saying use DCT? LCT? thanks for the reply.

4. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Which grows faster? ln(n) or n1/3?

5. Oct 19, 2009

### beaubulldog

ln(n) grows faster as n->inf...i'm still not seeing how this is pertinent to my problem however. Again, thanks.

6. Oct 19, 2009

### Office_Shredder

Staff Emeritus
You might want to double check that with a calculator. You probably went over in class how the exponential function grows faster than any power function, and the logarithm function grows slower; you might want to review your notes and check (I don't see how this problem is possible without making this observation)

7. Oct 19, 2009

### beaubulldog

I stand corrected. n^1/3 does indeed grow faster. How do I relate that fact to the sum of the sequence ((ln(n))^3)/n^3? Ohh....wait, so if i set up the inequality to use in the DCT, 1/n<(ln(n)/n)^3, then cube-root everything, multiply the n from the right side, I end up with n^(2/3)<ln(n)...Is that where you were going with this? in that case, it is a direct comparison with 1/n, proving that (ln(n))^3/n^3 diverges. Correct?

8. Oct 19, 2009

### Office_Shredder

Staff Emeritus
No, because your end result n2/3<ln(n) isn't true, so you can't conclude 1/n< ln(n)3/n3

If you know that ln(n)<n1/3 for big n, then ln(n)3<n.

So what is ln(n)3/n3 smaller than? Think of susbtituting something for ln(n)3

9. Oct 19, 2009

### beaubulldog

so since ln(n)^3 is less than n, that means that ln(n)^3/n^3 is less than n/n^3, so ln(n)^3/n^3 is less than 1/n^2, which is a convergent series...?

10. Oct 20, 2009

### Office_Shredder

Staff Emeritus
That's right. Keep in mind when you're wording your answer to be precise:

For large n, ln(n)3<n, so ln(n)3/n3<1/n2, so since convergence is not affected by the leading portions of the series (it doesn't matter what the values of ln(n)3/n3 when n is smaller) by the comparison test the series converges