• Support PF! Buy your school textbooks, materials and every day products Here!

Divergence/Convergence of the series (ln(n))^3/n^3

  • #1

Homework Statement


Does the following series converge or diverge? give reasons for your answers.
[tex]\Sigma[/tex](ln(n))^3/n^3

Homework 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.


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.
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
Compare ln(n) and n1/3
 
  • #3
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
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
Which grows faster? ln(n) or n1/3?
 
  • #5
ln(n) grows faster as n->inf...i'm still not seeing how this is pertinent to my problem however. Again, thanks.
 
  • #6
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
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
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
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
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
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
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
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
 

Related Threads for: Divergence/Convergence of the series (ln(n))^3/n^3

Replies
3
Views
13K
Replies
9
Views
7K
  • Last Post
Replies
5
Views
41K
  • Last Post
Replies
9
Views
8K
Replies
2
Views
20K
Replies
3
Views
2K
Replies
19
Views
5K
Replies
3
Views
3K
Top