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Series converging

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Show for what real numbers p and q ## \sum \frac{1}{n^{p}ln^{q}(n)} ## diverges or converges.

    2. Relevant equations



    3. The attempt at a solution

    I am kind of lost because it seems that with both subscripts p and q there are a bunch of cases you have to work through. My professor wasn't explicit in regards to what values p and q we had to consider.

    My number 1 question would be: do you consider p and q covering the same set of numbers? Such as p>1 and q>1? Or do you have to also consider 0<p<1 while q>1 as well? It just seems this would take forever.

    Anyways I think I have a solution for p,q > 1


    ## \frac{1}{n^{p}ln^{q}(n)} ## ≤ ## \frac{1}{n^{p}} ## for all n ≥ 3

    since ## \frac{1}{n^{p}} ## converges (p > 1) by comparison test ## \sum \frac{1}{n^{p}ln^{q}(n)} ## converges for p,q > 1
     
  2. jcsd
  3. Jun 6, 2013 #2

    Office_Shredder

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    You can extend your comparison test more, what if q=0 and p>1 for example?

    In general you are going to have to consider p and q being any number - there aren't that many actual different cases though
     
  4. Jun 8, 2013 #3
    ## \frac{1}{n^{p}ln^{q}(n)} \geq \frac{1}{nln(n)} ## for 0 ≤ p,q ≤ 1.

    and ## \frac{1}{nln(n)} ## is divergent by the integral test, thus ## \frac{1}{n^{p}ln^{q}(n)} ## diverges by comparison.

    Now if p,q < 0 then ## \sum \frac{1}{n^{p}ln^{q}(n)} ## becomes ## \sum n^{p}ln^{q}(n) ##

    which clearly diverges.

    Ok, so all of this together we have the series converging for p,q > 1 and diverging for p,q ≤ 1. So are the other cases I need to check are ones with p and q being in these two different ranges respectively? ie p ≤ 1 and q > 1 and vice versa?
     
    Last edited: Jun 8, 2013
  5. Jun 9, 2013 #4
    You saw that p,q=1 is a sort of boundary value. Then it's worthwhile to investigate it deeper.

    You could check what happens if you hold still p=1 and change q, then q=1 and vary p.
     
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