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Convergence of a Series

  1. Jul 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Test for convergence or divergence. Give a reason for your decision.



    2. Relevant equations

    [itex]\sum_{i=1}^{\infty} \frac{\sqrt{2n-1} \log{(4n + 1)}}{n(n + 1)}[/itex]

    3. The attempt at a solution

    I've tried to compare it to the series [itex]\sum_{i=1}^{\infty} \frac{\sqrt{2n-1} \log{(4n + 1)}}{n^2}[/itex] and show the latter converges. I have no idea how to show this. Although the limit of the sequence approaches 0 as n goes to infinite, that is not enough to guarantee convergence. The book says it converges.
     
    Last edited: Jul 15, 2011
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  3. Jul 15, 2011 #2

    micromass

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    Hi zonk! :smile:

    Do you know the Cauchy condensation test? The integral test? These two could prove useful here.

    What book are you using anyway?
     
  4. Jul 15, 2011 #3
    It's Apostol volume 1. I think Cauchy's test is a section or two after this. Yes he did teach the integral test in this section.
     
  5. Jul 15, 2011 #4

    micromass

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    OK, can you bring the series to something of the form

    [tex]\sum{\frac{\log(n)}{n^{2-\frac{1}{2}}}} [/tex]

    this is suitable for the integral test
     
  6. Jul 15, 2011 #5
    I can reduce it to:


    [itex]\sum_{i=1}^{\infty} \frac{\sqrt{2} \log{(4n + 1)}}{n^{(2 - \frac{1}{2})}}[/itex] and can factor sqrt(2) out, but I don't see how you can get the log argument that way.
     
    Last edited: Jul 15, 2011
  7. Jul 15, 2011 #6

    micromass

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    Well,

    [tex]4n+1\leq n^2[/tex]

    For large n. Thus

    [tex]\log(4n+1)\leq 2\log(n)[/tex]

    and you can factor the 2 out.
     
  8. Jul 15, 2011 #7
    Oh, thank you so much, I would have never figured this out without those hints.
     
  9. Jul 15, 2011 #8

    Ray Vickson

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    Let t(n) = nth term above. You could try to get a simple upper bound on t(n): sqrt(2n-1) < sqrt(2n), log(4n-1) < log(4n) and n(n+1) > n^2. Thus, t(n) < sqrt(2n)*log(4n)/n^2, which is of the form c*log(n)/n^(3/2). Convergence of sum log(n)/n^(3/2) is easier to show, and that implies convergence of sum t(n) [why?]

    RGV
     
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