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Homework Help: The divergence of a series

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Is the series [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] ([tex]\sqrt{k+1}[/tex] - [tex]\sqrt{k}[/tex])/k convergent or divergent?

    2. Relevant equations
    The Comparison Test:
    0<=ak<=bk
    1.The series [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] ak converges if the series [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] bk converges.
    2. The series [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] bk diverges if the series [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] ak diverges.

    3. The attempt at a solution
    I computed the equation until it looked like this: [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] ([tex]\sqrt{1/k + 1/k^2}[/tex] - 1/[tex]\sqrt{k}[/tex]) and then I tried to find some other series that would be smaller than the original but still diverge because my guess is that this series diverges. But if I take the series [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] ([tex]\sqrt{1/k^2}[/tex] - 1/[tex]\sqrt{k}[/tex]) the terms of the series become negative and the rules of the Comparison Test don't apply.
    Then I tried the series [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] ([tex]\sqrt{k+1}[/tex] - [tex]\sqrt{k-1}[/tex])/k and computed it to this: [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] ([tex]\sqrt{1/k + 1/k^2}[/tex] - [tex]\sqrt{1/k-1/k^2}[/tex]) but it's even harder to analyse than the original series.
    Any hints? I also tried series that are greater than the original but found all of them divergent so they weren't of any help.
     
  2. jcsd
  3. Jan 20, 2009 #2
    Personally I would do integral test. Your terms are decreasing, I believe; you know that

    [tex] \lim_{k=\infty} f(k) = 0 [/tex]

    So you'd have to evaluate

    [tex] \int_{1}^{\infty} \frac{\sqrt{k+1} - \sqrt{k}}{k} dk [/tex]

    The first part of that integral is tricky and I don't see a quick way to do it but I believe the integral evaluates to


    [tex] \int_{1}^{\infty} \frac{\sqrt{k+1} - \sqrt{k}}{k} dk = 2 - 2\sqrt{2} + ln\left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right|[/tex]
     
  4. Jan 20, 2009 #3
    The best I can come up with is that

    [tex] \frac{\sqrt{k+1}}{k} dk = \frac{k+1}{k\sqrt{k+1}} dk [/tex]

    So if we let

    [tex] u = \sqrt{k+1} \Rightarrow 2du = \frac{dk}{\sqrt{k+1}} [/tex]

    And we are left with integrating

    [tex] 2\frac{u^2}{u^{2} - 1} du [/tex]

    Which you can simplify fairly easily using polynomial division and then do partial fractions on what's left.

    Does that help?
     
  5. Jan 20, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can do a comparison test as well. First multiply numerator and denominator by sqrt(k+1)+sqrt(k). Now compare it with a p-series.
     
  6. Jan 22, 2009 #5
    Now I was able to compute the series [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] ([tex]\sqrt{k+1}[/tex] - [tex]\sqrt{k}[/tex])/k to [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] 1/k([tex]\sqrt{k+1}[/tex] + [tex]\sqrt{k}[/tex]).

    And from there I was able to tell that 1/k([tex]\sqrt{k+1}[/tex] + [tex]\sqrt{k}[/tex]) < 1/k(2[tex]\sqrt{k}[/tex]) . And because [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] 1/k(2[tex]\sqrt{k}[/tex]) converges, the Comparison Test states that [tex]\Sigma[/tex][tex]\stackrel{\infty}{k=1}[/tex] ([tex]\sqrt{k+1}[/tex] - [tex]\sqrt{k}[/tex])/k must converge too.

    Thank-you for your help!
     
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