1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    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


    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: The divergence of a series
  1. Divergent series (Replies: 4)

  2. Divergent series (Replies: 5)

  3. Series divergence (Replies: 3)

  4. Divergence of a series (Replies: 12)