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Covergence or Divergence?

  1. Jul 7, 2009 #1
    1. [tex]\sum(\sqrt{k^{2}+1}-\sqrt{k^{2}})[/tex] from K=0 to K=[tex]\infty[/tex]




    2. Hi all. I need some help here. I have to use a test to determine whether the sum series diverges or converges



    3. I thought it was the divergence test because I thought that the limit of the sum didn't approach zero, but I think I'm wrong. I need some help here. Thanks
     
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  3. Jul 7, 2009 #2

    Dick

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    The limit of the kth term does approach 0. But that doesn't mean it converges. Try multiplying by (sqrt(k^2+1)+sqrt(k^2))/(sqrt(k^2+1)+sqrt(k^2)) and simplify the algebra. Then give me your opinion about convergence.
     
  4. Jul 7, 2009 #3

    Office_Shredder

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    If you multiply

    [tex](\sqrt{k^2+1} - \sqrt{k^2})(\sqrt{k^2+1} + \sqrt{k^2}) = 1[/tex] So what you actually have is the series

    [tex] \sum \frac{1}{\sqrt{k^2+1} + \sqrt{k^2}})[/tex]

    Does that help?
     
  5. Jul 7, 2009 #4
    Thank you both. That helped a lot. I used the comparison test and the p-series test and it does converge.

    The real problem was similar in that instead of [tex]\sqrt{k^{2}+1}[/tex] - [tex]\sqrt{k^{2}}[/tex] it was [tex]\sqrt{k^{5}+10}[/tex] - [tex]\sqrt{k^{5}}[/tex]. I just didn't want it to feel like cheating. Thanks again.
     
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