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Convergence of p-series

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex]\sum\frac{1}{n^{p}}[/itex] converges for [itex]p>1[/itex] and diverges for [itex]p<1[/itex], [itex]p\geq0[/itex].

    3. The attempt at a solution

    (1) Diverges: I want to prove it diverges for [itex]1-p[/itex] and using the comparison test show it also diverges for p. [itex]\sum\frac{1}{n^{1-p}}=\sum\frac{1}{n^{1}n^{-p}}=\sum n^{p}/n=\sum n^{p-1}[/itex] for [itex]p<1[/itex]. [itex]\sum n^{p-1}[/itex] ...but this series converges? Where did I go wrong?
     
    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 21, 2012 #2

    vela

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    It's not very clear what you're thinking here. Can you elaborate a bit? What is less than 1?
     
  4. Jan 21, 2012 #3
    Sorry, typo -- p<1.
     
  5. Jan 21, 2012 #4

    vela

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    What exactly are you trying to prove? You refer to "it" but what exactly is "it"?
     
  6. Jan 21, 2012 #5
    Oh, sorry -- I want to prove that [itex]\sum\frac{1}{n^{p}}[/itex] diverges if [itex]\sum\frac{1}{n^{1-p}}[/itex] diverges using the Comparison Test.
     
  7. Jan 21, 2012 #6

    vela

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    If you're using the comparison test, you need to show that
    $$\frac{1}{n^p} \ge \frac{1}{n^{1-p}} = n^{p-1}$$Can you do that?
     
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