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Finding divergence/convergence by direct comparison test

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{n}[/itex]


    2. Relevant equations Direct comparison test



    3. The attempt at a solution Since the main factor in the equation is the exponent that would be changing as n goes to infinity, I know that from the p series as p > 1 the the series converges. So I know that I would be comparing the original equation to

    [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{2}[/itex]

    And I know that I need to show:

    0 [itex]\leq[/itex] [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{n}[/itex] [itex]\leq[/itex] [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{2}[/itex]

    but I don't know how to show

    [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{n}[/itex] [itex]\leq[/itex] [itex]\sum[/itex][itex]^{∞}_{1}[/itex]1/n[itex]^{2}[/itex]

    mathematically. Would I just blatantly say that the original term is smaller than the p series just by looking at it?
     
  2. jcsd
  3. Mar 24, 2012 #2

    fzero

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    The comparison test requires you to show that

    [tex] \left| \frac{1}{n^n} \right| \leq \left| \frac{1}{n^2} \right| [/tex]

    for sufficiently large [itex]n[/itex]. Try to find nice factors that you can multiply both sides of the equation to compare something to 1.
     
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