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Convergence of series

  1. Apr 28, 2014 #1
    1. The problem statement, all variables and given/known data

    I know that the harmonic series is divergent.But why [itex]\frac{-1}{n}[/itex] is also divergent?
    I've search for some test to test that, but I could not find a method for negative series.
    So how can I prove the series is divergent?:confused:

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 28, 2014 #2

    SammyS

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    [itex]\displaystyle (-1)\cdot\sum_{n=1}^\infty \left(\frac{1}{u}\right)=\sum_{n=1}^\infty \left(-\frac{1}{u}\right) [/itex]
     
  4. Apr 28, 2014 #3

    LCKurtz

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    If the partial sums of ##\sum\frac 1 n## diverge, then so do the partial sums of ##\sum -\frac 1 n=-\sum\frac 1 n##.
     
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