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Proof that series diverges ?

  1. May 21, 2012 #1
    [tex] \sum_{1}^{\infty} (-1)^n(1+\frac{1}{n})^n [/tex]

    I've tried the alternating series test but the "b_n" part converges to e.

    Can't think of any other test...
     
  2. jcsd
  3. May 21, 2012 #2

    tiny-tim

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    hi sid9221! :smile:

    hint: draw an against n on graph paper (as dots) …

    what does it look like? :wink:
     
  4. May 21, 2012 #3
    I know I've seen the plot somewhere (can't remember the technical name).

    It looks like a sin function but one that's increasing .

    I know it diverges just can't prove it.
     
  5. May 21, 2012 #4

    Mark44

    Staff: Mentor

    Use the "n-th term test for divergence." If the limit of the n-th term in the series is different from 0 or doesn't exist, the series diverges.
     
  6. May 21, 2012 #5
    How to you take the limit of (-1)^n.

    Wolfram alpha says e^2i 0 to Pi !?!!
     
  7. May 21, 2012 #6

    Ray Vickson

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    You are missing the whole point. If you have a series [itex] S = t_1 - t_2 + t_3 - t_4 + \cdots,[/itex] with all [itex] t_i > 0 \;( \text{or all } < 0),[/itex] you need [itex]|t_n| \rightarrow 0 [/itex] as [itex] n \rightarrow 0.[/itex] Of course the factors [itex] (-1)^n[/itex] do not have a limit, but that is not important.

    RGV
     
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