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Need to evaluate an inf. series

  1. Sep 19, 2006 #1
    For a series that is Sigma (n!)/(n^n), I did the ratio test and now I need to prove that limit (n^n)/(n+1)^n < 1, can someone give me a hand on how I can do this?
     
  2. jcsd
  3. Sep 19, 2006 #2

    shmoe

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    hint:the limit of [tex]\left(\frac{n}{n+1}\right)^{-n}[/tex] should be very familiar...
     
  4. Sep 20, 2006 #3
    Never seen it before...
     
  5. Sep 20, 2006 #4

    CRGreathouse

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    Caluclate its numeric value for n=1, n=2, n=10, and n=1000.
     
  6. Sep 20, 2006 #5

    shmoe

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    You've never seen:

    [tex]\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n[/tex]

    before?


    Okie, try a comparison of your original sum with [tex]\sum_{n=1}^\infty\frac{1}{2n^2}[/tex]
     
  7. Sep 20, 2006 #6
    1/e, you could've just said it... as for your second suggestion, how the hell are the two series similar? Are you joking?
     
  8. Sep 21, 2006 #7

    shmoe

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    I could have also done the whole problem for you, but what's the point of that? I make no apologies for expecting you to meet me well over half way.

    No joke. Though I did mean the 2 to be in the numerator to make it more obvious (it doesn't end up mattering though, the terms in your series decay very fast), namely [tex]\sum_{n=1}^\infty\frac{2}{n^2}[/tex].

    Break the terms of your series up into 'n' factors:

    [tex]\frac{n!}{n^n}=\left(\frac{1}{n}\right)\left(\frac{2}{n}\right)\ldots\left(\frac{n-1}{n}\right)\left(\frac{n}{n}\right)[/tex]

    Now compare with [tex]\frac{2}{n^2}[/tex].
     
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