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Convergence/Divergence of series

  1. Nov 25, 2007 #1
    Hello,

    I have to prove conv/div. for the following series:

    [tex]\sum\frac{(2n)!}{n^n}[/tex]

    I use the "ratio-test" and get the following:

    [tex]\lim_{n\to\infty} \frac{a_{n+1}}{a_{n}} = \lim_{n\to\infty} \frac{(2n+2)!}{(2n)!} \frac{n^n}{(n+1)^{n+1}} = \lim_{n\to\infty} \frac{(2n+2)(2n+1)}{(n+1)} (\frac{n}{1+n})^n = \infty \frac{1}{e} = \infty[/tex]

    This means the series diverges, however, the series should converge (I could find the finite sum online).

    Where is my mistake?

    Thank you!
     
  2. jcsd
  3. Nov 26, 2007 #2
    The last half terms can be written as {(n+l)/n}{(n+2)/n}{(n+3)/n}........{2n/n}, and the first n terms are just n!.
     
  4. Nov 26, 2007 #3
    Sorry, I do not quite understand what you mean or how this helps me. Could you elaborate on that?

    Also, I am interested in why the test I am using does not work out like it should or if I made an algebra mistake somewhere along the way.
     
  5. Nov 26, 2007 #4

    Gib Z

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    Using stirlings approximation to replace the factorial, I get the series diverges. Where did you find online its sum?
     
  6. Nov 26, 2007 #5

    Office_Shredder

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    What he's saying is that if you split it up, you get 1/n*1/n*1/n...*(2n)(2n-1)(2n-2)...(n+1)*n!

    So you put one n under each 2n-k and get

    2n/n*(2n-1)/n*(2n-2)/n....*(n+1)/n*n!

    As each (2n-k)/n>1, and n!>1, each term in the series is >1. So there's very little reason why it would converge
     
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