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Math Help - Comparison series

  1. #1
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    Comparison series

    The question is the sum from n= 1 to infinity of n!/(n^n).
    can someone tell me how to do this series...I don't know what to compare it with. I know what a factorial is, I just don't know what to do.

    Thanks
    John
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  2. #2
    MHF Contributor Calculus26's Avatar
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    do you have to use the comparison test? This is ideal for the ratio test.

    an = n!/(n^n)

    an+1 = (n+1)!/[(n+1)^(n+1)]

    an+1/an = (n+1)!n^n/[n!(n+1)^(n+1)]

    =(n+1)*n^n/[(n+1)^(n+1)]

    = [n/(n+1)]^n = [1/(1+1/n)]^n which converges to 1/e
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  3. #3
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    Thanks for the reply, but, yes, we have to use the comparison test.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by johntuan View Post
    Thanks for the reply, but, yes, we have to use the comparison test.
    Ever heard of Stirling's approximation?
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  5. #5
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