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Math Help - tough limit

  1. #1
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    tough limit

    Does anyone know how to show this limit?

    limit as n approaches infinity of the summation from k = 0 to n of [(n-k)/n]^n.

    The summation is just the gamma symbol with k = 0 on the bottom and n on the top. Apparently, the limit equals e/(e-1), but how would you show it with algebra?
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  2. #2
    Eater of Worlds
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    What it amounts to is whitttling it down to:

    \sum_{k=0}^{\infty}\frac{1}{e^{k}}=\frac{e}{e-1}

    Take note:

    \left(\frac{n-k}{n}\right)=\left(1-\frac{k}{n}\right)

    A take on the famous limit of e:

    \lim_{n\rightarrow{\infty}}\left(1-\frac{k}{n}\right)^{n}=\frac{1}{e^{k}}

    Now, if you take the sum of \frac{1}{e^{k}}

    \sum_{k=0}^{\infty}\frac{1}{e^{k}}=\frac{e}{e-1}
    Last edited by galactus; October 29th 2007 at 04:47 PM.
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