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Math Help - Test for Series Convergence/Divergence

  1. #1
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    Test for Series Convergence/Divergence

    Hi all, I have a series sum from n = 1 to infinity for n!/n^n.

    I've used the ratio test only to come up with a 1 which is obviously inconclusive. I tried the root test and also came up with a 1 though it's quite possible I mixed this up. I feel as though it has to do with either direct comparison or limit comparison but I'm drawing blanks and any help would be greatly appreciated.

    Zo
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  2. #2
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    Quote Originally Posted by zo1971so View Post
    Hi all, I have a series sum from n = 1 to infinity for n!/n^n.

    I've used the ratio test only to come up with a 1 which is obviously inconclusive. I tried the root test and also came up with a 1 though it's quite possible I mixed this up. I feel as though it has to do with either direct comparison or limit comparison but I'm drawing blanks and any help would be greatly appreciated.

    Zo
    let me guess ...

    you did the ratio test and came up with

    \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n

    and you think the limit is 1 , right?
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  3. #3
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    I didn't even recognize it as 1/e with it written as (n^n)/(n+1)^n

    Thanks a ton.

    Edit 2: You beat me to it, surprise = 0
    Last edited by zo1971so; April 29th 2010 at 05:19 PM. Reason: oops 1/e not e
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  4. #4
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    Quote Originally Posted by zo1971so View Post
    I didn't even recognize it as e with it written as (n^n)/(n+1)^n

    Thanks a ton.
    actually, the limit is \frac{1}{e} ...
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  5. #5
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    For those of us who are a little slow (i.e., me), I've written out a detailed proof

    We can use the ratio test to test for this series' convergence.

    L=\lim_{n\to\infty}\frac{\frac{\left(n+1\right)!}{  \left(n+1\right)^{n+1}}}{\frac{n!}{n^n}}
    =\lim_{n\to\infty}\frac{\left(n+1\right)!}{\left(n  +1\right)^{n+1}}\frac{n^n}{n!}
    =\lim_{n\to\infty}\frac{\frac{n+1}{n^n}}{\left(n+1  \right)^{n+1}}
    =\lim_{n\to\infty}\frac{n^n}{\left(n+1\right)^n}
    =\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n
    =\lim_{n\to\infty}\left(\frac{1}{\left(n+1\right)/n}\right)^n
    =\frac{1}{e}

    The ratio test specifies that if L<1 the series converges absolutely. Thus, this series converges absolutely.

    http://maths-help.info/questions/10
    Last edited by lovek323; April 29th 2010 at 06:08 PM. Reason: Added link
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