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Math Help - Evaluate The Following Limit. (2)

  1. #1
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    Evaluate The Following Limit. (2)

    Hello Evaluate: \lim_{n\to\infty} \frac{1}{n} [ (\frac{1}{n})^{9}+(\frac{2}{n})^{9} + (\frac{3}{n})^{9} + (\frac{4}{n})^{9} + .... + (\frac{n}{n})^{9} ]

    Please do not talk about Raiman Sums.
    because I did not study it at my college.
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  2. #2
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    Not sure

    This is outside my area, but it looks like an application for L'Hopital's Rule.
    Something to try unless someone else speaks up!
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  3. #3
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    Quote Originally Posted by Manx View Post
    This is outside my area, but it looks like an application for L'Hopital's Rule.
    Something to try unless someone else speaks up!
    it's a Riemann sum, plain and simple.

    since the OP doesn't want to hear it, I'll tell you ...

    the limit equals \int_0^1 x^9 \, dx
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TWiX View Post
    Hello Evaluate: \lim_{n\to\infty} \frac{1}{n} [ (\frac{1}{n})^{9}+(\frac{2}{n})^{9} + (\frac{3}{n})^{9} + (\frac{4}{n})^{9} + .... + (\frac{n}{n})^{9} ]

    Please do not talk about Raiman Sums.
    because I did not study it at my college.
    I mean, friend. It's Riemann sums. Plain and simple. You can do a laborious argument to show that it's bounded above and below by terms that approach one-tenth.

    Or, you can just note that this is \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\f  rac{k}{n}\right)^9=\lim_{n\to\infty}\frac{1}{n^{10  }}\sum_{k=1}^{n}k^9. This sum can be proved by induction to be \frac{1}{20} n^2 (n+1)^2 (n^2+n-1) (2 n^4+4 n^3-n^2-3 n+3). Take the limit and you will get one-tenth. Ta-da!
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    For the edification of the OP: Power Sum -- from Wolfram MathWorld
    Last edited by Jhevon; January 21st 2010 at 02:06 PM.
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