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  1. #1
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    Limit

    Challenge Problem:

    Evaluate $\displaystyle \lim_{n \to \infty} \Bigg( n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg) $



    Moderator Edit: Approved Challenge question.
    Last edited by mr fantastic; May 19th 2010 at 08:19 PM.
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  2. #2
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    Let me have a try


    $\displaystyle
    \lim_{n \to \infty} \Bigg( n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg)
    $

    $\displaystyle exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{1}{n^2} \sum_{j=1}^n j\ln{j} \right] \}\
    $

    $\displaystyle = exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{1}{n^2} \sum_{j=1}^n j(\ln{j} - \ln{n}) + \frac{\ln{n}}{n^2} \sum_{j=1}^n j \right] \}\$

    $\displaystyle = exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{(n+1)\ln{n}}{2n}\right] + \int_0^1 x\ln{x}~dx \}\ $

    $\displaystyle = exp \left( \int_0^1 x\ln{x}~dx \right) $

    $\displaystyle = \frac{1}{ \sqrt[4]{e}}$
    Last edited by simplependulum; May 19th 2010 at 03:34 AM.
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by simplependulum View Post
    Let me have a try


    $\displaystyle
    \lim_{n \to \infty} \Bigg( n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg)
    $

    $\displaystyle exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{1}{n^2} \sum_{j=1}^n j\ln{j} \right] \}\
    $

    $\displaystyle = exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{1}{n^2} \sum_{j=1}^n j(\ln{j} - \ln{n}) + \frac{\ln{n}}{n^2} \sum_{j=1}^n j \right] \}\$

    $\displaystyle = exp \{\ \lim_{n \to \infty}\left[ -\frac{1}{2} \left( 1 + \frac{1}{n} \right) \ln{n}+ \frac{(n+1)\ln{n}}{2n}\right] + \int_0^1 x\ln{x}~dx \}\ $

    $\displaystyle = exp \left( \int_0^1 x\ln{x}~dx \right) $

    $\displaystyle = \frac{1}{ \sqrt[4]{e}}$
    That's basically my solution, too.


    let $\displaystyle y = \ln \Bigg(n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg) $

    then $\displaystyle \lim_{n \to \infty} y = \lim_{n \to \infty} \ln \Bigg(n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg) $

    $\displaystyle = \lim_{n \to \infty} \Bigg( \ln n^{{-\frac{1}{2}(1+ \frac{1}{n})}} + \ln \Big( \prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg) $

    $\displaystyle = \lim_{n \to \infty} \Big(-\frac{n+1}{2n} \ln n + \frac{1}{n^{2}} \sum_{j=1}^{n} \ln j^{j} \Big) $

    $\displaystyle = \lim_{n \to \infty} \Big(- \frac{n(n+1)}{2n^{2}} \ln n + \frac{1}{n^{2}} \sum^{n}_{j=1} j \ln j \Big) $

    $\displaystyle = \lim_{n \to \infty} \Big( - \frac{1}{n^{2}} \sum^{n}_{j=1} j \ln n + \frac{1}{n^{2}} \sum^{n}_{j=1} j \ln j \Big) $

    $\displaystyle = \lim_{n \to \infty} \frac{1}{n^{2}} \sum_{j=1}^{n} (j \ln j - j \ln n) = \lim_{n \to \infty} \frac{1}{n^{2}} \sum_{j=1}^{n} j \ln \Big(\frac{j}{n} \Big) $

    $\displaystyle = \lim_{n \to \infty} \frac{1}{n} \sum^{n}_{j=1} \frac{j}{n} \ln \Big(\frac{j}{n} \Big) = \lim_{b \to 0} \int^{1}_{b} x \ln x \ dx $

    $\displaystyle = \lim_{b \to 0} \Big(\frac{x^{2}}{2} \ln x - \frac{x^{2}}{4}\Big) \Big|^{1}_{b} = - \frac{1}{4} $

    so $\displaystyle \lim_{n \to \infty} \Bigg(n^{-\frac{1}{2}(1+ \frac{1}{n})} \Big(\prod_{j=1}^{n} j^{j} \Big)^{\frac{1}{n^{2}}} \Bigg)= \lim_{n \to \infty} e^{y} = \frac{1}{\sqrt [4]{e}} $
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