Results 1 to 3 of 3

Math Help - Limit

  1. #1
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3

    Limit

    Challenge Problem:

    Evaluate  \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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Let me have a try


    <br />
\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)<br />

     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] \}\<br />

     = 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] \}\

     = 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 \}\

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

     = \frac{1}{ \sqrt[4]{e}}
    Last edited by simplependulum; May 19th 2010 at 03:34 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    Quote Originally Posted by simplependulum View Post
    Let me have a try


    <br />
\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)<br />

     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] \}\<br />

     = 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] \}\

     = 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 \}\

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

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


    let  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  \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)

     = \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)

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

     = \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)

      = \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)

     = \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)

     = \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

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

    so  \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}}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 8th 2010, 11:29 AM
  2. Replies: 1
    Last Post: February 5th 2010, 03:33 AM
  3. Replies: 16
    Last Post: November 15th 2009, 04:18 PM
  4. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 05:05 PM
  5. Replies: 15
    Last Post: November 4th 2007, 07:21 PM

Search Tags


/mathhelpforum @mathhelpforum