Results 1 to 2 of 2

Math Help - Limit

  1. #1
    Newbie
    Joined
    Dec 2011
    Posts
    19

    Limit

    Prove that \lim_{n \to \infty} \left( \frac{n!}{n^n}\right)^{1 \over n}=\frac{1}{e}.

    I wrote the limit as \exp{\lim_{n \to \infty} {1 \over n}\ln\left( \frac{n!}{n^n}\right)}

    What should I do next?? Can I use L hospital's rule (in any way)?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member sbhatnagar's Avatar
    Joined
    Sep 2011
    From
    New Delhi, India
    Posts
    200
    Thanks
    17

    Re: Limit

    1.
    k=\lim_{n \to \infty}\left( \frac{n!}{n^n}\right)^{\frac{1}{n}}= \exp{\left[ \lim_{n \to \infty}{\frac{1}{n}}}\ln\left( \frac{n!}{n^n}\right)\right]= \exp{\left[ \lim_{n \to \infty}{\frac{1}{n}}}\ln\left( \frac{1\cdot2\cdot3\cdots (n-1)\cdot n}{n\cdot n \cdot n \cdots n \cdot n}\right)\right]

    =\exp{\left[ \lim_{n \to \infty} {1 \over n}\sum_{r=1}^{n}\ln\left( \frac{r}{n}\right)\right]}

    2. It is known that : \lim_{n \to \infty}\frac{1}{n}\sum_{r=1}^{n}f\left( \frac{r}{n}\right)=\int_{0}^{1} f(x) \ dx. Therefore :

    k=\exp{\left( \int_{0}^{1}\ln(x) dx \right) }=\exp[{\left(  x\cdot\ln(x) -x\right)^1_0 }]=\exp{\left(-1-\lim_{x \to 0}x\ln(x) \right)}

    Finish it...
    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