Results 1 to 7 of 7

Math Help - [SOLVED] Convergence of an infinite series

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    [SOLVED] Convergence of an infinite series

    I've to show whether the following infinite series converges or not. The problem is that I tried to prove it's divergent by induction, by the root test and by trying to bound it from below. But I couldn't reach any conclusion.
    Here it comes : \sum_{n=1}^{+\infty} \frac{n!}{n^n}. I'm sure it's a famous one. It's very similar to the one that defines e^n (inversed).
    I'd like you to tell me what test should I try. Then if I'm not able to get a conclusion, I'll ask for further help. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by arbolis View Post
    I've to show whether the following infinite series converges or not. The problem is that I tried to prove it's divergent by induction, by the root test and by trying to bound it from below. But I couldn't reach any conclusion.
    Here it comes : \sum_{n=1}^{+\infty} \frac{n!}{n^n}. I'm sure it's a famous one. It's very similar to the one that defines e^n (inversed).
    I'd like you to tell me what test should I try. Then if I'm not able to get a conclusion, I'll ask for further help. Thanks in advance.
    Root test...just so you know it is (high-light below)

    convergent

    Or consider that for very large n we have that (high-light)
    n!<<n^n by stirlings approximation
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,388
    Thanks
    1476
    Awards
    1
    Does it help to know that \left( {\sqrt[n]{{\frac{{n!}}{{n^n }}}}} \right) \to \frac{1}{e}?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Or consider that for very large n we have that (high-light)
    n!<<n^n by stirlings approximation
    My intuition fails! I thought that n! would have been much greater than n^n. But from what I did, all seemed to show the contrary and I couldn't believe it.
    I was trying to show that if n^{n+1}+n \geq (n+1)^{n+1}, then it would diverges. I recognized it was false, but I wouldn't trust my result!
    Does it help to know that ?
    Of course! Since \frac{1}{e}<1, the series converges... Now I will try to prove it!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Plato showed you, but if you wanted you could just use the ratio test:

    \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!}=\f  rac{n^{n}}{(n+1)^{n}}

    {\rho}=\lim\frac{n^{n}}{(n+1)!}=\frac{1}{e}

    {\rho}<1...........convergent
    Last edited by galactus; July 10th 2008 at 01:41 PM. Reason: Sorry, I reckon this was rather redundant.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by galactus View Post
    Plato showed you, but if you wanted you could just use the ratio test:

    \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!}=\f  rac{n^{n}}{(n+1)^{n}}

    {\rho}=\lim\frac{n^{n}}{(n+1)!}=\frac{1}{e}

    {\rho}<1...........convergent
    Might as well jump on the band-wagon.

    \text{As}\quad{n\to\infty}\quad{n!\sim\sqrt{2\pi{n  }}n^ne^{-n}}


    \lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}\sim\lim_{n\to\infty}\sqrt[n]{\frac{\sqrt{2\pi{n}}n^ne^{-n}}{n^n}}=\frac{1}{e}\lim_{n\to\infty}(2\pi{n})^{\  frac{1}{2n}}

    so using the connection of the root and ratio test

    \lim_{n\to\infty}(2\pi{n})^{\frac{1}{2n}}=\sqrt{\l  im_{n\to\infty}(2\pi{n})^{\frac{1}{n}}}=\sqrt{\lim  _{n\to\infty}\frac{2\pi(n+1)}{2\pi{n}}}=\sqrt{1}=1

    \therefore\sqrt[n]{\lim_{n\to\infty}\frac{n!}{n^n}}=\frac{1}{e}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    We certainly know it converges. Now find what it converges to.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: October 18th 2009, 06:47 AM
  2. [SOLVED] Infinite series convergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 30th 2008, 09:19 AM
  3. [SOLVED] Convergence of an infinite series
    Posted in the Calculus Forum
    Replies: 8
    Last Post: November 29th 2008, 01:38 PM
  4. [SOLVED] Infinite Series Convergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 4th 2008, 06:48 PM
  5. [SOLVED] Infinite series convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 12th 2008, 12:02 AM

Search Tags


/mathhelpforum @mathhelpforum