Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - limit equal to 0?

  1. #1
    Member disclaimer's Avatar
    Joined
    Dec 2007
    Posts
    77

    limit equal to 0?

    Hello everyone;

    This might be a silly question, but what is the limit of \frac{\pi\sqrt[n]{n!}}{n} when n tends to infinity? To my mind, it would be 0. What do you think?

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Can you use the inequality \left( {n - 1} \right)! \le n^n e^{ - n} e \le n! to prove the following?
    \lim _{n \to \infty } \left( {\frac{{\sqrt[n]{{n!}}}}{{n }}} \right) = \frac{1}{e}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Actually, this limit is not 0.

    \lim_{n\rightarrow{\infty}}\frac{(n!)^{\frac{1}{n}  }}{n}=\frac{1}{e}

    So, with the Pi, your limit is \frac{\pi}{e}

    EDIT: Oops, Plato beat me.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Plato View Post
    Can you use the inequality \left( {n - 1} \right)! \le n^n e^{ - n} e \le n! to prove the following?
    \lim _{n \to \infty } \left( {\frac{{\sqrt[n]{{n!}}}}{{n }}} \right) = \frac{1}{e}
    where did that inequality come from?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by galactus View Post
    Actually, this limit is not 0.

    \lim_{n\rightarrow{\infty}}\frac{(n!)^{\frac{1}{n}  }}{n}=\frac{1}{e}

    So, with the Pi, your limit is \frac{\pi}{e}

    EDIT: Oops, Plato beat me.
    O wait, i did this problem already. i should have known this (TPH would be so disappointed in me)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    where did that inequality come from?
    Here is a slightly different version.
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member disclaimer's Avatar
    Joined
    Dec 2007
    Posts
    77
    It's been bugging me for quite long, I wanted to determine if a series <br />
\sum_{n=1}^{\infty}\left(\frac{\pi}{n}\right)^nn!<br />
is convergent or divergent.

    By d'Alembert's criterion:

    a_{n}=\left(\frac{\pi}{n}\right)^{n}(n)!
    a_{n+1}=\left(\frac{\pi}{n+1}\right)^{n+1}(n+1)!
    \lim_{n\rightarrow{\infty}}\left|\frac{a_{n+1}}{a_  n}\right|=\lim_{n\rightarrow{\infty}}\frac{\left(\  frac{\pi}{n+1}\right)^{n+1}(n+1)!}{\left(\frac{\pi  }{n}\right)^nn!}=\lim_{n\rightarrow{\infty}}\frac{  \left(\frac{\pi}{n}\right)^n\left(\frac{\pi}{n}\ri  ght)n!(n+1)}{\left(\frac{\pi}{n}\right)^nn!}=
    =\lim_{n\rightarrow{\infty}}\left[\pi\left(\frac{n}{n+1}\right)^n\right]=\lim_{n\rightarrow{\infty}}\left[\pi\left(1+\frac{1}{n}\right)^n\right]^{-1}=\pi{e}^{-1}=\frac{\pi}{e}

    Then \frac{\pi}{e}>1, so \sum_{n=1}^{\infty}\left(\frac{\pi}{n}\right)^nn! is divergent.

    By Cauchy's criterion:

    \lim_{n\rightarrow{\infty}}\sqrt[n]{a_n}=\lim_{n\rightarrow{\infty}}\sqrt[n]{\left(\frac{\pi}{n}\right)^nn!}=\lim_{n\rightarro  w{\infty}}\frac{\pi}{n}\sqrt[n]{n!}=\lim_{n\rightarrow{\infty}}\frac{\pi\sqrt[n]{n!}}{n}=?

    That's where I was left puzzled. Were the limit equal to zero, then by Cauchy's criterion the series would be convergent. Now that I know it's also \frac{\pi}{e}, I have no more doubts.

    Thanks a lot for your replies.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    where did that inequality come from?
    I haven't bothered to look it up, but it's likely a variation on Stirling's formula for n! (where n is very large.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    What are e, \forall, ln?

    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by SengNee View Post
    What are e, \forall, ln?

    "e" is "Napier's constant," the number 2.718281828459045235360287471352662497757... This number has a lot of special properties, which I'll let you look up.

    \forall is a Math hieroglyph for the words "for all." It is most commonly used in set theory.

    "ln()" is a logarithmic function, for example: f(x) = \log_e(x) = ln(x), where e is Napier's number above.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    You can use the simple fact. If a_n is a sequence of non-zero terms such that a_{n+1}/a_n \to L then |a_n|^{1/n}\to L also.

    (This was an excercise in section 12 # 14 in Ross book, if I remember correcly).

    Thus, if you let a_n = \frac{n!}{n^n} then \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n!}{n^n} \to \frac{1}{e}.

    That means, (a_n)^{1/n} = \frac{1}{n}(n!)^{1/n} \to \frac{1}{e}.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by disclaimer View Post
    Hello everyone;

    This might be a silly question, but what is the limit of \frac{\pi\sqrt[n]{n!}}{n} when n tends to infinity? To my mind, it would be 0. What do you think?

    Thanks in advance.
    There is a nice trick with factorials, which is like killing a fly with a cannon, but it works well over here.

    Sitriling approximation says you can replace n! by \sqrt{2\pi} \cdot \sqrt{n} \cdot n^n \cdot e^{-n}.

    That means,
    \frac{\pi (n!)^{1/n}}{n} \to  \frac{\pi \left[ \sqrt{2\pi} \cdot \sqrt{n} \cdot n^n \cdot e^{-n} \right]^{1/n}}{n} \to \frac{\pi}{e}.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    Quote Originally Posted by topsquark View Post
    "e" is "Napier's constant," the number 2.718281828459045235360287471352662497757... This number has a lot of special properties, which I'll let you look up.

    \forall is a Math hieroglyph for the words "for all." It is most commonly used in set theory.

    "ln()" is a logarithmic function, for example: f(x) = \log_e(x) = ln(x), where e is Napier's number above.

    -Dan
    Can you show me some properties of e and how to use \forall in set theory?
    Last edited by SengNee; January 13th 2008 at 08:50 AM. Reason: Typing error...
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by SengNee View Post
    Can you sow me some properties of e and how to use \forall in set theory?
    Here are some properties of e.
    • \lim ~ \left( 1 + \frac{1}{n} \right)^n = e
    • 1+\frac{1}{1!}+\frac{1}{2!}+... = \sum_{n=0}^{\infty} \frac{1}{n!} = e
    • \int_1^e \frac{dt}{t} = 1 --> this is what I use as the definition of e.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    Quote Originally Posted by ThePerfectHacker View Post
    Here are some properties of e.
    • \lim ~ \left( 1 + \frac{1}{n} \right)^n = e
    • 1+\frac{1}{1!}+\frac{1}{2!}+... = \sum_{n=0}^{\infty} \frac{1}{n!} = e
    • \int_1^e \frac{dt}{t} = 1 --> this is what I use as the definition of e.


    \sum_{n=0}^{\infty} \frac{1}{n!} =1+\frac{1}{1!}+\frac{1}{2!}+... = e

    or

    \sum_{n=0}^{\infty} \frac{1}{n!} =0+1+\frac{1}{1!}+\frac{1}{2!}+... = e


    ???
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: July 2nd 2011, 12:35 PM
  2. When a limit is equal to a number...
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: September 28th 2010, 07:50 PM
  3. Why does this limit equal e?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 29th 2009, 04:37 PM
  4. Replies: 7
    Last Post: November 16th 2009, 05:07 PM
  5. strange limit equal to second derivative?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 23rd 2009, 06:32 PM

Search Tags


/mathhelpforum @mathhelpforum