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Math Help - Factorial limits

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Factorial limits

    Someone told me this wasn't correct?

    Could anyone clarify please

    I wanted to show that \lim_{n\to\infty}\frac{3^nn!}{(2n)!}=0
    two ways

    Method one

    Since \frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!},\forall{n}>  0

    and also since as n gets sufficiently large there \exists\text{ an }N\backepsilon\forall{n>N}\frac{3^nn!}{(2n)!}\leq\  frac{n^n}{(2n)!}


    So from there I set up my inequalities

    \frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!}\leq\frac{n^  n}{(2n)!}

    and \therefore\lim_{n\to\infty}\frac{1}{(2n)!}\leq\lim  _{n\to\infty}\frac{3^nn!}{(2n)!}\leq\lim_{n\to\inf  ty}\frac{n^n}{(2n)!}


    Knowing the left and right integrals we see that

    0\leq\lim_{n\to\infty}\frac{3^nn!}{(2n)!}\leq{0}

    \therefore\lim_{n\to\infty}\frac{3^nn!}{(2n)!}=0

    by the squeeze theorem

    Method two

    Since as n\to\infty

    n!\sim{\sqrt{2\pi{n}}n^ne^{-n}}

    and (2n)!\sim{\sqrt{4\pi{n}}(2n)^{2n}e^{-2n}}=2\sqrt{\pi{n}}4^n(n^n)^2(e^{-n})^2




    \therefore\lim_{n\to\infty}\frac{3^nn!}{(2n)!}=\li  m_{n\to\infty}\frac{3^n\sqrt{2}\sqrt{\pi{n}}n^ne^{-n}}{2\sqrt{\pi{n}}4^n(n^n)^2(e^{-n})^2}=\lim_{n\to\infty}\frac{3^ne^n}{\sqrt{2}4^nn  ^n}




    and just to completely be thorough I included this step




    \lim_{n\to\infty}\frac{3^ne^n}{\sqrt{2}4^nn^n}=\fr  ac{1}{\sqrt{2}}\cdot\lim_{n\to\infty}\frac{3^n}{4^  n}\cdot\lim_{n\to\infty}\frac{e^n}{n^n}=\frac{1}{\  sqrt{2}}\cdot{0}\cdot{0}=0


    could somone point me in the direction of my mistake? I am sure it is a wrong assumption

    Thanks...Mathstud
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello


    Method one

    Since \frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!},\forall{n}>  0

    and also since as n gets sufficiently large there \exists\text{ an }N\backepsilon\forall{n>N}\frac{3^nn!}{(2n)!}\leq\  frac{n^n}{(2n)!}
    You should show that 3^nn!<n^n.
    So from there I set up my inequalities

    \frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!}\leq\frac{n^  n}{(2n)!}

    and \therefore\lim_{n\to\infty}\frac{1}{(2n)!}\leq\lim  _{n\to\infty}\frac{3^nn!}{(2n)!}\leq\lim_{n\to\inf  ty}\frac{n^n}{(2n)!}


    Knowing the left and right integrals
    You should also show that \lim_{\infty}\frac{n^n}{(2n)!}=0


    we see that

    0\leq\lim_{n\to\infty}\frac{3^nn!}{(2n)!}\leq{0}

    \therefore\lim_{n\to\infty}\frac{3^nn!}{(2n)!}=0

    by the squeeze theorem
    I did not notice anything wrong... (by the way, you don't need \frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!} to apply the squeeze theorem since \frac{3^nn!}{(2n)!}>0)

    Method two

    [...]
    It is correct too.
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  3. #3
    Super Member wingless's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Someone told me this wasn't correct?
    and also since as n gets sufficiently large there \exists\text{ an }N\backepsilon\forall{n>N}\frac{3^nn!}{(2n)!}\leq\  frac{n^n}{(2n)!}
    This is not true.

    3^n n! \geq n^n for all n>0.
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  4. #4
    Super Member wingless's Avatar
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    \lim_{n\to\infty} \frac{3^n n!}{(2n)!}

    \lim_{n\to\infty} \frac{\overbrace{(3\cdot 3\cdot 3\cdot...\cdot 3)}^{\text{n terms}}\cdot \overbrace{(\not 1\cdot \not 2 \cdot \not 3 \cdot...\cdot \not n)}^{\text{n terms}}}{\underbrace{[\not1\cdot \not 2 \cdot \not 3 \cdot...\cdot \not n \cdot (n+1) \cdot (n+2) \cdot ...\cdot (2n)]}_{\text{2n terms}}}

    \lim_{n\to\infty} \frac{\overbrace{(3\cdot 3\cdot 3\cdot...\cdot 3)}^{\text{n terms}}}{\underbrace{[(n+1)\cdot(n+2)\cdot(n+3)\cdot...\cdot(2n)]}_{\text{n terms}}}

    \lim_{n\to\infty} \underbrace{\frac{3}{n+1}\cdot\frac{3}{n+2}\cdot\f  rac{3}{n+3}\cdot...\cdot \frac{3}{2n}}_{\text{n terms}}

    Each term will approach 0 while n approaches infinity. So we can say that \lim_{n\to\infty} \frac{3^n n!}{(2n)!} = 0
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hello


    You should show that 3^nn!<n^n.

    You should also show that \lim_{\infty}\frac{n^n}{(2n)!}=0


    I did not notice anything wrong... (by the way, you don't need \frac{1}{(2n)!}\leq\frac{3^nn!}{(2n)!} to apply the squeeze theorem since \frac{3^nn!}{(2n)!}>0)

    It is correct too.
    Quote Originally Posted by wingless View Post
    \lim_{n\to\infty} \frac{3^n n!}{(2n)!}

    \lim_{n\to\infty} \frac{\overbrace{(3\cdot 3\cdot 3\cdot...\cdot 3)}^{\text{n terms}}\cdot \overbrace{(\not 1\cdot \not 2 \cdot \not 3 \cdot...\cdot \not n)}^{\text{n terms}}}{\underbrace{[\not1\cdot \not 2 \cdot \not 3 \cdot...\cdot \not n \cdot (n+1) \cdot (n+2) \cdot ...\cdot (2n)]}_{\text{2n terms}}}

    \lim_{n\to\infty} \frac{\overbrace{(3\cdot 3\cdot 3\cdot...\cdot 3)}^{\text{n terms}}}{\underbrace{[(n+1)\cdot(n+2)\cdot(n+3)\cdot...\cdot(2n)]}_{\text{n terms}}}

    \lim_{n\to\infty} \underbrace{\frac{3}{n+1}\cdot\frac{3}{n+2}\cdot\f  rac{3}{n+3}\cdot...\cdot \frac{3}{2n}}_{\text{n terms}}

    Each term will approach 0 while n approaches infinity. So we can say that \lim_{n\to\infty} \frac{3^n n!}{(2n)!} = 0
    To Flyingsquirrel
    Thank you very much!

    As for your comment on the limit, I know we both know it is 0; but to show it see below


    Using the same trick of asymptotic equivlaence

    \lim_{n\to\infty}\frac{n^n}{(2n)!}=\lim_{n\to\inft  y}\frac{e^{2n}}{2\sqrt{\pi{n}}4^nn^n}=0

    To Wingless

    Thank you very much. As for your first comment on the falseness of that inequality, I really appreciate it, its weird I am like inequality impaired . I will be more careful next time.

    As for your proof, I thought about this but it wasn't as aesthetically pleasing.
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    To Wingless

    Thank you very much. As for your first comment on the falseness of that inequality, I really appreciate it, its weird I am like inequality impaired . I will be more careful next time.

    As for your proof, I thought about this but it wasn't as aesthetically pleasing.
    You may be right about that. OK, then let me say:

    \left(\frac{3}{2n}\right)^n \leq \left ( \frac{3}{n+1}\cdot\frac{3}{n+2}\cdot\frac{3}{n+3}\  cdot...\cdot \frac{3}{2n} \right ) \leq \left(\frac{3}{n}\right)^n

    \lim_{n\to\infty} \left(\frac{3}{n}\right)^n = \lim_{n\to\infty} \left(\frac{3}{2n}\right)^n = 0
    ( You can find this easily by a cheap l'hopital trick, turn (3/n)^n into e^(n ln(3/n)... )

    I hope it's clear now =)
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by wingless View Post
    You may be right about that. OK, then let me say:

    \left(\frac{3}{2n}\right)^n \leq \left ( \frac{3}{n+1}\cdot\frac{3}{n+2}\cdot\frac{3}{n+3}\  cdot...\cdot \frac{3}{2n} \right ) \leq \left(\frac{3}{n}\right)^n

    \lim_{n\to\infty} \left(\frac{3}{n}\right)^n = \lim_{n\to\infty} \left(\frac{3}{2n}\right)^n = 0
    ( You can find this easily by a cheap l'hopital trick, turn (3/n)^n into e^(n ln(3/n)... )

    I hope it's clear now =)
    Yeah thanks! I would never use L'hopital's on something like this, its probably easiest with the stirlings formula, it isnt as fun anyways
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  8. #8
    Super Member wingless's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Yeah thanks! I would never use L'hopital's on something like this, its probably easiest with the stirlings formula, it isnt as fun anyways
    Ain't this the spirit of a true mathematician or what?
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  9. #9
    Super Member PaulRS's Avatar
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    Ok, note that \sqrt[n]{n!}\sim{\frac{n}{e}} (*) ( use Stirling's formula if you want, here you have a proof of (*): http://www.mathhelpforum.com/math-he...-sequence.html )

    We have: \lim \frac{\sqrt[n]{n!}}{\sqrt[n]{(2n)!}}=\lim \frac{\frac{n}{e}}{\left(\frac{2n}{e}\right)^2}=0

    Thus: \lim \sqrt[n]{\frac{3^n\cdot{n!}}{(2n)!}}=0 and that can only happen if \lim \frac{3^n\cdot{n!}}{(2n)!}=0

    Try proving that <br />
\sum\limits_{n = 0}^\infty  {\tfrac{{\left( {n!} \right)}}<br />
{{\left( {2n} \right)!}} \cdot x^n }  = 2 \cdot x \cdot \int_0^1 {\left( {t - t^2 } \right) \cdot e^{x \cdot \left( {t - t^2 } \right)} dt}  + \int_0^1 {e^{x \cdot \left( {t - t^2 } \right)} dt} <br />
    Hint: Beta Function
    Last edited by PaulRS; May 22nd 2008 at 05:39 PM.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Ok, note that \sqrt[n]{n!}\sim{\frac{n}{e}} (*) ( use Stirling's formula if you want, here you have a proof of (*): http://www.mathhelpforum.com/math-he...-sequence.html )

    We have: \lim \frac{\sqrt[n]{n!}}{\sqrt[n]{(2n)!}}=\lim \frac{\frac{n}{e}}{\left(\frac{2n}{e}\right)^2}=0

    Thus: \lim \sqrt[n]{\frac{3^n\cdot{n!}}{(2n)!}}=0 and that can only happen if \lim \frac{3^n\cdot{n!}}{(2n)!}=0

    Try proving that <br />
\sum\limits_{n = 0}^\infty {\tfrac{{\left( {n!} \right)}}<br />
{{\left( {2n} \right)!}} \cdot x^n } = 2 \cdot x \cdot \int_0^1 {\left( {t - t^2 } \right) \cdot e^{x \cdot \left( {t - t^2 } \right)} dt} + \int_0^1 {e^{x \cdot \left( {t - t^2 } \right)} dt} <br />
    Hint: Beta Function
    Where is the hint?

    And it appears to be a very convoluted form of the gamma or beta function..one of the two
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