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Just a quick limit question!

  1. Dec 20, 2005 #1
    (This isn't homework, just a curiousity derived from another problem)

    Well, this is probably quite simple...:shy:

    For any natural 'k', what is the

    [tex]\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {kn} \right)!}}
    {{n^{kn} }} [/tex]

    Last edited: Dec 20, 2005
  2. jcsd
  3. Dec 20, 2005 #2


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    Homework Helper

    I love it!

    For k=0, it's 1. For k>0, we have

    [tex]\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {kn} \right)!}}{{n^{kn} }} = \sqrt{2\pi} \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {kn} \right)}^{nk+\frac{1}{2}}e^{-nk}}{{n^{kn} }} = \sqrt{2k\pi} \mathop {\lim }\limits_{n \to \infty } \sqrt{n}\left( \frac{k}{e}\right) ^{nk} = \left\{\begin{array}{cc}0,&\mbox{ if }
    k=1,2\\ \infty, & \mbox{ if } k>e\end{array}\right.[/tex]

    since by Stirling's approximation: for [itex]n \gg 1[/itex],

    [tex]n! \sim \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}[/tex].
  4. Dec 20, 2005 #3


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    Homework Helper

    Assuming you have knowledge of the Gamma function, try this one:

    Prove that [itex]\forall n,k\in\mathbb{Z} ^{+}[/itex],

    [tex]\mathop {\lim }\limits_{N \to \infty } \frac{\left[ \Gamma \left( 1+ \frac{k}{N}\right) \right] ^{n}}{\Gamma\left( 1+ \frac{nk}{N}\right)} =1[/tex]

    Hint: Use infinite products!
  5. Dec 20, 2005 #4
    Thanks; when I mentioned "curiousity derived from another problem"
    I was trying to find (from the product)

    [tex]\mathop {\lim }\limits_{n \to \infty } \prod\limits_{i = 1}^{kn} {\frac{i}
    {n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {kn} \right)!}}
    {{n^{kn} }} [/tex]
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