What is the limit of (kn)! / n^(kn) as n approaches infinity?

[bomba923]

(This isn't homework, just a curiosity derived from another problem)

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

For any natural 'k', what is the

\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {kn} \right)!}}<br /> {{n^{kn} }}

?

 

[benorin]

I love it!

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

\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,&amp;\mbox{ if }<br /> k=1,2\\ \infty, &amp; \mbox{ if } k&gt;e\end{array}\right.

since by Stirling's approximation: for n \gg 1,

n! \sim \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}.

 

[benorin]

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

Prove that \forall n,k\in\mathbb{Z} ^{+},

\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

Hint: Use infinite products!

 

[bomba923]

Thanks; when I mentioned "curiousity derived from another problem"
I was trying to find (from the product)

\mathop {\lim }\limits_{n \to \infty } \prod\limits_{i = 1}^{kn} {\frac{i}<br /> {n}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {kn} \right)!}}<br /> {{n^{kn} }}