# Just a quick limit question!

1. Dec 20, 2005

### bomba923

(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

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {kn} \right)!}} {{n^{kn} }}$$

?

Last edited: Dec 20, 2005
2. Dec 20, 2005

### benorin

I love it!

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

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

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

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

3. Dec 20, 2005

### 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!

4. Dec 20, 2005

### bomba923

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

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