# Limit of factorial functions

1. Mar 4, 2013

### Zetison

1. The problem statement, all variables and given/known data
Show that the folowing holds:
$$\lim_{n\to\infty} \frac{(n!)^2 4^n}{(2n)!} = \infty$$

2. Relevant equations
It can be shown that
$$\frac{(n!)^2 4^n}{(2n)!} = \prod_{k=1}^n \frac{4k}{k+n}$$

3. The attempt at a solution
If I can proove that
$$\lim_{n\to\infty} \ln\left[\frac{(n!)^2 4^n}{(2n)!} \right] = \infty$$
then I am done since the logarithm is a strictly increasing function.

By using the "Relevant equation" I have
$$\ln\left[\frac{(n!)^2 4^n}{(2n)!} \right] = \ln\left[\prod_{k=1}^n \frac{4k}{k+n}\right] = \sum_{k=1}^n \ln\left(\frac{4}{1+\frac{n}{k}}\right)$$

But from here I don't get any further...

Last edited: Mar 4, 2013
2. Mar 4, 2013

### Dick

Can't you just use Stirling's approximation on the factorials in the original problem?

3. Mar 4, 2013

### hapefish

I approached it a little differently than you - here was my thinking:

in the numerator distribute the 2's so that you get two products of only even numbers: $\frac{(2*4*6*...*2n)^2}{2n!}$ This would allow you to cancel out the even terms in the denominator and get $\frac{2*4*6*...*2n}{1*3*5*...*2n-1}$. I then Identify this as $\prod_{k=1}^n \frac{2k}{2k-1}$. From here you should be able to use logs to convert this product to a sum and use the integral test (or perhaps something else) to show that the sum must diverge.

Good Luck!

4. Mar 4, 2013

### Ray Vickson

If you don't like to use Stirling's approximation, you can use a modification of it to get both lower and upper bounds: for any integer $k \geq 1$ we have
$$\sqrt{2 \pi}k^{k + \frac{1}{2}}e^{-k} < k! < \sqrt{2 \pi}k^{k + \frac{1}{2}}e^{-k+ \frac{1}{12k}}$$
This works well even for small k; for example, for k = 1 it gives
.9221370087 < 1 < 1.002274449 and for k = 2 it gives 1.919004351 < 2 < 2.000652047 .

For a simple proof, see Feller, "Introduction to Probability Theory and Its Applications, Vol I, Wiley (1967).

Last edited: Mar 4, 2013
5. Mar 4, 2013

### Zetison

Thanks for the replies! The method hapefish proposed worked very vell. The sum diverges by the comparison test with the harmonic series. I also solved the problem using stirlings formula, thanks a lot guys! :)