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! :)