Convergence of Factorial Functions at Infinity

[Zetison]

Homework Statement

Show that the folowing holds:
\lim_{n\to\infty} \frac{(n!)^2 4^n}{(2n)!} = \infty

Homework Equations

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

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...

 

[Dick]

Homework Statement

Show that the folowing holds:
\lim_{n\to\infty} \frac{(n!)^2 4^n}{(2n)!} = \infty

Homework Equations

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

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...

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

 

[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!

 

[Ray Vickson]

Homework Statement

Show that the folowing holds:
\lim_{n\to\infty} \frac{(n!)^2 4^n}{(2n)!} = \infty

Homework Equations

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

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...

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).

 

[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! :)