- #1
Zetison
- 35
- 0
Homework Statement
Show that the folowing holds:
[tex]\lim_{n\to\infty} \frac{(n!)^2 4^n}{(2n)!} = \infty[/tex]
Homework Equations
It can be shown that
[tex]\frac{(n!)^2 4^n}{(2n)!} = \prod_{k=1}^n \frac{4k}{k+n}[/tex]
The Attempt at a Solution
If I can proove that
[tex]\lim_{n\to\infty} \ln\left[\frac{(n!)^2 4^n}{(2n)!} \right] = \infty[/tex]
then I am done since the logarithm is a strictly increasing function.
By using the "Relevant equation" I have
[tex]\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)[/tex]
But from here I don't get any further...
Last edited: