Lower bound for the falling factorial $(2n)_{n}$

Question

I'm looking for a lower bound for the falling factorial $$(2n)_{n}:= \frac{(2n)!}{n!}$$

I saw on Wikipedia that $n! > \sqrt{2{\pi}n}(\frac{n}{e})^n$ . So I assume that a possible lower bound could be $$ \frac{\sqrt{4{\pi}n}\left(\frac{2n}{e}\right)^{2n}}{\sqrt{2{\pi}n}(\frac{n}{e})^n}={\sqrt{2}}\left(\frac{2n}{e}\right)^{2n}\left(\frac{n}{e}\right)^{-n}$$

but I am not sure if this is a proper lower bound because the denominator of the LHS depends on $n$ and may be closer to $n!$ than the numerator is to $(2n)!$ making the quotient greater that the true value for $(2n)_n$. I also don't even know if the bound cited in Wikipedia is correct.

Thanks for your help.

Answer 1

We have, through summation by parts:

$$ \log\frac{(2n)!}{n!} = \sum_{k=1}^{n}\log(n+k)= n\log(2n)-\sum_{k=1}^{n-1} k \log\left(1+\frac{1}{n+k}\right) $$ and since $\log(1+z)<z$ for positive $z$ it follows that: $$ \log\frac{(2n)!}{n!} \geq n\log(2n)-\sum_{k=1}^{n-1}\frac{k}{n+k}=n\log(2n)-(n-1)+n\sum_{k=1}^{n-1}\frac{1}{n+k} $$ or: $$ \log\frac{(2n)!}{n!} \geq n\log(2n)-n+1+n(H_{2n-1}-H_{n}). $$

As an alternative, you can use Stirling's inequality:

$$ \left(\frac{n}{e}\right)^n \sqrt{2\pi n}\exp\left(\frac{1}{12n+1}\right)\leq n! \leq \left(\frac{n}{e}\right)^n \sqrt{2\pi n}\exp\left(\frac{1}{12n}\right). $$

Answer 2

By Stirling's approximation$$\sqrt{2\pi}\frac{n^{(n+1)/2}}{e^{n}}\leq n!\leq\frac{n^{(n+1)/2}}{e^{n-1}}$$ so$$\frac{\left(2n\right)!}{n!}\geq\frac{\sqrt{2\pi}\frac{\left(2n\right)^{(2n+1)/2}}{e^{2n}}}{\frac{n^{(n+1)/2}}{e^{n-1}}}=\frac{\sqrt{2\pi}}{e}\frac{\left(2n\right)^{(2n+1)/2}}{e^{2n}}\frac{e^{n}}{n^{(n+1)/2}}=\frac{\sqrt{2\pi}}{e}2^{(2n+1)/2}n^{n/2}e^{-n}.$$

Answer 3

This is a rough estimate, however it's simpler than the other ones.

Since $$\lim_{n \to \infty} \left( \frac{4n}{e} \right)^{-n} \frac{(2n)!}{n!} = \sqrt{2}$$ You can approximate $$\frac{(2n)!}{n!} \sim \sqrt{2} \left( \frac{4n}{e} \right)^{n}$$ So, for example, you can say $$\frac{(2n)!}{n!} \geq \left( \frac{4n}{e} \right)^{n}$$ for $n$ large enough.