Finding $\lim_{n \to \infty} \dfrac{n^n}{(2n)!}$

Question

Struggling to apply Squeeze THM to find this limit. Specifically, I need a sequence which is always larger than the one in the problem, but which can easily be derived from the middle sequence.

Answer 1

In the denominator $(2n)!= (2n)(2n-1) \dots (n+1)\cdot n!$. Each of the factors from $(n+1)$ to $(2n)$ are larger than $n$; there are $n$ of these factors. So you can show that this sequence is less than $1/n!$.

Answer 2

$$ \frac{\frac{(n+1)^{n+1}}{(2(n+1))!}}{\frac{n^n}{(2n)!}} = \frac{\left(\frac{n+1}n\right)^n}{2(2n+1)}<\frac3{4n} $$ Now, by induction, $$\frac{n^n}{(2n)!}\le K\frac{3^n}{4^n(n-1)!}$$ for some $K>0$.

Answer 3

Basically,

$$ (2n)!=2n\,(2n-1)\,(2n-2)\cdots\,(n+1)\,n!\ \geq\ n^n\,n! $$ because each of the factor before $n!$ is g.e.q. than $n$, and there are $2n-n=n$ of these factors.

So: $$ n^n\frac{1}{(2n)!}\leq\ n^n\frac{1}{n^nn!} $$

and therefore your sequence is squeezed.

Answer 4

Use Stirling approximation for large $n$:

$$(2n)! = \sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}$$

So

$$\lim_{n\to +\infty} \frac{n^n}{(2n)!} = \lim_{n\to +\infty}\frac{n^n\cdot e^{2n}}{2\sqrt{\pi n} (2n)^{2n}} = \frac{1}{2\sqrt{\pi}}\left(\frac{e^2}{4}\right)^{n}\frac{1}{\sqrt{n}n^n} \leq \frac{C}{\sqrt{n}n^n}$$

For some $C > 0$ with $C < \sqrt{n}n^n$ for $n\to +\infty$

Which means that more or less the sequence is less than $\frac{1}{n!}$