Find $\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}.$ Find $$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}.$$
I have tried the following:
$$(2n-1)!!=\frac{(2n)!}{2^{2n}n!}$$
$$(2n)!!=2^nn!$$
$$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!}=\lim_{n\rightarrow \infty}\frac{(2n)!}{2^{2n}(n!)^2}$$
Now using Stirling approximation,
$$n!=\sqrt {2\pi n}\left(\frac{n}{e}\right)^{n}$$ 
$$(2n)!=\sqrt {2\pi 2n}\left(\frac{2n}{e}\right)^{2n}$$
$$\lim_{n\rightarrow \infty}\frac{(2n)!}{2^{2n}(n!)^2}=\lim_{n\rightarrow \infty}\frac{\sqrt{2\pi}\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{2^{2n}(\sqrt{2\pi})^2(\sqrt{n})^2\left(\frac{n}{e}\right)^{2n}}$$
$$=\frac{1}{\sqrt{2\pi}}\lim_{n\rightarrow \infty}\frac{\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{2^{2n}(\sqrt{n})^2\left(\frac{n}{e}\right)^{2n}}$$
How to proceed with solving this limit?
 A: We have
$$ \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^n}\binom{2n}{n} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)=\frac{1}{2}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right) \tag{A}$$
and by squaring both sides
$$ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)=\frac{1}{4n}\prod_{k=1}^{n-1}\left(1+\frac{1}{4k(k+1)}\right)\tag{B} $$
from which:
$$ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\leq \frac{1}{4n}\prod_{k=1}^{n-1}\exp\left(\frac{1}{4k}-\frac{1}{4(k+1)}\right)\leq \frac{e^{1/4}}{4n}\tag{C} $$
implying that the wanted limit is zero.
A: $$\frac{1}{\sqrt{2\pi}}\lim_{n\rightarrow \infty}\frac{\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{2^{2n}(\sqrt{n})^2\left(\frac{n}{e}\right)^{2n}}=\frac{1}{\sqrt{2\pi}}\lim_{n\rightarrow \infty}\frac{\sqrt{2n}\left(\frac{2n}{e}\right)^{2n}}{(\sqrt{n})^2\left(\frac{2n}{e}\right)^{2n}} =\frac{1}{\sqrt{2\pi}}\lim_{n\rightarrow \infty}\frac{\sqrt{2n}}{(\sqrt{n})^2}$$
$$=\frac{1}{\sqrt{\pi}}\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}$$
A: We have
$$ \frac{(2n-1)!!}{(2n)!!}  = \frac{(2n-1)!!}{((2n)(2n-1))!!}\le \frac{(2n-1)!!}{(2n)((2n-1)!!)}=\frac{1}{2n}$$
Now we have upper bound on our equation. Note that as $n \to \infty$, our upper bound tends to 0.
$$\lim_{n\rightarrow \infty}\frac{(2n-1)!!}{(2n)!!} \le0$$
Hence our answer is 0
