Proving an inequality involving factorials: $\frac{1}{2\sqrt{n}}\le \frac{(2n)!}{(n!)^2}\cdot\frac{1}{2^{2n}}\le\frac{1}{\sqrt{2n+1}}$ I had a problem in probability and I've solved the first part. For the second part I need to prove the inequalities :
$$\frac{1}{2\sqrt{n}}\le \frac{(2n)!}{(n!)^2}\cdot\frac{1}{2^{2n}}\le\frac{1}{\sqrt{2n+1}}$$
I tried using Stirling's formula but it looks too complicated in the end. Any help would be appreciated.
 A: $$\begin{eqnarray*}\frac{1}{4^n}\binom{2n}{n}=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)&=&\sqrt{\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)}\\&=&\frac{1}{2\sqrt{n}}\sqrt{\prod_{k=2}^{n}\left(1-\frac{1}{(2k-1)^2}\right)^{-1}}\\(\text{Wallis product})\quad&=&\frac{1}{\sqrt{\pi n}}\sqrt{\prod_{k>n}\left(1+\frac{1}{4k(k+1)}\right)^{-1}}\end{eqnarray*}$$
is trivially $\leq\frac{1}{\sqrt{\pi n}}$ but also
$$\begin{eqnarray*}\phantom{\frac{1}{4^n}\binom{2n}{n}aaaaaaaaa}&\geq &\frac{1}{\sqrt{\pi n}}\sqrt{\prod_{k>n}\exp\left(\frac{1}{4(k+1)}-\frac{1}{4k}\right)}\\(\text{Telescopic})\quad&=&\frac{1}{\sqrt{\pi n\exp\frac{1}{4n}}}\end{eqnarray*}$$
and the double inequality can be improved up to:
$$\boxed{ \frac{1}{\sqrt{\pi\left(n+\frac{1}{3}\right)}}\leq\frac{1}{4^n}\binom{2n}{n}\leq\frac{1}{\sqrt{\pi\left(n+\frac{1}{4}\right)}}.}$$
A: It can be solved directly by induction.
Note that
$$
(n!)^22^{2n}=(2\cdot 4\cdot 6 \cdot\cdots \cdot 2n)^2,
$$
the middle expression can be rewritten as
$$
\prod_{k=1}^n\left(1-\frac{1}{2k}\right).
$$
By indcution, for the right inequality, it suffices to show that
$$
\frac{1}{\sqrt{2n+1}}\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+3}}.
$$
This is equivalent to
$$
(2n+1)(2n+3)<(2n+2)^2,
$$which is trivial.
For the left inequality, it suffices to show that
$$
\frac{1}{2\sqrt{n}}\frac{2n+1}{2n+2}>\frac{1}{2\sqrt{n+1}}.
$$
This is equivalent to 
$$
(2n+1)^2>2n(2n+2),
$$which is also trivial.
