Show that $\frac1{\sqrt{(n+\frac12) \pi}} \le\frac{1\cdot 3\cdot 5 ... (2n-1)}{2\cdot 4\cdot 6 ... (2n)} \le \frac1{\sqrt{n \pi}} $ Show that,
if $n$ is a positive integer,
$$\frac1{\sqrt{(n+\frac12) \pi}}
\le\frac{1\cdot 3\cdot 5 ... (2n-1)}{2\cdot 4\cdot 6 ... (2n)}
\le \frac1{\sqrt{n \pi}} .
$$
This result is in a current issue
of a MAA magazine,
and I thought it would be interesting to see
how many proofs could be found.
I know it may be a duplicate,
but none were suggested when I entered it.
 A: Since $0\leqslant \sin t\leqslant 1 $ over $[0,\pi/2]$ we have that  $$\int_0^{\pi/2}\sin^{2n+1}t dt\leqslant \int_0^{\pi/2}\sin^{2n}tdt\leqslant \int_0^{\pi/2}\sin^{2n-1}t dt$$
Now, it is not hard to show that if $I_k= \displaystyle\int_0^{\pi/2}\sin^{k}t$ then $I_k=\dfrac{k-1}{k}I_{k-2}$, by integrating by parts. Using $I_1=1$ and $I_0=\dfrac{\pi}2$, we get that
$$\int_0^{\pi/2}\sin^{m}tdt=\begin{cases}\dfrac{4^n}{2n+1}\dbinom{2n}n^{-1},&m=2n+1\text{ odd}\\\dfrac 1 {4^n}\dbinom{2n}n\dfrac \pi 2,&m=2n\text{ even}\end{cases}$$
Substituting in the above, one gets 
$$2n\left(\frac 1 {4^n}\binom {2n}n\right)^2\leqslant \frac{2}\pi\leqslant  (2n+1)\left(\frac 1 {4^n}\binom {2n}n\right)^2$$
which becomes$$
  \frac{1}{{2n}}{\left( {\frac{1}{{{4^n}}}\binom{{2n}}{n}} \right)^{ - 2}} \geqslant \frac{\pi }{2} \geqslant \frac{1}{{2n + 1}}{\left( {\frac{1}{{{4^n}}}\binom{{2n}}{n}} \right)^{ - 2}} $$ and in turn gives $$
  \frac{1}{{\pi n}} \geqslant {\left( {\frac{1}{{{4^n}}}\binom{{2n}}{n}} \right)^2} \geqslant \frac{1}{\pi }\frac{1}{{n + 1/2}}  $$
As desired.
