Prove $ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $ . Prove
$$
\frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}}
$$
Can this be done by induction using the pi function.
If no, why not. 
 A: Let $$A=\frac12\cdot\frac34\cdot\frac56\cdot...\frac{2n-1}{2n}$$
$$B=\frac23\cdot\frac45\cdot\frac67\cdot...\frac{2n}{2n+1}$$
Then $$A<B$$
Then $$A^2<AB=\frac1{2n+1}$$
$$A=\frac12\cdot\frac34\cdot\frac56\cdot...\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$$
A: Hint. If you want to prove it by induction, you may observe that, for $n\ge1$,
$$
(2n+2)^2=4n^2+8n+4>4n^2+5n+3=(2n+1)(2n+3)
$$ giving
$$
\frac1{2n+2}<\frac1{\sqrt{2n+1}\cdot \sqrt{2n+3}}, \qquad n\ge1,
$$ which is equivalent to the inductive step:
$$
\frac1{\sqrt{2n+1}}\cdot \frac{2n+1}{2n+2}<\frac1{\sqrt{2n+3}}.
$$
A: An interesting way may be the following: if we define
$$ a_n = \int_{0}^{\pi/2}\sin^n(x)\,dx \tag{1}$$
we clearly have that $\{a_n\}_{n\geq 1}$ is a decreasing sequence. On the other hand, integration by parts gives:
$$ a_{2n} = \frac{\pi}{2}\binom{2n}{n}\frac{1}{4^n},\qquad a_{2n+1}=\frac{1}{2n+1}\cdot\left(\binom{2n}{n}\frac{1}{4^n}\right)^{-1}\tag{2} $$
and we want to give an upper bound to
$$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2n)!!^2} = \binom{2n}{n}\frac{1}{4^n}\tag{3} $$
so we may just exploit $a_{2n+1}>a_{2n+2}$, leading to:
$$\frac{1}{2n+1}\left(\binom{2n}{n}\frac{1}{4^n}\right)^{-1}>\frac{\pi}{2}\binom{2n+2}{n+1}\frac{1}{4^{n+1}}=\frac{\pi}{2}\binom{2n}{n}\frac{1}{4^n}\cdot\frac{2n+1}{2n+2} \tag{4}$$
or to:
$$\left(\binom{2n}{n}\frac{1}{4^n}\right)^2< \color{red}{\frac{2}{\pi}\cdot\frac{2n+2}{(2n+1)^2}}\tag{5}$$
that is much stronger than the wanted inequality.
