<s>[INST] 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. 
 [/INST] 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.
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