How to prove that $\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$ How to prove that $$\displaystyle\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$$
Thanks in advances.

Rewrite a partial product as
$$\prod_{n=0}^N \frac{(4 n+2)^3 (4 n+4)}{(4 n+1)(4 n+2)(4 n+3) (4 n+4)} = \frac{2^{3 N+3} (2 N+1)!!^3 4^{N+1} (N+1)!}{(4 N+4)!} $$
Use the fact that 
$$(2 N+1)!! = \frac{(2 N+1)!}{2^N N!} $$
and
$$M! \approx M^M e^{-M} \sqrt{2 \pi M} \quad (M \to \infty)$$
The rest is careful bookkeeping, and making sure you use the fact that
$$\lim_{M \to \infty} \left ( 1+\frac1{M} \right )^M = e$$
the sought-after result follows.
