# How to prove that $\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$

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

#### Solutions Collecting From Web of "How to prove that $\prod_{n=0}^\infty \frac{(4n+2)^2}{(4n+1)(4n+3)}=\sqrt{2}$"

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.

From the Weierstrass product for the cosine function we have:
$$\cos z = \prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2\pi^2}\right)$$
and by taking $z=\frac{\pi}{4}$ it follows that:
$$\prod_{n=0}^{+\infty}\left(1-\frac{1}{(4n+2)^2}\right)=\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$$
whose LHS is just the reciprocal of our product.

Hint: Rewrite the infinite product expression in terms of the $\Gamma$ function, and then apply Euler’s famous reflection formula.