TITLE: Evaluate this infinite product: $\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)$
QUESTION [8 upvotes]: $$\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)\;=\;\text{???}$$

I took the LCM and split the numerator as $(n+2)(n-2)$ and then took the product of the numerator and the denominator separately but I was not able to get the answer from that so can you please help me in what to do next.

REPLY [4 votes]: A different solution. You can use the fact that
$$\sin(\pi x)=\pi x \prod_1^\infty \left(1-\frac{x^2}{n^2} \right).$$
So your product is e qual to
$$\lim_{x \to 2} \frac{\sin(\pi x)}{\pi x (1-x^2)(1-x^2/4)}=1/6$$