TITLE: Evaluating the infinite product $\prod\limits_{k=2}^\infty \left ( 1-\frac1{k^2}\right)$
QUESTION [23 upvotes]: Evaluate the infinite product
$$\lim_{ n\rightarrow\infty }\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ).$$

I can't see anything in this limit , so help me please.

REPLY [3 votes]: From the Weiestrass factorization of sine function $$\frac{\sin\left(\pi x\right)}{\pi x}=\prod_{k\geq1}\left(1-\frac{x^{2}}{k^{2}}\right)
 $$ we get $$\frac{\sin\left(\pi x\right)}{\pi x\left(1-x^{2}\right)}=\prod_{k\geq2}\left(1-\frac{x^{2}}{k^{2}}\right)
 $$ hence $$\prod_{k\geq2}\left(1-\frac{1}{k^{2}}\right)=\lim_{x\rightarrow1}\frac{\sin\left(\pi x\right)}{\pi x\left(1-x^{2}\right)}=\color{red}{\frac{1}{2}}.$$