Evaluating the infinite product $\prod\limits_{k=2}^\infty \left ( 1-\frac1{k^2}\right)$ 
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.
 A: Let $g(k)=\frac{k-1}{k}$.  Then this product is:
$$\prod_{k=2}^n \frac{g(k)}{g(k+1)}$$
which is a telescoping product, and thus equal to $\frac{g(2)}{g(n+1)}$
A: $$ 1-\frac{1}{k^2}  = \frac{(k-1)(k+1)}{k^2}$$
$$ \prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right )  = \underbrace{\frac{2-1}{2}}_{\text{from first}} \times \underbrace{\frac{(n+1)}{n}}_{\text{from last}}$$
A: $$\left(1-\frac1{(k-1)^2}\right)\left(1-\frac1{(k)^2}\right)\left(1-\frac1{(k+1)^2}\right)$$
$$=\frac{k(k-2)(k-1)(k+1)k(k+2)}{(k-1)^2k^2(k+1)^2}=\frac{(k-2)(k+2)}{(k-1)(k+1)}$$
Observe that $\left(1-\frac1{(k)^2}\right)$ is cancelled out by the previous & the next term except for the extreme terms, the 1st & the last term leaving behind the 1st part of the 1st term $=\frac{2-1}{2}$ and the 2nd part of the last term $=\frac{n+1}n$
A: Hint: The "typical" term is $\dfrac{k-1}{k}\dfrac{k+1}{k}$. Express the first few terms in this way, and observe the nice cancellations.
For example, here is the product of the first seven terms:
$$\frac{1}{2}\frac{3}{2}\frac{2}{3}\frac{4}{3}\frac{3}{4}\frac{5}{4}\frac{4}{5}\frac{6}{5}\frac{5}{6}\frac{7}{6}\frac{6}{7}\frac{8}{7}\frac{7}{8}\frac{9}{8}=\frac{9}{2\cdot 8}.$$
A: Note that $$1-\frac1{k^2}=\left(1-\frac1k\right)\left(1+\frac1k\right)=\frac{k-1}{k}\frac{k+1}{k}=\frac{a_k}{a_{k-1}}$$ with $a_k= \frac{k+1}k$, hence this is a telescoping product, i.e.
$$ \prod_{k=2}^n\left(1-\frac1{k^2}\right)=\frac{a_2}{a_1}\frac{a_3}{a_2}\cdots\frac{a_n}{a_{n-1}}=\frac{a_n}{a_1}=\frac{n+1}{2n}.$$
A: 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 $$\prod_{k\geq2}\left(1-\frac{1}{k^{2}}\right)=\lim_{x\rightarrow1}\frac{\sin\left(\pi x\right)}{\pi x\left(1-x^{2}\right)}=\color{red}{\frac{1}{2}}.$$
