How to solve this : $ \prod^{\infty}_{n=2}(1-\frac{1}{n^2}) $ How to  solve this : 
$$ \prod^{\infty}_{n=2}(1-\frac{1}{n^2}) $$
We  can write it as follows : 
$$ \prod^{\infty}_{n=2}\frac{(n-1)(n+1)}{n^2} $$
How can we proceed further please guide.. thanks..
 A: HINT:
$$\text{If }T_r=\frac{(r-1)(r+1)}{r^2},$$
$$T_{r+1}=\frac{r(r+2)}{(r+1)^2}\text{ and } T_{r-1}=\frac{(r-2)(r)}{(r-1)^2}$$
Clearly, the denominator of any $n$th term will be cancelled by the numerators of the previous & the next term for $n\ge3$
Putting $n=2,3,\cdots,n-2,n-1,n$
$$\prod_{2\le r\le n}\frac{(1\cdot3)(2\cdot4)(3\cdot5)\cdots(n-3)(n-1)(n-2)(n)(n-1)(n+1)}{(2\cdot2)(3\cdot3)(4\cdot4)\cdots(n-2)(n-2)(n-1)(n-1)\cdot n\cdot n}=\frac{n+1}{2n}$$
A: First, consider the partial product
$$
\begin{align}
\prod_{k=2}^n\left(1-\frac1{k^2}\right)
&=\prod_{k=2}^n\frac{k-1}{k}\prod_{k=2}^n\frac{k+1}{k}\\
&=\frac{\prod\limits_{k=1}^{n-1}k}{\prod\limits_{k=2}^nk}
\frac{\prod\limits_{k=3}^{n+1}k}{\prod\limits_{k=2}^nk}\\
&=\frac1n\frac{n+1}{2}
\end{align}
$$
then take limits,
$$
\prod_{k=2}^\infty\left(1-\frac1{k^2}\right)=\frac12
$$
A: Just a different technique.
Use the formula $$\sin x=x\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)$$ to get $$\prod_{n\ge 2}\left(1-\frac{1}{n^2}\right)=\lim_{x\rightarrow \pi}\frac{\sin x}{x\left(1-\frac{x^2}{\pi^2}\right)}$$ Then use L'Hospital to get the answer $\displaystyle \frac{1}{2}$.
