<s>[INST] Is $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ Question is to check if  $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$
we have $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\infty}\frac{n+1}{n}\frac{n-1}{n}=(\frac{3}{2}.\frac{1}{2})(\frac{4}{3}.\frac{2}{3})(\frac{5}{4}.\frac{3}{4})...$
In above product we have for each term $\frac{a}{b}$ a term $\frac{b}{a}$ except for $\frac{1}{2}$.. So, all  other terms gets cancelled and we left with $\frac{1}{2}$.
So, $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\frac{1}{2}$.
I would be thankful if some one can assure that this explanation is correct/wrong?? 
I am solving this kind of problems for the first time so, it would be helpful if some one can tell if there are any other ways to do this..
Thank you
 [/INST] In fact, Euler discovered that
$$\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty (1-z^2/n^2)$$
which we can rearrange to 
$$­\frac{\sin \pi z}{\pi z (1-z^2)} =  \prod_{n=2}^\infty (1-z^2/n^2).$$
By comparison of both sides at $z=1$, your product is $1/2$.
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