<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] Your explanation as is is not sufficient. The part where you say that everything but $1/2$ gets cancelled needs more rigorous verification.
To demonstrate, let's "prove" that the infinite product $1\cdot1\cdot1\cdot\ldots$ equals $1/2$. Indeed,
$$
    1\cdot 1 \cdot 1 \cdot \ldots =
    \left(\frac{1}{2} \cdot 2\right) \cdot \left(\frac{1}{2} \cdot 2\right) \cdot \left(\frac{1}{2} \cdot 2\right)
    \cdot \cdots.
$$
The $2$ in the first factor cancels the $\frac{1}{2}$ in the second, the $2$ in the second factor cancels the $\frac{1}{2}$ in the third, and so on. Everything but the very first $\frac{1}{2}$ gets canceled, so the infinite product equals $\frac{1}{2}$.
This is clearly wrong. For a correct explanation you should look at partial products, like in user17762's answer.
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