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
 A: 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.
A: 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 $1/2$.
A: Obviously, the product of any positive numbers that less than 1 is still less than one.
A: One of the easiest ways to deal with infinite sums/products, is to stop at a finite value, say $N$, and look at what happens to the finite sum/product and then let $N \to \infty$. (In fact, this is the typical way infinite sums/products are to be understood/interpreted.)
Hence, in your case, let us look at
\begin{align}
S_N & = \prod_{n=2}^N \left(1-\dfrac1{n^2}\right) = \dfrac{1}{2}\cdot\dfrac{3}{2}\cdot \dfrac23 \cdot \dfrac43 \cdot \dfrac34 \cdot \dfrac54 \cdots \dfrac{N-2}{N-1} \cdot \dfrac{N}{N-1} \cdot \dfrac{N-1}{N} \cdot \dfrac{N+1}N\\
& = \dfrac12 \cdot \dfrac{N+1}N = \dfrac{N+1}{2N}
\end{align}
Now let $N \to \infty$ to conclude that
$$\prod_{n=2}^{\infty} \left(1-\dfrac1{n^2}\right) = \dfrac12$$
