TITLE: Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$
QUESTION [54 upvotes]: While trying some problems along with my friends we had difficulty in this question.

True or False: The value of the infinite product $$\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)$$ is $1$.

I couldn't do it and my friend justified it by saying that since the terms in the product have values less than $1$, so the value of the product can never be $1$. I don't know whether this justification is correct or not. But i referred to Tom Apostol's Mathematical Analysis book and found a theorem which states, that 

The infinite product $\prod(1-a_{n})$ converges if the series $\sum a_{n}$ converges.

This assures that the above product converges. Could anyone help me in finding out where it converges to? And,

Does there exist a function $f$ in $\mathbb{N}$ ( like $n^{2}$, $n^{3}$) such that $\displaystyle \prod\limits_{n=1}^{\infty} \Bigl(1-\frac{1}{f(n)}\Bigr)$ has the value $1$?

REPLY [2 votes]: You could use the infinite product for the sine function which says that:
$$\sin(x)=x\prod_{n=1}^\infty (1-\frac{x^2}{\pi^2n^2})$$
Or:
$$\frac{\sin(x)}{x(1-\frac{x^2}{\pi^2})}=\prod_{n=2}^\infty (1-\frac{x^2}{\pi^2n^2})$$
Then, for your problem, you can just add $\lim_{x\rightarrow\pi}$ to the equation and solve.