How this :$ \prod_{n=2}^{+\infty}(1-\frac {1}{n^2})=\frac12$?

Question

I have found this problem

$$ \prod_{n=2}^{+\infty}\left(1-\frac {1}{n^2}\right)=\frac12,$$

in a book entitled INFINITE SERIES AND PRODUCTS but I'm not able to get it's partial sum. However wolfram alpha assumed it equal $\frac 12$. My key idea is to use Euler product but I don't know how. Is there a proof for that?

Answer 1

Note that

$$1-\frac1{n^2}=\frac{(n-1)(n+1)}{n^2}$$

so the partial product telescopes to

$$\prod_{n=2}^N\left(1-\frac1{n^2}\right)=\frac{N+1}{2N}.$$

Answer 2

Hint: Write $1-\frac{1}{n^2}=\frac{n^2-1}{n^2}=\frac{(n-1)(n+1)}{n^2}$

Answer 3

Your product is $$\left(\frac12\frac32\right)\left(\frac23\frac43\right)\left(\frac34\frac54\right)\left(\frac 45\frac65\right)\cdots =\frac12\left(\frac32\frac23\right)\left(\frac43\frac34\right)\left(\frac54\frac45\right)\cdots.$$

Answer 4

Just to be obstuse:

$\prod\limits_{n=2}^k(1-\frac 1{n^2}) = $

$\prod\limits_{n=2}^k\frac {n^2 - 1}{n^2}=\prod\limits_{n=2}^k\frac {(n-1)(n+1)}{n^2}=$

$\frac {\prod\limits_{n=2}^k (n-1) \prod\limits_{n=2}^k(n+1)}{(\prod\limits_{n=2}^k n)^2}=$

$\frac {\prod\limits_{n=1}^{k-1} n \prod\limits_{n=3}^{k+1}n}{(\prod\limits_{n=2}^k n)^2}=$

$\frac {(1*2\prod\limits_{n=3}^{k-1}n)([\prod\limits_{n=3}^{k-1}n]k(k+1))}{(1*2(\prod\limits_{n=3}^{k-1}n)k)^2}=$

$\frac {(1*2)(k(k+1))}{(1*2k)^2}=$

$\frac {k+1}{2k} = \frac 12 + \frac 1{2k}$

which in itself is kind of interesting.

....

But anyhow so

$\prod\limits_{n=2}^\infty(1-\frac 1{n^2}) = \lim\limits_{k\to\infty}\prod\limits_{n=2}^k(1-\frac 1{n^2})=\lim\limits_{k\to\infty} (\frac 12 + \frac 1{2k}) = \frac 12$