$\displaystyle \prod_{n=2}^{\infty}(1-\frac{1}{n^2})=\lim_{n\to\infty}(1-\frac{1}{4})(1-\frac{1}{9})\dots (1-\frac{1}{n^2})$
Thank you!
Let's write that out:
$\displaystyle {1 \cdot 3 \over 2^2} \cdot {2 \cdot 4 \over 3^2} \cdot {3 \cdot 5 \over 4^2} \cdot {4 \cdot 6 \over 5^2} \cdot ... \cdot {(n-1) (n+1) \over n^2}$
Do you see numbers that cancel?
Remembering the 'infinite product'...
$\displaystyle \frac{\sin \pi x}{\pi x}= \prod_{n=1}^{\infty} (1- \frac{x^{2}}{n^{2}})$ (1)
... Your 'infite product' seems to be [applying l'Hopital rule] ...
$\displaystyle \prod_{n=2}^{\infty} (1- \frac{1}{n^{2}}) = \lim_{x \rightarrow 1} \frac{\sin \pi x}{\pi x (1-x^{2})} = \frac{1}{2}(2)$
Marry Christmas from Serbia
$\displaystyle \chi$ $\displaystyle \sigma$