$\prod_{n=2}^{\infty} \left(1- \dfrac1 {n^2}\right)= ?$
Attempt:
Simplified it to $\dfrac{(n-1)(n+1)}{n^2}$ , and then wrote some terms to observe the cancelllation pattern, but that didn't help, how to solve it then?
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Are you allowed to use Euler's infinite product formula for sine? $$\sin x = x \prod_{n=1}^{\infty} \left(1-\frac{x^2}{n^2 \pi^2}\right)$$
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$$\prod_{n=2}^{m} \left(1- \dfrac1 {n^2}\right)= \frac{1}{2}\frac{\Gamma(m)\Gamma(m+2)}{\Gamma(m+1)^2}$$ – gammatester May 31 '18 at 10:45