Proving $\sqrt{2}=\prod_{n=1}^{\infty }\left(1+\frac{0.75}{4n^2-1}\right)$

Question

Proving $$\sqrt{2}=\prod_{n=1}^{\infty }\left(1+\frac{0.75}{4n^2-1}\right)$$ By using the numerical calculation I saw that the convergence of product series is slow, so I need the proving. thanks.

Answer 1

Use the infinite product representation: $\displaystyle \dfrac{\sin \pi x}{\pi x} = \prod\limits_{n=1}^{\infty} \left(1-\frac{x^2}{n^2}\right)$ to rewrite the product $$\displaystyle \prod_{n=1}^{\infty }\left(1+\frac{3/4}{4n^2-1}\right) = \prod_{n=1}^{\infty }\frac{\left(1-\frac{1}{16n^2}\right)}{\left(1-\frac{1}{4n^2}\right)} = \frac{2\sin \frac{\pi}{4}}{\sin \frac{\pi}{2}} = \sqrt{2}$$