Recordamos la formula de la factorización del seno de x entre x
http://rinconmatematico.com/foros/index.php?topic=112219.0\( \frac{\text{sen}(x)}{x}=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\left(1-\frac{x^2}{3^2\pi^2}\right)\left(1-\frac{x^2}{4^2\pi^2}\right)... \)
\( \frac{\text{sen}(x)}{x}=\displaystyle\prod_{k=1}^{\infty}{\left(1-\frac{x^2}{k^2\pi^2}\right)} \)
Evaluamos en \( x=\frac{\pi}{2} \)
\( \frac{\text{sen}\left(\frac{\pi}{2}\right)}{\frac{\pi}{2}}=\displaystyle\prod_{k=1}^{\infty}{\left(1-\frac{\left(\frac{\pi}{2}\right)^2}{k^2\pi^2}\right)} \)
\( \frac{1}{\frac{\pi}{2}}=\displaystyle\prod_{k=1}^{\infty}{\left(1-\frac{\frac{\pi^2}{2^2}}{k^2\pi^2}\right)} \)
\( \frac{2}{\pi}=\displaystyle\prod_{k=1}^{\infty}{\left(1-\frac{\pi^2}{2^2k^2\pi^2}\right)} = \displaystyle\prod_{k=1}^{\infty}{\left(1-\frac{\pi^2}{2^2k^2\pi^2}\right)} = \displaystyle\prod_{k=1}^{\infty}{\left(1-\frac{1}{2^2k^2 }\right)} \)
\( \frac{2}{\pi}=\displaystyle\prod_{k=1}^{\infty}{\left( 1-\frac{1}{(2k)^2 }\right)}=\displaystyle\prod_{k=1}^{\infty}{ \frac{(2k)^2-1}{(2k)^2 }}= \displaystyle\prod_{k=1}^{\infty}{\frac{(2k-1)(2k+1)}{(2k)(2k) }} \)
\( \frac{2}{\pi}= \left(\frac{1\times 3}{2\times 2}\right)\left(\frac{3\times 5}{4\times 4}\right)\left(\frac{5\times 7}{6\times 6}\right)... \)
