¿Es esto correcto?

Question

Dado este producto infinito

$$\lim_{n \to \infty}{2\over 3}\cdot{6\over 5}\cdot{10\over 11}\cdots{4n+2\over 4n+2+(-1)^n}\cdots=\sqrt{2-\sqrt{2}}\tag1$$

Donde $n\ge0$

Experimental en el producto infinito calculador Mostrar $(1)=\sqrt{2-\sqrt{2}}$

¿Cómo podemos mostrar?

Answer 1

En el mismo espíritu que Cumplir Taraviya la respuesta de escribir $$\prod_{n=0}^{\infty}\frac{4n+2}{4n+2+(-1)^n}=\prod_{k=0}^{\infty}\frac{(8k+2)(8k+6)}{(8k+3)(8k+5)}$$ Considerando el producto parcial $$P_m=\prod_{k=0}^{m}\frac{(8k+2)(8k+6)}{(8k+3)(8k+5)}=\sqrt{\pi }\frac{ \sec \left(\frac{\pi }{8}\right)\, \Gamma \left(2 m+\frac{5}{2}\right)}{4^{m+1}\,\Gamma \left(m+\frac{11}{8}\right)\, \Gamma \left(m+\frac{13}{8}\right)}$$ Now, using Stirling approximation for large values of $m$ $$\log \left(\frac{\Gamma \left(2 m+\frac{5}{2}\right)}{\Gamma \left(m+\frac{11}{8}\right) \Gamma \left(m+\frac{13}{8}\right)}\right)=2 m \log (2)+\log \left(2 \sqrt{\frac{2}{\pi }}\right)+\frac{3}{64 m}+O\left(\frac{1}{m^2}\right)$$ Truncating to $S\left(\frac{1}{m}\right)$ we then have $$\frac{\Gamma \left(2 m+\frac{5}{2}\right)}{\Gamma \left(m+\frac{11}{8}\right) \Gamma \left(m+\frac{13}{8}\right)}\approx 4^m\times 2 \sqrt{\frac{2}{\pi }}$$ making $$P_\infty=\frac{\sec \left(\frac{\pi }{8}\right)}{\sqrt{2}}=2 \sin \left(\frac{\pi }{8}\right)=\sqrt{2-\sqrt{2}}$$

Answer 2

\begin{align} \prod_{n=0}^{\infty}\frac{4n+2}{4n+2+(-1)^n}&=\prod_{n=0}^{\infty}\frac{8n+2}{8n+3}\frac{8n+6}{8n+5}\\ &=\prod_{n=0}^{\infty}\frac{8n+2}{8n+3}\frac{8n+4}{8n+3}\frac{8n+3}{8n+4}\frac{8n+6}{8n+5}\frac{8n+4}{8n+5}\frac{8n+5}{8n+4}\\ &=\prod_{n=0}^{\infty}\left(\frac{8n+2}{8n+3}\frac{8n+4}{8n+3}\right)\left(\frac{8n+6}{8n+5}\frac{8n+4}{8n+5}\right)\left(\frac{8n+5}{8n+4}\frac{8n+3}{8n+4}\right)\\ &=\prod_{n=0}^{\infty}\left(\frac{8n+2}{8n+3}\frac{8n+4}{8n+3}\right)\prod_{n=0}^{\infty}\left(\frac{8n+6}{8n+5}\frac{8n+4}{8n+5}\right)\prod_{n=0}^{\infty}\left(\frac{8n+5}{8n+4}\frac{8n+3}{8n+4}\right)\\ &=\prod_{n=0}^{\infty}\left(1-\frac{1}{(8n+3)^2}\right)\prod_{n=0}^{\infty}\left(1-\frac{1}{(8n+4)^2}\right)\prod_{n=0}^{\infty}\left(1-\frac{1}{(8n+5)^2}\right)\\ \end {Alinee el}

Ahora tenga en cuenta que utilizando las propiedades de la función Gamma - esto me mantuvo ocupado pero bastante divertida - tenemos\begin{align} \prod_{n=0}^{\infty}\left(1-\frac{1}{(8n+q)^2}\right) &=\frac{(q^2-1)\Gamma\left(\frac{q+8}{8}\right)^2}{q^2\Gamma\left(\frac{q+7}{8}\right)\Gamma\left(\frac{q+9}{8}\right)}\\ \end {Alinee el}

Por lo tanto se convierte en su límite:\begin{align} l&=\frac{8\Gamma\left(\frac{11}{8}\right)^2}{9\Gamma\left(\frac{10}{8}\right)\Gamma\left(\frac{12}{8}\right)}\frac{15\Gamma\left(\frac{12}{8}\right)^2}{16\Gamma\left(\frac{11}{8}\right)\Gamma\left(\frac{13}{8}\right)}\frac{24\Gamma\left(\frac{13}{8}\right)^2}{25\Gamma\left(\frac{12}{8}\right)\Gamma\left(\frac{14}{8}\right)}\\ &=\frac45\frac{\Gamma\left(\frac{11}{8}\right)}{\Gamma\left(\frac{10}{8}\right)}\frac{\Gamma\left(\frac{13}{8}\right)}{\Gamma\left(\frac{14}{8}\right)}\\ &=\frac45\frac{\frac{3}{8}\Gamma\left(\frac{3}{8}\right)}{\frac{2}{8}\Gamma\left(\frac{2}{8}\right)}\frac{\frac{5}{8}\Gamma\left(\frac{5}{8}\right)}{\frac{6}{8}\Gamma\left(\frac{6}{8}\right)}\\ &=\frac{\Gamma\left(\frac{3}{8}\right)}{\Gamma\left(\frac{2}{8}\right)}\frac{\Gamma\left(\frac{5}{8}\right)}{\Gamma\left(\frac{6}{8}\right)}\\ &=\frac{\frac{\pi}{\sin\frac38\pi}}{\frac{\pi}{\sin\frac28\pi}}\\ &=\frac{\sqrt{2}/2}{\sqrt{2+\sqrt2}/2}=\sqrt{2-\sqrt2} \end {Alinee el}

Answer 3

$$\prod_{i=0}^{\infty}\frac{4i+2}{4i+2+(-1)^i}=\Big(\prod_{k=0}^{\infty}\frac{4(2k)+2}{4(2k)+2+(-1)^{(2k)}}\Big)\Big(\prod_{k=0}^{\infty}\frac{4(2k+1)+2}{4(2k+1)+2+(-1)^{(2k+1)}}\Big)$$ $$=\Big(\prod_{k=0}^{\infty}\frac{8k+2}{8k+3}\Big)\Big(\prod_{k=0}^{\infty}\frac{8k+6}{8k+5}\Big)$$

Vea si usted puede sacar de aquí.