Is this correct ${2\over 3}\cdot{6\over 5}\cdot{10\over 11}\cdots{4n+2\over 4n+2+(-1)^n}\cdots=\sqrt{2-\sqrt{2}}?$ Given this infinite product

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

Where $n\ge0$
Experimental on the infinite product calculator show $(1)=\sqrt{2-\sqrt{2}}$
How can we show that?
 A: In the same spirit as Meet Taraviya's answer, write 
$$\prod_{n=0}^{\infty}\frac{4n+2}{4n+2+(-1)^n}=\prod_{k=0}^{\infty}\frac{(8k+2)(8k+6)}{(8k+3)(8k+5)}$$ Considering the partial product 
$$P_m=\prod_{k=0}^{m}\frac{(8k+2)(8k+6)}{(8k+3)(8k+5)}=\sqrt{\pi }\frac{  \sec \left(\frac{\pi }{8}\right)\, \Gamma \left(2
   m+\frac{5}{2}\right)}{4^{m+1}\,\Gamma \left(m+\frac{11}{8}\right)\, \Gamma
   \left(m+\frac{13}{8}\right)}$$ 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 $O\left(\frac{1}{m}\right)$  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 $$P_\infty=\frac{\sec \left(\frac{\pi }{8}\right)}{\sqrt{2}}=2 \sin \left(\frac{\pi }{8}\right)=\sqrt{2-\sqrt{2}}$$
A: $$\prod_{i=0}^{\infty}\frac{4i+2}{4i+2+(-1)^i}=\Big(\prod_{k=0}^{\infty}\frac{4(2k)+2}{4(2k)+2+(-1)^{(2k)}}\Big)\Big(\prod_{k=0}^{\infty}\frac{4(2k+1)+2}{4(2k+1)+2+(-1)^{(2k+1)}}\Big)$$
$$=\Big(\prod_{k=0}^{\infty}\frac{8k+2}{8k+3}\Big)\Big(\prod_{k=0}^{\infty}\frac{8k+6}{8k+5}\Big)$$
See if you can take off from here.
