Problem
Evaluate the infinite product
$$\frac{2}{\sqrt{2}}\cdot \frac{2}{\sqrt{2+\sqrt{2}}}\cdot \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$$
Attempt
For convenience，let's rewrite the limit. Denote$$x_1=\sqrt{2},~~~x_{n+1}=\sqrt{2+x_n}(n=1,2,\cdots),$$ then $$\lim_{n \to \infty}\left(\frac{2}{x_1}\cdot \frac{2}{x_2}\cdots \frac{2}{x_n}\right)$$ is what we want.
It's easy to obtain that，$\{x_n\}$ is increasing with a greater $n$，and convergent to $2$. Hence $$x_1<x_2<\cdots x_n<2.$$Therefore$$\frac{2}{x_1}>\frac{2}{x_2}>\cdots>\frac{2}{x_n}>1.$$ Can we go on from here?
Use $\cos2A=2\cos^2A-1,$
$$\dfrac2{\sqrt2}=\dfrac2{2\cos\dfrac\pi4}\text{ and }2+\sqrt2=2\left(1+\cos\dfrac\pi4\right)=4\cos^2\dfrac\pi8$$
$$\implies\dfrac2{\sqrt{2+\sqrt2}}=\dfrac2{2\cos\dfrac\pi8}$$
$\dfrac1{\cos\dfrac\pi4}\dfrac1{\cos\dfrac\pi8}=\dfrac{2\sin\dfrac\pi8}{\cos\dfrac\pi4\sin\dfrac\pi4}=\dfrac{4\sin\dfrac\pi8}{\sin\dfrac\pi2}=?$
$\dfrac1{\cos\dfrac\pi4}\dfrac1{\cos\dfrac\pi8}\dfrac1{\cos\dfrac\pi{16}}=2^3\sin\dfrac\pi{2^4}$
$$\implies\prod_{r=2}^n\cos\dfrac\pi{2^r}=2^{n-1}\sin\dfrac\pi{2^n}$$
$$\implies\lim_{n\to\infty}\prod_{r=2}^n\cos\dfrac\pi{2^r}=\dfrac\pi2\cdot\lim_{n\to\infty}\dfrac{\sin\dfrac\pi{2^n}}{\dfrac\pi{2^n}}=?$$