Prove the convergence of the infinite product involving nested radicals

Question

How to show that $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot ... =\prod\limits_{n\in \mathbb{N}} \frac{1}{2} ~ \overbrace{\sqrt{2+\sqrt{2+...+\sqrt{2}}}}^{n}$ converges ?

My approach : Let $A_n =\overbrace{\sqrt{2+\sqrt{2+...+\sqrt{2}}}}^{n}$ ; now $~~ \prod\limits_{n\in\mathbb{N}} \frac{A_n }{2} $ converges if and only if $\sum\limits_{n\in \mathbb{N}}\log \frac{A_n }{2} $ converges.

By the ratio-test of the sum above we have : $\lim\limits_{n\to\infty} \bigg|\dfrac{\log \frac{A_{n+1} }{2}}{\log \frac{A_n }{2}}\bigg|=1$ which doesn't prove anything ,so how to continue from here ?

Answer 1

It's $$\lim_{n\rightarrow+\infty}\prod_{k=1}^n\cos\frac{\pi}{2^{k+1}}$$ and use a telescopic product with $\sin2\alpha=2\sin\alpha\cos\alpha$.

I got $\frac{2}{\pi}$ as answer.