<s>[INST] Evaluate $\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\cdots$ Evaluate 
$$
\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\frac{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}{4}\cdots .
$$
First, it is clear that terms tend to $1$. 
It seems that the infinity product is not 0. This is related to the post Sequence $x_{n+1}=\sqrt{x_n+a(a+1)}$.
 [/INST] If we set $a_0=1/2$ and define
$$
a_k=\frac{1+\sqrt{a_{k-1}}}{2}\tag{1}
$$
then the product sought is
$$
\prod_{k=0}^\infty\ a_k\tag{2}
$$
Since
$$
\cos(2x)=2\cos^2(x)-1\tag{3}
$$
we have that
$$
a_k=\cos^2(2^{-k}x_0)\tag{4}
$$
satisfies $(1)$ with $x_0=\frac\pi4$. Then
$$
\sin(2x)=2\sin(x)\cos(x)\tag{5}
$$
implies by telescoping product that
$$
\begin{align}
\prod_{k=0}^\infty\ a_k
&=\prod_{k=0}^\infty\frac{\sin^2(2^{-k+1}x_0)}{4\sin^2(2^{-k}x_0)}\\
&=\lim_{n\to\infty}\left(\prod_{k=0}^n\sin^2(2^{-k+1}x_0)\middle/\prod_{k=1}^{n+1}4\sin^2(2^{-k+1}x_0)\right)\\
&=\lim_{n\to\infty}\left(\frac{\sin^2(2x_0)}{4^{n+1}\sin^2(2^{-n}x_0)}\right)\\
&=\frac{\sin^2(2x_0)}{4x_0^2}\\
&=\frac4{\pi^2}\tag{6}
\end{align}
$$
</s>