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 $x_{n+1}=\sqrt{x_n+a(a+1)}$.

Define the sequence $x_n$ by $x_0=\dfrac{1}{2}$ and $x_{n+1}=\dfrac{1}{2}+\dfrac{\sqrt{x_n}}{2}$,
and let $y_n=x_0x_1\cdots x_{n}$. The question is to evaluate
$\lim\limits_{n\to\infty}y_n$.
It is easy to see by induction that $0<x_n<1$ for every $n$, so we can define
$$\theta_n=\arccos(\sqrt{x_n})$$
So that
$$\cos^2(\theta_{n+1})=x_{n+1}=\frac{1+\cos\theta_n}{2}=\cos^2\left(\frac{\theta_n}{2}\right).$$
Thus
$\theta_{n+1}=\dfrac{\theta_n}{2}$. This shows that $\theta_n=2^{-n}\theta_0=\dfrac{\pi}{2^{n+2}}$.
Now, noting that $\cos(\theta_{k+1})=\dfrac{\sin(2\theta_{k+1})}{2\sin\theta_{k+1}}
=\dfrac{\sin\theta_{k}}{2\sin\theta_{k+1}}$, we conclude that
$$
x_{k}=\frac{1}{4}\frac{\sin^2\theta_{k-1}}{\sin^2\theta_{k}}
$$
Thus
$$
y_n=\prod_{k=0}^{n}x_k=\frac{1}{2}\prod_{k=1}^{n}\left(\frac{1}{4}\frac{\sin^2\theta_{k-1}}{\sin^2\theta_{k}}\right)=\frac{1}{2^{2n+1}} \frac{\sin^2\theta_{0}}{\sin^2\theta_{n}}
$$
Finally,
$$
y_n=\frac{1}{2^{2n+2}\sin^2(2^{-n-2}\pi)}
$$
So, $$\lim_{n\to\infty}y_n=\frac{4}{\pi^2 },$$
which is the desired limit.$\qquad  \square$
