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 $x_{n+1}=\sqrt{x_n+a(a+1)}$.
 A: 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
$$
\begin{align}
\prod_{k=0}^\infty\ a_k
&=\prod_{k=0}^\infty\frac{\sin^2(2^{-k+1}x_0)}{4\sin^2(2^{-k}x_0)}\\
&=\lim_{n\to\infty}\left(\prod_{k=0}^n\sin^2(2^{-k+1}x_0)\middle/\prod_{k=1}^{n+1}4\sin^2(2^{-k+1}x_0)\right)\\
&=\lim_{n\to\infty}\left(\frac{\sin^2(2x_0)}{4^{n+1}\sin^2(2^{-n}x_0)}\right)\\
&=\frac{\sin^2(2x_0)}{4x_0^2}\\
&=\frac4{\pi^2}\tag{6}
\end{align}
$$
A: Consider the corresponding finite product containing $n+1$ factors. Multiplying the last  factor (with $n$ square roots) by
$$2-\underbrace{\sqrt{2+\sqrt{2+\ldots}}}_{x_n},$$
the product telescopes to $4^{-n}$. 
On the other hand, since
$x_n^2-2=x_{n-1}$,
writing $x_n=2\cos\varphi_n$ we get $2\varphi_n=\varphi_{n-1}$, and therefore $\varphi_n=\frac{\varphi_1}{2^{n-1}}=\frac{\pi}{2^{n+1}}$. This allows to make an estimate
$$2-x_n=2-2\cos\varphi_n\approx \frac{\pi^2}{4}4^{-n},$$
which finally gives the answer:
$$\boxed{\displaystyle\lim =\frac{4}{\pi^2}}$$
A: 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.$\qquad  \square$
