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.$\qquad \square$
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 $$ \begin{align} \prod_{k=0}^\infty\ a_k &=\prod_{k=0}^\infty\frac{\sin^2(2^{-k+1}x_0)}{4\sin^2(2^{-k}x_0)}\\ &=\lim_{n\to\infty}\left(\prod_{k=0}^n\sin^2(2^{-k+1}x_0)\middle/\prod_{k=1}^{n+1}4\sin^2(2^{-k+1}x_0)\right)\\ &=\lim_{n\to\infty}\left(\frac{\sin^2(2x_0)}{4^{n+1}\sin^2(2^{-n}x_0)}\right)\\ &=\frac{\sin^2(2x_0)}{4x_0^2}\\ &=\frac4{\pi^2}\tag{6} \end{align} $$
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: $$\boxed{\displaystyle\lim =\frac{4}{\pi^2}}$$
