<s>[INST] Value of the product: $ \sqrt{2} \sqrt{2 - \sqrt{2}} \sqrt{2 - \sqrt{2 - \sqrt{2}}} \sqrt{2 - \sqrt{2 - \sqrt{2-\sqrt{2}}}} \cdots $ =? Let the recursive sequence
$$ a_0 = 0, \qquad a_{n+1} = \sqrt{2-a_n},\,\,n\in\mathbb N.
$$
T
Can we find the value of the product
$$
\prod_{n=1}^{\infty}{a_n}?
$$
Well, from here I don't seem to follow. I can understand that there would be some good simplification and the product will hopefully telescope but I'm lacking the right algebra. I also thought of finding a recurrence solution probably from the corresponding DE but that didn't follow as well.
 [/INST] Another approach is the following one: if we assume $ a_n = 2\cos(\theta_n) $ it follows that
$$ \cos(\theta_{n+1})=\sqrt{\frac{1-\cos\theta_n}{2}} = \sin\left(\frac{\theta_n}{2}\right) = \cos\left(\frac{\pi-\theta_n}{2}\right)\tag{1} $$
from which we have $\theta_{n+1}=\frac{\pi-\theta_n}{2}$ and, by induction:
$$ \theta_{n+k} = \frac{\pi}{3}-(-1)^k\frac{\pi}{3\cdot 2^k}+(-1)^k\frac{\theta_n}{2^k}.\tag{2}$$
Since $\theta_0=\frac{\pi}{2}$,
$$ \theta_k = \frac{\pi}{3}+(-1)^k \frac{\pi}{6\cdot 2^k},\qquad \color{red}{a_k = \cos\left(\frac{\pi}{6\cdot 2^k}\right)-(-1)^k\sqrt{3}\sin\left(\frac{\pi}{6\cdot 2^k}\right)}\tag{3} $$
but since $2\cos\theta_n = \frac{\sin(2\theta_n)}{\sin(\theta_n)}$ and $\sin(\pi-\theta)=\sin(\theta)$, we also have a telescopic product.
In particular:
$$ a_1\cdot a_2\cdot\ldots\cdot a_n = \frac{\sin(2\theta_1)}{\sin(\theta_n)} \tag{4}$$
hence:

$$ \prod_{n\geq 1} a_n = \frac{\sin(2\theta_1)}{\sin(\lim_{n\to +\infty}\theta_n)} = \frac{1}{\sin\frac{\pi}{3}}=\color{red}{\frac{2}{\sqrt{3}}}.\tag{5}$$

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