What's the limit of $\sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2 + ...}}}}}} $? Let's look at the continued radical 
$ R = \sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2 + ...}}}}}} $
whose signs are defined as $ (+, -, +, -, -, + ,-, -, -,...)$, similar to the sequence $101001000100001...$, where
$1 = +$
$0 = - $
This radical seems to converge to a constant approximately equal to $1.567883...$. 
The question is: Is it possible to find this limit $R$ in closed form?
Remark: In the article "On the periodic continued radicals of 2 and generalization for Vieta’s product", it is proved that a periodic sequence of signs composed of nested square roots of two converges to $2\sin(q\pi)$ for some rational number $q$. I have tried with non periodic sequences of plus and minus, and they also converge to numbers between $0$ and $2$. if this radical has a closed form, It can be the sine of an irrational multiple of $\pi$, since both are transcendental numbers.
 A: For any function $f$ and sequence of functions $f_1, f_2, \cdots, f_n$, let 
$$\mathop{\bigcirc}_{k=1}^n f_k \stackrel{def}{=} f_1 \circ f_2 \circ \cdots \circ f_n
\quad\text{ and }\quad
f^{\circ n} \stackrel{def}{=} \mathop{\bigcirc}_{k=1}^n f = \underbrace{f\circ f \circ \cdots \circ f}_{n \text{ times}}
$$
be a short hand of composing the functions in given order.
Consider following maps:
$$\begin{align}
\psi :&\quad [0,1] \ni \theta \quad\mapsto\quad 2\cos\left(\frac{\pi}{2}\theta\right) \in [0,2]\\
\phi_{\pm} :&\quad [0,2] \ni x\quad \mapsto\quad \sqrt{ 2 \pm x  } \in [0,2]
\end{align}
$$
The infinite radical at hand can be interpreted as picking an arbitrary $x \in [0,2]$ and study following limit:
$$\lim_{n\to\infty} \left( \mathop{\bigcirc}_{k=1}^n \phi_{+}\circ \phi_{-}^{\circ k}\right)(x)$$
As functions, it is not hard to verify following equalities
$$\begin{align}
\psi^{-1}\circ\phi_{+}\circ\psi \;=&\quad [0,1] \ni \theta &\mapsto&\quad \frac{\theta}{2} \in [0,1]\tag{*1a}\\
\psi^{-1}\circ\phi_{-}\circ\psi \;=&\quad [0,1] \ni \theta &\mapsto&\quad 1 - \frac{\theta}{2} \in [0,1]\tag{*1b}
\end{align}
$$
Let $\alpha = -\frac12.\,$ Apply $(*1b)$ $k$ times followed by $(*1a)$, we get
$$\begin{align}
\left(\psi^{-1}\circ\phi_{+}\circ\phi_{-}^{\circ k}\circ\psi\right)(\theta)
&=
\frac12\left(1 + \alpha + \alpha^2 + \cdots + \alpha^{k-1} + \alpha^k\theta\right)
= \frac12\left(\frac{1 - \alpha^k}{1 - \alpha} + \alpha^k \theta\right)\\
&= \frac12\left(\frac23(1-\alpha^k) + \alpha^k\theta\right)
= \frac13 + \alpha^{k+1}\left(\frac23 - \theta\right)
\end{align}
$$
From this, we find
$$\left(\psi^{-1}\circ \left(\mathop{\bigcirc}_{k=1}^n \phi_{+}\circ\phi_{-}^{\circ k}\right)\circ\psi\right)(\theta)
= \frac13 + 
\alpha^2\left[
\frac13 - \alpha^3\left[
\cdots
\left[
\frac13 - \alpha^{n+1}\left[\frac23-\theta\right]
\right]
\cdots
\right]
\right]\\
= \frac13 \left [ 1 + \sum_{k=2}^n(-1)^k \alpha^{\frac{k(k+1)}{2}-1} \right] + (-1)^{n+1} \alpha^{\frac{(n+1)(n+2)}{2}-1}\left(\frac23-\theta\right)
$$
Since $|\alpha| < 1$, the $\theta$ dependent term in above expression drops off exponentially.
Independent of choice of $\theta \in [0,1]$, we have
$$\lim_{n\to\infty}
\left(\psi^{-1}\circ \left(\mathop{\bigcirc}_{k=1}^n \phi_{+}\circ\phi_{-}^{\circ k}\right)\circ\psi\right)(\theta)
= \frac13 \left[ 1 - 2\sum_{k=2}^\infty(-1)^k \alpha^{\frac{k(k+1)}{2}}\right]$$
This means the infinite radical is well defined and has value
$$2\cos\left[\frac{\pi}{6}\left(1 - 2\sum_{k=2}^\infty(-1)^k \alpha^{\frac{k(k+1)}{2}}\right)\right]$$
Numerically, this expression evaluates to $\approx 1.567883223337111$, matching the number mentioned in question.
A: Let $R$ be the iterated square root in question
One easily check that any finite expression of the form
$$\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm\cdots}}}$$
is between $0$ and $2$. So $0\le R\le 2$.
Suppose that $a=2\cos t$. Then
$$\sqrt{2+a}=\sqrt{2(1+\cos t)}=\sqrt{4\cos^2(a/2)}=2\cos\frac a2$$
and
$$\sqrt{2-a}=\sqrt{2(1-\cos t)}=\sqrt{4\sin^2(a/2)}=2\sin\frac a2
=2\cos\left(\frac\pi2-\frac a2\right).$$
Therefore
$$\begin{align}
\cos^{-1}\frac R2&=\frac12\cos^{-1}\frac12\sqrt{2-\sqrt {2+\sqrt{2
-\sqrt{2-\sqrt{2-\cdots}}}}}\\
&=\frac{\pi}4-\frac14\cos^{-1}
\frac12\sqrt {2+\sqrt{2
-\sqrt{2-\sqrt{2-\cdots}}}}\\
&=\frac{\pi}4-\frac18\cos^{-1}
\frac12\sqrt{2
-\sqrt{2-\sqrt{2-\cdots}}}\\
&=\frac{\pi}4-\frac{\pi}{16}+\frac1{16}\cos^{-1}
\frac12
\sqrt{2-\sqrt{2-\cdots}}\\
\end{align}
$$
etc. So we can get a series for $\cos^{-1}(R/2)$. I think it might be
something like
$$\frac\pi2\sum_{n=0}^\infty\frac{(-1)^n}{2^{n(n+1)/2}}$$
which is some sort of theta function.
A: First aid.
$ \sqrt {2_1 - \sqrt {2_2 \pm \sqrt {2_3 \pm \ldots \pm \sqrt {2_n }}}} = 2 \sin \left ( \dfrac {90 ^ \circ (2a + 1)} {2 ^ n} \right ) $
$ \sqrt {2_1 + \sqrt {2_2 \pm \sqrt {2_3 \pm \ldots \pm \sqrt {2_n }}}} = 2 \cos \left ( \dfrac {90 ^ \circ (2a + 1)} {2 ^ n} \right ) $
If $ n \le 2  $ then $  a = 0 $
If $ n \ge 2  $ then $  0 \le a \le 2 ^ {n - 2} - 1 $
If $ n, a $ are known, then the signs $ s_t = \pm 1 $ lying between the terms $ 2_2 $ and $ 2_n $ are calculated from the relationships
$ r = round(a / 2 ^ {n - t}) $
$ s_t = (- 1) ^ r ,  t = 2, 3, \ldots , n - 1 $
if we match $ + = 0, - = 1 $ , then the binary digits which are among the terms $ 2_2 $ and $ 2_n $ form a number $ b $ , which is closely related to the $ a $ parameter:
● Binary representations of $ a, b $ always have the same number of digits.
● Each value of $ a $ is assigned to a single value of $ b $, regardless of the $ n $ value.
More in https://new-trigonometry.quora.com/
