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.

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.
