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.
