Evaluating $\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \cdots}}}}$ 
$$x =\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \sqrt{5+ \sqrt{6 - \cdots}}}}}}$$
Find $x$.

I am not sure how to proceed.
Is this a sort of Arithmetico-Geometric Progression? Will this converge at a point?
Any help would be sincerely appreciated.


I doubt a closed form exists for this number, but I'll show convergence and give some bounds on the values.
Let $f_m^n = \sqrt{m\pm\sqrt{(m+1)\mp \sqrt{\cdots\pm \sqrt n}}} $
be the the version where we go from $m$ to $n$ (don't take the $\pm$'s too seriously). We have the recursive expression
$$
f_m^n = \begin{cases}
\sqrt{m + (-1)^{m+1}f_{m+1}^n} & m<n \\
\sqrt n & m=n
\end{cases}
$$
Realness and general bounds:
Let $3 \le m < n$ with $n$ large enough. I claim that
$$
f_m^n - \sqrt{m-1} \in \begin{cases}
[0.5, 0.7] & m \textrm{ odd} \\
[-0.4, -0.6] & m \textrm{ even}
\end{cases}
$$
These constants can surely be refined, and maybe the ranges can be made to depend on $m$ as well to sharpen it further. Furthermore, it shouldn't be hard to show that $f_m^\infty-\sqrt m \to \pm \frac12$ for $m\to \infty$.
In particular everything is real in these cases (which is easy to check). We can show it by backwards induction on $m$. You can check that the base case $m=n-1$ holds for $n\ge8$.
For $m<n-1$ we must check for odd $m$ that
$$
f_m^n \in \left[ \sqrt{m + \sqrt{m} - 0.4}, \sqrt{m + \sqrt{m} - 0.6} \right]
\subset \left[ \sqrt{m-1} + 0.5, \sqrt{m-1} + 0.7 \right]
$$
and for even $m$ that
$$
f_m^n\in \left[ \sqrt{m - \sqrt{m} - 0.7}, \sqrt{m - \sqrt{m} - 0.5} \right]
\subset \left[ \sqrt{m-1} - 0.6, \sqrt{m-1} - 0.4 \right]
$$
These are quadratic inequalities, so we can run the numbers and see that they all hold for $m\ge3$.
Bounds on $f_1$:
Assume for now that $f_m^n$ converges as $n\to\infty$, which I'll prove below. Then the same bounds as above hold for $f_m := \lim_{n\to\infty}f_m^n$. Let's get some better bounds on $f_3$. We have
$$
f_3 = \sqrt{3+\sqrt{4-f_5}}
\in \left[\sqrt{3+\sqrt{4-\sqrt{4}-0.7}}, \sqrt{3+\sqrt{4-\sqrt{4}-0.5}}\right]
\approx [2.035, 2.055]
$$
We could get increasingly better bounds by using a higher term than $f_5$. Assuming the principal squareroot for complex numbers, we get approximately
$$
f_2 = \sqrt{2-f_3}
\in [0.186i, 0.235i]
$$
This finally gives us bounds on $f_1$. If $f_1=x+iy$, then $f_1$ lies on the curve $x^2-y^2=1$ and between the values approximately $1.00430 + 0.0928i$ and $1.00681 + 0.117i$.
Convergence:
We want to use that
$$
\lvert \sqrt{a+x} - \sqrt a \rvert \le \frac{|x|}{2}
$$
or
$$
\sqrt{a+x} = \sqrt a + \varepsilon\left(\frac{|x|}{2}\right)
$$
where $\varepsilon(x)$ means some number that is smaller than $x$ in absolute value.
This holds for all $x$ as long as $a\ge4$. We only need to show convergence for one value of $m$, so choose $m_0$ large enough that $(f_m^n)^2 \ge 4$ for all $n \ge m \ge m_0$ (we can do this using the bounds above). We unravel the radical from the inside out and get inductively (with the base case $k=0$ evident):
$$ \begin{split}
f_{n-k}^n 
&= \sqrt{(n-k) \pm f_{n-k+1}^{n}} \\
&= \sqrt{(n-k) \pm f_{n-k+1}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{k-1}}\right)} \\
&= \sqrt{\left(f_{n-k}^{n-1}\right)^2 + \varepsilon\left(\frac{\sqrt n}{2^{k-1}}\right)} \\
&= f_{n-k}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{k}}\right) \\
\end{split}
$$
In particular
$$
f_{m_0}^n = f_{m_0}^{n-1} + \varepsilon\left(\frac{\sqrt n}{2^{n-m_0}}\right)
$$
which means that
$$
f_{m_0} = \sum_{n=m_0}^\infty \varepsilon\left(\frac{\sqrt n}{2^{n-m_0}}\right)
$$
which is absolutely convergent.
