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.

This is just an empirical answer. Consider

$$f(n) = \sqrt{t+\sqrt{t+1 - \sqrt{t+2 + \sqrt{t+3 - \cdots}}}}$$

We need to find $f(1)$. But

\begin{align} f(1) &= \sqrt{1+\sqrt{2 - f(3)}}\\ f(3) &= \sqrt{3+\sqrt{4 - f(5)}}\tag{Similarly moving on }\\ f(n) &= \sqrt{n+\sqrt{n+1 - f(n+2)}}\\ \end{align}

After squaring and rearranging, we get

$$f(n+2) = n+1 - (f(n)^2 - n)^2$$

I used the wolfram alpha recurrence solver to solve the recurrence and it gave me this:

$$\begin{array}{|c|c|c|c|} \hline n& 0 & 1 & 2 & 3 & 4 \\ \hline g(n) & 0 & 1 & 1 & 2& 2\\ \hline \end{array}$$

I don't know if this helps, but $x = 1$.

The computations will involve complex numbers. Starting at stage $4$. So $$ \sqrt{1+\sqrt{2-\sqrt{3+\sqrt{4}}}} \approx 1.03 + i\;0.24 $$ Always choosing the "principal" square root, the limit seems to be $1.005667607+ i\; 0.1066177058$ .

added
In fact, $\sqrt{3+\dots}$ converges to a real value, but $\sqrt{2-\dots}$ and $\sqrt{1+\dots}$ to non-real values.