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One night after hectic schedule at college, I was standing in front of black board with a chalk in my hand (to relax my mind!).

A thought of nested roots came to my mind and I wrote: $\sqrt{2 +\sqrt{2 +\sqrt{2 + \ldots}}} = 2$
As I had `learnt’ from some olympiad book in class 8, the trick to solve such problems.
Let, $\sqrt{2 +\sqrt{2 +\sqrt{2+ \ldots}}} = x$
As we have infinite number of terms on left hand side, if we keep one of the two’s aside, we can rewrite it as:\\ $\sqrt{2 + x} = x$ $\Rightarrow 2 + x = x^2$
Solving this quadratic we will get $x = 2, -1$
Clearly $x\geq 0$, thus $x = 2$.
Now thrilled by this I proceeded to evaluate nested roots for other arithmetic operations in similar fashion: $\sqrt{2 -\sqrt{2 -\sqrt{2 - \ldots}}} = 1$ $\sqrt{2\sqrt{2\sqrt{2\ldots}}} = 2$ $\sqrt{\frac{2}{\sqrt{\frac{2}{\sqrt{\frac{2}{\vdots}}}}}} = \sqrt{2}$
Now I knew that no one knows the answer for: $1-1+1-1+1-1\ldots = ??$
so, inspired by this I wrote: $\sqrt{1-\sqrt{1 +\sqrt{1-\ldots}}} = ??$
But then one of my friend suggested:

Since we are dealing with infinite terms we can say $\sqrt{1-\sqrt{1 +x}} = x$ $\Rightarrow 1-\sqrt{1 +x} = x^2$ $\Rightarrow x^4 - 2x^2 - x = 0$ $\Rightarrow x(x^3 - 2x - 1) = 0$ $\Rightarrow x(x+1)(x^2-x-1) = 0$ $\Rightarrow x = 0, -1, \frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}$
But I can’t say which one out of $0\quad \& \quad \frac{1+\sqrt{5}}{2}$ is solution.
I know that flaw in this argument is similar to that in saying: $1-1+1-1+1-1\ldots = 0$, by pairing $+1\quad \& \quad -1$.
But I found the idea itself flawless and tried to apply it to my starting expression: $\sqrt{2 +\sqrt{2 + x}} = x$ $\Rightarrow 2 + \sqrt{2 + x} = x^2$ $\Rightarrow x^4 - 4x^2 - x +2 = 0$ $\Rightarrow (x-2)(x+1)(x^2 + x -1) = 0$ $\Rightarrow x = 2, - 1, \frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}$
Unlike last case I can decide between $2\quad \& \quad \frac{-1+\sqrt{5}}{2}$  [Thanks to Sagar Shrivastava]

Since, $\sqrt{2 +\sqrt{2 +\sqrt{2+ \ldots}}} > \sqrt{2} > \frac{-1+\sqrt{5}}{2}$

Thus if I take out more and more 2’s out of square root, I will get more and more values of $x$ and among them we will get only one appropriate value of $x$.
But still I can’t figure out the correct value of $\sqrt{1-\sqrt{1 +\sqrt{1-\ldots}}}$ since I don’t have much experience with ‘series’ at this stage, but will surely investigate this question in detail by end of this semester.
Find the value of : $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\ldots}}}}$  (See this article by B. Sury)