Many Radicals!

Let's look at the related problem of finding $y=\sup S$ . It's easy to see that, for any given $n$ , the largest possible number of this type is found when all $\epsilon_k$ are $+1$ , that is $y_n=\sqrt{3+\sqrt{3+\sqrt{3+\cdots +\sqrt3}}}$ (where the expression contains $n$ $3$ s). As ever, we have to be cautious around convergence. To make this sequence easier to handle, note that $y_1=\sqrt3$ and $y_{n+1}=\sqrt{3+y_n}$

$y_n$ is always positive. Define $f(t)=\sqrt{3+t}$ ; then $f(t)$ has a fixed point at $t_c=\frac12 +\frac{\sqrt{13}}{2}$ . Since $|f'(t_c)|<1$ , this is an attracting fixed point; in other words, whatever positive start value is chosen for $y_0$ , the limit as $n \to \infty$ of $y_n$ will always be $t_c$ .

So $y=\sup S=t_c$ .

To solve the actual problem, note that the smallest (positive) number we can possibly make for a given $n$ has the form $x_n=\sqrt{3{\color{#D61F06}-}\sqrt{3+\sqrt{3+\cdots+\sqrt3}}}$ (again with $n$ $3$ s in the expression). In the limit, this is just $x=\sqrt{3-t_c}$ which works out to be $x=\sqrt{\frac12 \left(5-\sqrt{13} \right)} =0. {\color{#D61F06} 834999} 6\ldots$

We can introduce the following notation $x_n(\epsilon_1, \epsilon_2,..., \epsilon_n)=\epsilon_n \sqrt{3+\epsilon_{n-1} \sqrt{3+...+\epsilon_2 \sqrt{3+\epsilon_1 \sqrt{3}}}}, \;\;\;\;\;\;\;(*)$ for any non-negative integer number $n,$ and any finite sequence $(\epsilon_k)_{1\leq k \leq n},$ where $\epsilon_k=\pm 1$ for any integer k satisfying $1\leq k\leq n.$ Of course, $S$ is the set of all possible numbers of form (*). It easy to prove by Mathematical Induction that all these number are real. We are going to use also the following notation $y_n=x_n(1,1,1,...,1),$ where the last expression contains $n$ ones. We are going to divide the proof into three parts.

Part 1. To prove that the sequence $y_n$ has the limit $r=\frac{1+\sqrt{13}}{2}$ and $y_n <r$ for any non-negative integer $n.$

We can see that $y_{n+1}^2= 3+y_{n},$ and using mathematical induction and this recursion, we can prove that the sequence $(y_n)$ is increasing and is bounded above by the number $r.$ Since this sequence is increasing, upper-bounded and all its members are positive, then it has positive limit $l$ and from the recursion (taking limits on both sides) we obtain that $l$ satisfies the equation $l^2=3+l,$ therefore $l=\frac{1\pm \sqrt{13}}{2}.$ When you picked the negative sign in front of the radical you get a negative value that cannot happen, therefore, $l=\frac{1+ \sqrt{13}}{2}=r.$ Of course, it is clear that as the sequence $(y_n)$ is increasing and its limit is $r,$ then $y_n<r$ for all values of $n.$

Part 2. To prove that $|x_n(\epsilon_1,..., \epsilon_n)|\leq y_n$ for all non-negative integer $n.$

This is another thing that is easy to prove by Mathematical Induction and, of course, equality is reached when $\epsilon_k=1$ for any $k$ .

Part 3. To prove that the largest positive number positive integer $c$ such that $|x|>c$ for all $x\in S$ is $\sqrt{3-r}.$

Let $n$ be whatever integer number such that $n\geq 1.$ If $n=1$ then $|x_1(\epsilon_1)|=y_1=\sqrt{3}> c$ If $n>1,$ then $|x_n(\epsilon_1,..., \epsilon_n)|=\sqrt{3+x_{n-1}(\epsilon_1,..,\epsilon_{n-1})}\geq \sqrt{3-|x_{n-1}(\epsilon_1,..,\epsilon_{n-1})|}.\;\;\;\;\;\;\;(**)$ Using Part 2 , and Part 1 , we have that $|x_{n-1}(\epsilon_1,..,\epsilon_{n-1})|\leq y_{n-1}< r$ Then $\sqrt{3-|x_{n-1}(\epsilon_1,..,\epsilon_{n-1})|}>\sqrt{3- r},$ and this together with (**) implies that $|x_n|>\sqrt{3-r}=c.$ To see that $\sqrt{3-r}$ is the largest positive number satisfying the previous inequality it is enough to see that the numbers $\sqrt{3-y_n}$ belong to S and $\lim_{n \rightarrow \infty} \sqrt{3-y_n}=\sqrt{3-r}=c.$

Then $c= \sqrt{3-r}=\sqrt{3-\frac{1+\sqrt{13}}{2}}.$ Then the answer to this problem is $\boxed{834999}.$