Evaluate the general infinite square root $$x = \sqrt{n\sqrt{n\sqrt{n}} \cdots}$$
I see that:
$$x = \sqrt{nx}$$
$$x^2 -nx = 0$$
Them:
$$x(x - n) = 0 \implies x \in \{0, n\}$$
How should I reject the $x = 0$ solution?  (any level proof is fine, analysis, calculus etc...)
 A: Define
$$x_m:=\overbrace{\sqrt{n\sqrt {n\ldots\sqrt n}}}^{m\;\text{times}}\;\;,\;\;\;\;m,n\in\Bbb N$$
then
$$x_m\le x_{m+1}\iff x_m\le\sqrt{nx_m}\iff x_m^2\le nx_m\iff x_m(x_m-n)\le0\iff x_m\le n$$
The last inequality can be proved by induction:
$$x_1=\sqrt n\le n\;\;\;\color{green}\checkmark$$
$$x_m=\sqrt{nx_{m-1}}\stackrel{\text{Ind. Hyp.}}\le\sqrt{nn}=n$$
We get thus that the sequence $\;\{x_m\}_{m\in\Bbb N}\;$ is monotonic increasing , and thus it converges (in the wide sense) to its supremum $\;\alpha\;$ (and we can already forget of zero), which is finite by the above, and then using arithmetic of limits:
$$\alpha\xleftarrow[m\to\infty]{} x_m=\sqrt{nx_{m-1}}\xrightarrow[m\to\infty]{}\sqrt{n\alpha}\implies \alpha^2=n\alpha\implies\alpha=n$$
A: One thing I like to think about is that this is essentially iterating the function
$$f(x)=\sqrt{nx}$$
and trying to ask what the value of $f^{\infty}$ is. This isn't a particularly well-defined notion, however, we can make it one by defining
$$f^{\infty}(x)=\lim_{n\rightarrow\infty}f^n(x).$$
Now, it's obvious that $f^{\infty}(x)$ must be a fixed point of $f$ if it exists - and you've proven that $0$ and $n$ are the only fixed points. Moreover, notice that, for $0<x<n$ we have $0<x<f(x)<n$, meaning that $f^n(x)$ is an increasing sequence, bounded above, and thus has a limit. Since it is always positive, that limit is positive, and the only positive fixed point is $n$. Similarly, we have that if $n<x$, then $n<f(x)<x$, thus $f^n(x)$ would be a decreasing sequence, bounded below by $n$ and hence, by a similar argument, converges to $n$. So, we can say for any $x>0$ that $f^{\infty}(x)=n$. This is a compelling reason to say that the expression:
$$\sqrt{n\sqrt{n\sqrt{n\ddots}}}$$
is actually $n$. So, unless we have a zero, hidden under those infinitely many radicals and canceling everything (note $f^{\infty}(0)=0$), the limit of the relevant sequence is $n$.
However, notice an important point here that the limit of iterating does depend on initial conditions (even if we might prefer to start iterating with $x>0$), so when we ask for a value like $f(f(f(\ldots)))$, we have to be somewhat careful. That is, we sort of have to imagine that we actually are looking at an expression like:
$$\sqrt{n\sqrt{n\sqrt{n\ddots x}}}$$
where the $x$ comes within infinitely many radicals and actually influences things. Here, it's rather trivial, but it can lead to strange questions otherwise.
A: You should not reject the solution $x=0$. On the other hand you should reject the question. Infinite expressions are meaningless in general, and can only sometimes be accepted if they can unambiguously translated into a limit expression (which is then subject to usual considerations of convergence).
Infinite expressions with ellipses "inside", like the one in the question, cannot be transformed unambiguously into limits, since all they suggest is an operation to iterate, but not the starting value. The expression in the question just suggests iteration of the function $x\mapsto \sqrt{nx}$, but says nothing about a starting point of this iteration; if there were such a starting point it would be at the end of the infinite nesting, but as we all know there is no such thing as the end of an infinite sequence (etymologically, "infinite" means without end, or bound). So the expression is asking to iterate that function indefinitely without starting anywhere. It turns out that if one starts at $0$ one remains there; if one starts in a positive number one converges to$~n$ (provided that $n\geq0$, which was however not given in the question), and if one starts in a negative number one runs into trouble right away. The only objective reason to prefer the second option is the cozy feeling that convergence on a large open set of starting values procures us. If one would take $n<0$, a similar argument would lead to preferring the first option, just to avoid trouble.
Note that the situation would be different if the ellipses were on the outside, as in $$\ldots \sqrt{n\sqrt{n\sqrt{n\sqrt{n\times5}}}},$$
as now it would more or less clear that you are asking about $\lim_{k\to\infty}f^k(5)$ where $f:x\mapsto\sqrt{nx}$.
A: Assuming that $x=0$ is also a solution with $n \neq 0$. Then you would have the relation $0 = \sqrt{n \sqrt{n \sqrt {n ...}}}$. Now one can take the square on both sides and it follows that either $n$ is zero or the iterated square root is zero. When $n=0$ you are done; else you continue proving the rejection of 0 by squaring again on both sides and compare solutions. Since this procedure is repeadet infinitely many, one concludes that $n=0$ which is a contradiction.
