Ambiguous or informal ellipses I was thinking about the infinite expression:
$$\sqrt{6+\sqrt{6+\sqrt{6+...}}}$$
It seems to me a bit informal or underspecified. You could state it better by defining an infinite series:
$$a_n = \sqrt{6+a_{n-1}}$$
And then ask about $\lim_{n\rightarrow\infty}{a_n}$. But then you have to wonder what $a_0$ is. Although anyway it's not too hard to figure out that if $a_0 \geq -6$, then the limit is 3. I fully understand this is the typical treatment and has been covered before. But where does this $a_0 \geq -6$ come from? Is there a way to evaluate it without making any assumptions like this?
It also dawned on me that maybe you could also define this sequence going "outside-in" instead of "inside-out". For instance:
$$b_n = b_{n-1}^2-6$$
But this seems a bit more unruly. It diverges for values other than $3$ and $-2$.
So I guess my question is: is that original expression just underdefined, and requires a small "leap" to formalize? Or is there a better way to treat that kind of thing strictly? Are there examples of similar expressions where the "informality" is troublesome - like maybe there's more than one way to formalize it that leads to different answers?
 A: Actually it is not ambiguous. The map $f:x\mapsto \sqrt{6+x}$ is a contraction of the metric space $[-1,7]$, due to $f'(x)=\frac{1}{2\sqrt{6+x}}$, so by the Banach fixed point theorem for any $x\in[-1,7]$ the sequence
$$ x, f(x), f(f(x)), f(f(f(x))),\ldots $$
converges towards the unique fixed point of $f$ in $[-1,7]$, i.e. $3$. Summarizing
$$ \sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=3.$$
A: The use of "$\cdots$" in this way will always indicate a limit. On the face of it, it indicates repeating some process an infinite number of times -- which is of course impossible. You can only take the limit of a sequence of finite performable operations (or determine that such a limit does not exist). So the expression $$\sqrt{6 + \sqrt{6 + \sqrt{6 + \ldots}}}$$
can only mean the limit of the sequence $s_n$ (if it exists), where
$$s_0 =\sqrt{6}$$
$$s_1 = \sqrt{ 6 + \sqrt{6}} = \sqrt{6 + s_0}$$
$$s_2 = \sqrt{ 6 + \sqrt{ 6 + \sqrt{6}}} = \sqrt{ 6 + s_1}$$
and so
$$s_n = \sqrt{6 + s_{n-1}}$$
So in other words, your expression means unambiguously
$$\lim_{n\to\infty}s_n$$
if this limit exists.
