Vtyjkyuk

I was thinking about the infinite expression:

$$sqrt6+sqrt6+sqrt6+...$$

It seems to me a bit informal or underspecified. You could state it better by defining an infinite series:

$$a_n = sqrt6+a_n-1$$

And then ask about $lim_nrightarrowinftya_n$. 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?

Actually it is not ambiguous. The map $f:xmapsto sqrt6+x$ is a contraction of the metric space $[-1,7]$, due to $f'(x)=frac12sqrt6+x$, so by the Banach fixed point theorem for any $xin[-1,7]$ 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
$$ sqrt6+sqrt6+sqrt6+ldots=3.$$

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 $$sqrt6 + sqrt6 + sqrt6 + ldots$$
can only mean the limit of the sequence $s_n$ (if it exists), where

$$s_0 =sqrt6$$
$$s_1 = sqrt 6 + sqrt6 = sqrt6 + s_0$$
$$s_2 = sqrt 6 + sqrt 6 + sqrt6 = sqrt 6 + s_1$$
and so
$$s_n = sqrt6 + s_n-1$$
So in other words, your expression means unambiguously
$$lim_ntoinftys_n$$
if this limit exists.