Limit of a sequence $\sqrt{1+\sqrt{\underbrace{1+ \ldots +\sqrt{1}}_{n}}}$ $\sqrt{1+\sqrt{\underbrace{1+ \ldots +\sqrt{1}}_{n}}}$
I think the limit is $2$ for intuitive reasons.
Now, to prove it, I need to use the definition of a sequence's limit.
So, I need to solve the following inequality
$|\sqrt{1+\sqrt{\underbrace{1+ \ldots +\sqrt{1}}_{n}}}-2| < \epsilon$
for an arbitrary $\epsilon$.
But I'm not sure it is possible.
One implication of the inequality is 
$\sqrt{1+\sqrt{\underbrace{1+ \ldots +\sqrt{1}}_{n}}} < \epsilon + 2$
What I found out is the same as
$ \underbrace{( \ldots ((}_{n-1}\epsilon + 2\underbrace{)^2 - 1)^2 \ldots )^2}_{n-1} - 2 > 0$
But I'm not sure it is solvable in this form.
 A: Hints: Suppose the limit exists, and denote it by $L$. Then notice that $L=\sqrt {1+L}$, from which you get $L^2-L-1=0$ and you can find the roots to obtain candidates for the limit. If you can then argue why the limit does in fact exist, and you can eliminate one of the limits, then you will have found the answer. In any case, $2$ does not solve the equation, so the limit is certainly not $L=2$. (There are better ways of showing that a sequence converges than playing with $\varepsilon $. Ask yourself, is the sequence monotone? is it bounded?) 
A: Write out the sequence recursively. You can start with $a_1 = 1$ and define $a_{n+1} = \sqrt{a_n + 1}$.  You can check rather quickly that the sequence is bounded above and increasing so it must have a limit $L$. Take limits in your recursive formula to find $L = \sqrt{L + 1}$.

Added in response to comments:
If $a_n \to L$, then $a_n + 1 \to L + 1$, and as long as $a_n + 1 \ge 0$, $\sqrt{a_n + 1} \to \sqrt{L + 1}$.
If $a_n \to L$, then $a_{n+1} \to L$ since these are the terms of the same sequence.
Since $a_n \ge 1$ for all $n$, you can take the limit on both sides of $a_{n+1} = \sqrt{a_n + 1}$ to get $L = \sqrt{L + 1}$.
A: The map $f: [0,\infty) \to [0,\infty)$ given by
$$
f(x) =  \sqrt{1 + x}
$$
is a contraction mapping because when $x < y$,
$$
\begin{align*}
f(y) - f(x)
&= \int_x^y \left[ \frac{d}{dx} \sqrt{1 + x} \right] \; dx \\
&= \int_x^y \left[ \frac{1}{2\sqrt{1+x}} \right] \;dx \\
&\le \int_x^y \frac1{2 \sqrt{1}} \; dx \\
&= \frac12 (y - x).
\end{align*}
$$
It follows by the Banach fixed-point theorem that $f$ has a unique fixed point $a$ and the sequence $x, f(x), f(f(x)), \ldots$ converges to $a$ for any $x$.
What is the fixed point? Set $a = \sqrt{1 + a} \implies a^2 = 1 + a$.
Of the two solutions to this quadratic, only one lies in the correct range $[0,\infty)$.
A: Based on the statement on the problem and my interpretation of the "$\dots$," it appears that all of the previous answers given are not correct.  The answers seem to assume an embedded square root, but I don't see that here.  The sequence as written seems to be nothing more than $a_n=\sqrt{1+\sqrt{n}}$, which clearly diverges.
