The convergence of $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$ I would like to know if this sequence converges $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$.
I know this sequence is increasing monotone, but I couldn't prove it's bounded.
Thanks
 A: Let $l=\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$, then $l=\sqrt {2+l}$.
From this we have $l^2-l-2=0$
Can you solve it from here? It has to be bounded if there's a finite solution in the limit as $n\to\infty$ and the sequence is monotone.
A: As written it's not clear this is a sequence at all but I'm assuming your sequence is
$\sqrt{2}$, $\sqrt{2 + \sqrt{2}}$, $\sqrt{2 + \sqrt{2+ \sqrt{2}}}$, $\sqrt{2 + \sqrt{2+ \sqrt{2+\sqrt{2}}}}$, $\sqrt {2+\sqrt {2+\sqrt {2+ \sqrt{2 +\ldots}}}}$.
Then it should be obvious that each entry in the sequence is positive and the sequence increases each time.
Given this the final entry in the infinite sequence would be $x = \sqrt{2 + x}$ which we can easily solve.
$$\begin{align}
x &= \sqrt{2+x} \\
x^2 &= 2 + x\\
x^2 - x -2 &= 0\\
x &= \dfrac{1 \pm \sqrt{1+8}}{2} \\
x &= \dfrac{4}{2} = 2
\end{align}$$
The sequence has a lower bound of $\sqrt{2}$ and an upper bound of $2$ increasing monotonically.
A: Let $a_n = \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{n\;\text{terms}}$ and $\epsilon_n = 2-a_n$. It is clear 
$$a_1 = \sqrt{2} \quad\implies\quad \epsilon_1 = 2-\sqrt{2} \in(0,1).$$
Notice
$$\epsilon_{n+1} = 2 - \sqrt{2+a_n} = 2 - \sqrt{4-\epsilon_n}
= \frac{\epsilon_n}{2 + \sqrt{4-\epsilon_n}}
\quad\implies\quad \epsilon_{n+1} \in \left(0,\frac{\epsilon_n}{2}\right)
$$
We find 
$\displaystyle\;|\epsilon_n| < \frac{\epsilon_1}{2^{n-1}} < \frac{1}{2^{n-1}}
\to 0\;$ as $n \to \infty$. As a result, $\;a_n \to 2\;$ as $\;n \to \infty$.
A: Agreeing with what the others have said, I would like to add that there is in fact a simple explicit formula for the terms of this sequence:
$$
\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{\vphantom{\large A}2\,}\,}\,}\,}\
=\
2\cos\left(\vphantom{\Large A}\pi \over 2^{n + 1}\right)
$$
where the square root sign appears $n$-times. In particular, the sequence clearly converges to $2$.
Proof:
For $n=1$, the claim is true, since $\cos(\pi/4)=\sqrt{2}/2$. By the half-angle formula $$2\cos(x/2)=\sqrt{2+2\cos(x)}$$
Therefore
$$\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}=\sqrt{2+2\cos\left(\frac{\pi}{2^n}\right)}=2\cos\left(\frac{\pi}{2^{n+1}}\right)$$
where in the left square root expressions there are $n$ square roots and in the first equality we have used the induction hypothesis that the claim holds for $n-1$.
(From my answer to this question.)
A: Suppose $x\lt 2$
Consider $y=\sqrt {2+x}$ so that $y^2=2+x\lt4$ and $y$ being positive we have $y\lt 2$
That ought to enable you to prove a bound.
