Discuss the convergence of the sequence: $a_1=1,a_{n+1}=\sqrt{2+a_n} \quad \forall n \in \mathbb{N}$ Computing the first few terms $$a_1=1, a_2=\sqrt{3}=1.732....,a_3=1.9318....,a_4=1.9828...$$ I feel that $(a_n)_{n\in \mathbb{N}}$ is bounded above by 2, although I have no logical reasoning for this. Since,  $(a_n)_{n\in \mathbb{N}}$ is monotone increasing sequence, it must converge by monotone convergence theorem, and converge to 2. 
Can anyone help me to make this more formal? Besides, I would really appreciate if anyone could shed some light on how to find the bound and limit of such sequences (that are not in closed form but in recursion). 
 A: $$
a_{n+1}-2 = \sqrt{2+a_n}-2 
$$
$$
a_{n+1}-2 = \frac{a_n-2}{\sqrt{2+a_n}+2} 
$$
$$
|a_{n+1}-2| \le \frac{1 }{2}.|a_n-2|
$$
$$
|a_{n}-2| \le {(\frac{1 }{2})}^n.|a_0-2|
$$
 hence
$$
a_n \to 2
$$
A: Hints:
$$a_1\le a_2\;\;\text{and}\;\;a_{n+1}:=\sqrt{2+a_n}\stackrel{\text{induction}}\le\sqrt{2+a_{n+1}}=:a_{n+2}$$
$$a_1\le 2\;\;\text{and}\;\; a_{n+1}:=\sqrt{2+a_n}\stackrel{\text{induction}}\le\sqrt{2+2}=2$$
The above shows your sequence is a monotone ascending one and bounded above, so its limit exists, say it is $\;w\;$, and now a little arithmetic of limits:
$$w\leftarrow a_{n+1}=\sqrt{2+a_n}\rightarrow\sqrt{2+w}$$
so $\;w=\ldots\ldots?$
A: Here is an alternate way to approach this problem. 
Set $f(x) = \sqrt{2 + x}$, then we notice $f(x) : [0,\infty)\rightarrow[0,\infty)$ 
and $f'(x) = \frac{1}{2\sqrt{2 + x}} \le \frac{1}{2\sqrt{2}}<1$ for $x\in (0,\infty)$ 
So by the contraction mapping principle $f$ has a unique fixed point in $(0,\infty)$ and any iteration will converge to to this fixed point. This rigorously justifies just solving the equation 
$$x = \sqrt{2 + x}$$ to get the limit of the sequence $a_n$.
A: Note that $(2-\sqrt{2+x})  (2+\sqrt{2+x} )= 2-x$. We assume $x \ge 0$ in the following.
In particular, $\sqrt{2+x} \le 2$ iff $x \le 2$. 
Then, if $x \le 2$, we have $2-\sqrt{2+x} = \frac{1}{2+\sqrt{2+x}} (2-x) \le \frac{1}{2+\sqrt{2}}(2-x)$.
Hence $2-a_{n+1} \le \frac{1}{2+\sqrt{2}} (2-a_n)$, from which we get $0 \le 2-a_n \le \frac{1}{(2+\sqrt{2})^{n-1}} (2-a_1)$. Hence $a_n \uparrow 2$.
