Can someone help me with proving the convergence of a sequence written in this form? From this exercise:

If $s_1 = \sqrt{2}$, and $$s_{n+1} = \sqrt{2+\sqrt{s_n}} \quad (n = 1,2,3,\dots),$$ prove that $\{s_n\}$ converges, and that  $s_n< 2$ for $n = 1,2,3,\dots$.

in the book Principles of Mathematical Analysis, I find this book very obscure about many concepts, so I really need some help.
Can someone help me with proving the convergence of a sequence written in this form?
 A: *

*Show by indzction, that $s_n<2$ for all $n$.

*Show by induction that $s_{n+1} > s_n$ for all $n$.
From 1. and 2. it follows then , that $(s_n)$ is convergent.
A: If such a limit exists we must have $$l=\sqrt{2+\sqrt l}$$or $$l^2=2+\sqrt l$$define $e_n=s_n-l$. We show that $e_n\to 0$. We have $$s_{n+1}=\sqrt{2+\sqrt{s_n}}$$therefore $$e_{n+1}{=\sqrt{2+\sqrt{s_n}}-l\\=\dfrac{2+\sqrt{s_n}-l^2}{\sqrt{2+\sqrt{s_n}}+l}\\=\dfrac{2+\sqrt{s_n}-l^2}{\sqrt{2+\sqrt{s_n}}+l}\\=\dfrac{\sqrt{s_n}-\sqrt l}{\sqrt{2+\sqrt{s_n}}+l}\\=\dfrac{e_n}{(\sqrt{2+\sqrt{s_n}}+l)(\sqrt s_n+\sqrt l)}}$$therefore $$|e_{n+1}|=|\dfrac{e_n}{(\sqrt{2+\sqrt{s_n}}+l)(\sqrt s_n+\sqrt l)}|=\dfrac{|e_n|}{(\sqrt{2+\sqrt{s_n}}+l)(\sqrt s_n+\sqrt l)}\le \dfrac{|e_n|}{l\sqrt l}\le\dfrac{|e_1|}{(l\sqrt l)^{n}}$$since both $l$ and $\sqrt l$ are non-negative and $$\sqrt{2+\sqrt{s_n}}+l>l\\\sqrt s_n+\sqrt l>\sqrt l$$also the last inequality is attained by iteratively applying the one before that i.e.$$|e_{n+1}|<\dfrac{|e_n|}{l\sqrt l}<\dfrac{|e_{n-1}|}{(l\sqrt l)^2}<\cdots<\dfrac{|e_1|}{(l\sqrt l)^n}$$ which means that $|e_{n}|\to 0$ or $e_n\to 0$
A: Because nobody mentioned this method, not even in the question I flagged as duplicate ...
Proposition 1. $0<s_n <2$
By induction. $0<s_1=\sqrt{2}<2$. Let's assume (induction hypothesis) 
$$0<s_n<2 \Rightarrow 
0<\sqrt{s_n}<\sqrt{2}\Rightarrow \\
0<2<2+\sqrt{s_n}<2+\sqrt{2}<4 \Rightarrow \\
0<\sqrt{2+\sqrt{s_n}}<2 \Rightarrow \\
0<s_{n+1}<2$$

Proposition 2. $\{s_n\}$ is ascending.
We have $s_{n+1}=f(s_n)$ where $f(x)=\sqrt{2+\sqrt{x}}$. But $f'(x)=\frac{1}{4 \sqrt{2 + \sqrt{x}} \sqrt{x}}>0$, so $f(x)$ is ascending for $x>0$. We also have that $s_2=\sqrt{2+\sqrt{\sqrt{2}}}>\sqrt{2}=s_1$, thus 
$$f(s_2)\geq f(s_1) \iff s_3\geq s_2$$
and by induction
$$f(s_n)\geq f(s_{n-1}) \iff s_{n+1}\geq s_{n}$$

As a result, we have a monotone and bounded sequence, so it converges ... to the solution of 
$$L=\sqrt{2+\sqrt{L}}$$
