real analysis - 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?

Answer

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 $e_n\to 0$