TITLE: Baby Rudin ("Principles of Mathematical Analysis"), chapter 3 exercise 3
QUESTION [5 upvotes]: If $s_1=\sqrt 2$,
and $$s_{n+1}=\sqrt {2+\sqrt{s_n}}$$
Prove that ${s_n}$ is converging, and that $s_n<2$ for all $n=1,2\ldots$
My attempt to use mathematical induction: if $s_n<2$ then I use
$s_{n+1}=\sqrt {2+\sqrt s_n}$, to prove $s_{n+1}<\sqrt {2+2}=2$.
I guess $s_n$ is an increasing sequence, I can use the $s_n$ is bounded and increasing, and then prove that $s_n$ converges.

REPLY [2 votes]: Wesley's lemma 1 nicely shows that $s_n$ is bounded above.
Rephrasing:
Show that $s_n$ is increasing:
1) $n=1$: $s_2 > s_1√$
2) Hypothesis : $s_{n+1} \ge s_n.$
Step  $n+1:$
$s_{n+2} =\sqrt{2+√s_{n+1}} {Hyp. \atop \ge} \sqrt{2+√s_n} =s_{n+1}.$
Used: $f(x)=√x$ is an increasing function.