calculus - Baby Rudin ("Principles of Mathematical Analysis"), chapter 3 exercise 3

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.

Answer

To give a name to your idea, you want to use the monotone convergence theorem; a bounded and increasing sequence converges to the supremum of that set, indeed a standard approach would be to use induction.

Lemma: $s_n$ is bounded

Base case:
$n=1$

$$s_1= \sqrt{2} <2$$

Suppose now for arbitrary $k$

$$s_k< 2$$

Now observe as the inductive step:
$$s_{k+1}=\sqrt{2 + \sqrt{s_k}}< \sqrt{2+2}=2$$

By the principle of mathematical induction we have found that $2$ is a bound for all $n\in \mathbb N$.

Lemma: $s_n$ is increasing

Base case:
$$s_2-s_1>0$$
$$\sqrt{2 + \sqrt{\sqrt{2}}}- \sqrt{2}>\sqrt{2 + 0}- \sqrt{2}=0$$

Now suppose for arbitrary $k$ we have:
$$s_{k}-s_{k-1}>0 \implies s_k > s_{k-1}\implies \sqrt{s_k}> \sqrt{s_{k-1}}$$

We now make our inductive step:

$$s_{k+1}-s_k=\sqrt{2 + \sqrt{s_k}}-\sqrt{2 + \sqrt{s_{k-1}}}>\sqrt{2 + \sqrt{s_{k-1}}}-\sqrt{2 + \sqrt{s_{k-1}}}=0$$
Hence we have an increasing sequence for all $n \in \mathbb N$ this completes the proof and our sequence is indeed convergent $\square$.