Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = \sqrt{2}$, and $s_{n+1} = \sqrt{2 + \sqrt{s_n}}$, what is the limit of this sequence?

Here’s Prob. 3, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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

My effort:

We can show that $\sqrt{2} \leq s_n \leq 2$ for all $n = 1, 2, 3, \ldots$. [Am I right?]

Then we can also show that $s_n < s_{n+1}$ for all $n = 1, 2, 3, \ldots$. [Am I right?]

But how to calculate the exact value of the limit? Where does this sequence occur in applications?

Solutions Collecting From Web of "Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = \sqrt{2}$, and $s_{n+1} = \sqrt{2 + \sqrt{s_n}}$, what is the limit of this sequence?"

But how to calculate the exact value of the limit?

Hint. You have shown that the limit does exist, then it has to satisfy
$$
l= \sqrt{2 + \sqrt{l}}, \qquad \sqrt{2}<l\le2,
$$ by squaring twice one gets
$$
l^4-4l^2-l+4=0
$$ then, by solving the quartic equation with Ferrari’s method, one gets

$$
l=\frac{1}{3\sqrt[3]{2}} \left(79+3 \sqrt{249}\right)^{1/3}+\frac{1}{3\sqrt[3]{2}} \left(79-3 \sqrt{249}\right)^{1/3}-\frac13
$$

observing that
$$
l= 1.83117720\cdots.
$$

If the limit $s$ of $s_n$ exists, then
$$
s=\lim s_n=\lim s_{n+1}=\lim \sqrt{2+\sqrt{s_n}}=\lim \sqrt{2+\sqrt{s}}.
$$
Hence, the limit satisfies the equation
$$
s=\sqrt{2+\sqrt{s}}.
$$
Thus,
$$
s^2=2+\sqrt{s}\qquad\text{or equivalently}\qquad (s^2-2)^2=s.
$$
Thus the limit is the unique solution of $s^4-4s^2-s+4=0$ which lies in the interval $[\sqrt{2},2]$.

To solve this equation, first observe that $s=1$ is a solution, and hence
$$
s^4-4s^2-s+4=(s-1)(s^3+s^2-3s-4).
$$
We can use this method to solve $\,s^3+s^2-3s-4=0,\,$ and obtain that
$$
s=-\frac13+\frac{1}{\sqrt[3]{54}} \Big(\left(79+3 \sqrt{249}\right)^{1/3}+ \left(79-3 \sqrt{249}\right)^{1/3}\Big).
$$

For the first part of your question, not answered above:

If $s_1=\sqrt{2}$ and
$$
s_{n+1}=\sqrt{2+\sqrt{s_n}}
$$
prove 1) that $\{ s_n \}$ converges and 2) that $s_n<2$ for any $n\in \mathbb{N}$..
First we show $\{ s_n\}$ is increasing by induction.

Base case:
$$
2>0\Rightarrow 2+\sqrt{2}>\sqrt{2}\Rightarrow \sqrt{2+\sqrt{2}}>\sqrt{2}
\Rightarrow s_2>s_1
$$
Inductive hypothesis: Suppose $s_n>s_{n-1}$, then
$$
\sqrt{s_n}>\sqrt{s_{n-1}}\Rightarrow
2+\sqrt{s_{n}}>2+\sqrt{s_{n-1}}\Rightarrow
\sqrt{2+\sqrt{s_{n}}}>\sqrt{2+\sqrt{s_{n-1}}}
\Rightarrow s_{n+1}>s_{n}
$$
Now, showing part 2 to be true will allow us to conclude part 1, as increasing
sequences bounded above converge.
Base case: $\sqrt{2}<2$.
Inductive hypothesis: Suppose $s_n<2$. Then
$$
\sqrt{2+\sqrt{s_{n-1}}}<2\Rightarrow 2+\sqrt{2+\sqrt{s_{n-1}}}<4\Rightarrow
\sqrt{2+\sqrt{2+\sqrt{s_{n-1}}}}<2\Rightarrow \sqrt{2+s_n}<2\\
\stackrel{\text{since $s_n>1\Rightarrow s_n>\sqrt{s_{n}}>0$}}{\Rightarrow}\sqrt{2+
\sqrt{s_n}}<\sqrt{2+s_n}<2\Rightarrow s_{n+1}<2
$$
and we can conclude.