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? 
 A: 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
$$
l= 1.83117720\cdots.
$$
A: 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.
A: 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
$$
s=-\frac13+\frac{1}{\sqrt[3]{54}} \Big(\left(79+3 \sqrt{249}\right)^{1/3}+ \left(79-3 \sqrt{249}\right)^{1/3}\Big).
$$
