<s>[INST] 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? 
 [/INST] 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.
</s>