Prove sequence $S_n$ converges If $S_1 = \sqrt{2}$, and
  
  $S_{n+1} = \sqrt{2 + \sqrt{S_n}}$ (n = 1,2,3....),
  
  prove that $\{S_n\}$ converges, and that $S_n &lt; 2$ for all $n \in \Bbb{N}$
  
  This is one the questions from Principles of Mathematical Analysis by Rudin. I am not sure how to proceed with the solution.
Hints:

1) $S_n &lt; 2.$ Clearly, $S_1&lt;2$. Suppose that $S_k &lt;2$. Then $S_{k+1}=\sqrt{2 + \sqrt{S_k}} &lt;\sqrt{2 + 2}=2.$

2) $S_n \le S_{n+1}$.  Clearly, $S_1&lt;S_2$. Suppose that $S_{k-1} &lt;S_k$. Then $$S_{k}=\sqrt{2 + \sqrt{S_{k-1}}} &lt;\sqrt{2 +\sqrt{S_{k}}}=S_{k+1}.$$

3) So its limit exists, say $S$. So, $S=\sqrt{2+\sqrt{S}}.$
Well if we prove that $S_n&lt;2$ for $n=1,2,3,\ldots$ then the first part comes easily as $S_n$ is a sequence in $\mathbb{R}$ which is complete. Since $S_n$ is a monotone increasing sequence with an upper bound (namely 2), it converges. We can prove $S_n&lt;2$ for $n \ge 1$ via induction as follows:

$\textbf{I:}$ $ S_1 &lt; 2 $, i.e. $\sqrt{2} &lt; 2$, can be checked by direct computation.

$\textbf{II:}$ Suppose $S_{n-1}&lt;2$ but that $S_n \ge 2$. But then we have $\sqrt{2 + \sqrt{S_{n-1}}} \ge 2 \Rightarrow \sqrt{S_{n-1}} \ge 2$, a contradiction. Thus $S_n &lt; 2$.

$\textbf{III:}$ Thus $S_n&lt;2$ for all $n = 1,2,3,\ldots$.