Proof Verification of Baby Rudin 3.3 I am currently studying Rudin's Foundations of Real Analysis. I wrote a solution to Rudin's problem 3.3, and wanted to get some input on my proof. I appreciate any suggestions on the technicalities, style, and clarity.
Problem:

If $s_1=\sqrt2$, and 
  $s_{n+1} = \sqrt{2+\sqrt{s_n}}$ where $n = 1, 2, 3, ...$ prove that {${s_n}$} converges, and that $s_n<2$ for $n = 1, 2, 3, ...$

My proof:
We first show $s_n<2$ for $n = 1, 2, 3, ...$ by induction.
When $n = 1$, it is clear that $\sqrt2 < 2$. 
If $s_{n-1}$ < 2, then 
$0<\sqrt{2+\sqrt{s_{n-1}}} < 2$
$\leftrightarrow 2+\sqrt{s_{n-1}}<4$ 
$\leftrightarrow \sqrt{s_{n-1}} < 2$
$\leftrightarrow s_{n-1} < 4$ 
But $s_{n-1} < 2$ by induction hypothesis. Thus $s_n<2$ for $n = 1, 2, 3, ...$ as desired.
We next show that {$s_{n}$} is monotonically increasing by induction.
It is clear that $s_1\leq s_2$.
We need to show $s_n \leq s_{n+1}$ for all $n \geq 2$. Assume $s_{n-1} \leq s_{n}$. Then,
$s_n \leq s_{n+1}$
$\leftrightarrow 0 < \sqrt{2+\sqrt{s_{n-1}}} \leq \sqrt{2+\sqrt{s_{n}}}$
$\leftrightarrow 2+\sqrt{s_{n-1}} \leq 2+\sqrt{s_{n}}$
$\leftrightarrow \sqrt{s_{n-1}} \leq \sqrt{s_{n}}$
$\leftrightarrow s_{n-1} \leq s_{n}$
But $s_{n-1} \leq s_{n}$ by the induction hypothesis. Thus the sequence is monotonically increasing.
Since {$s_n$} is bounded above and monotonically increasing, it converges.
 A: *

*For the first proof, Also, I would show a series of forwards implications rather than the string if "if and only if" statements you are using, since your proof doesn't require both directions. 


In this spirit, your first inductive proof, upper-bounding $s_{n}$ by $2$, could be re-phrased as follows: 
We claim $s_{n} < 2$ for all $n$ by induction. 
For $n = 1$ this is clear. Assume $s_{n-1} < 2$. Then since the square root function is monotonically increasing on the non-negative reals,
\begin{eqnarray}
\sqrt{s_{n-1}} < 2 \\
2 + \sqrt{s_{n-1}} < 4 \\
\sqrt{2 + \sqrt{s_{n-1}}} < 2 
\end{eqnarray}
Since $s_n = \sqrt{2 + \sqrt{s_{n-1}}}$ this completes the inductive step. 
I made a point of saying the square-root function is monotonic. This is fairly pedantic, and I don't think is really needed, but given that your argument implicitly uses this fact, it doesn't hurt.


*Your second proof has what I believe to be a typo.



We need show $s_n \leq s_{n−1}$ for all $n \geq 2$. Assume $s_n \leq s_{n-1}$. 

At first glance, I wasn't sure if this was a typo or if you meant to do a proof by contradiction. I think what you meant to show was $s_n \leq s_{n+1}$. In that case, replacing $s_{n-1}$ with $s_{n+1}$ where appropriate should be fine, and stylistically I would say the same things as regarding your first proof. 
