TITLE: Baby Rudin exercise 3.3 solution, possible typo in solutions manual?
QUESTION [5 upvotes]: Okay so I'm working through the exercises in Rudin and after checking my solutions manual for 3.3, I found something that seems like it can't be true. 
Here is the original question in rudin: 
Exercise 3.3:
Let $s_1= \sqrt 2 $
and $s _{n+1} =\sqrt{2 +\sqrt s_n}$   ($n=1,2,3....$)
prove that $s_n$ converges and that $s_n$ < 2 for $n=1,2,3....$
So I believe we need to proceed by showing the sequence is bounded above and increasing monotonically. The solutions manual I'm using did something very strange which doesn't seem to be true. They wrote:
$${s_n}^2 -2 =s_{n-1}$$
But how could this be true considering:
$${s_n}^2 =s_{n-1} +2 $$
$${s_n} =\sqrt{2+s_{n-1}}$$
is not true since the problem gave $$s_{n+1} =\sqrt{2+\sqrt{s_n}\\}$$
Is my thought process correct?
Now I'm seeing something confusing in another manual as well. This is true correct
I think:
$$s_2 = \sqrt{2+\sqrt{\sqrt{2}}}$$ correct?
but Im seeing 
$$s_2 = \sqrt{2+\sqrt{2}}$$
Which one is correct?

REPLY [3 votes]: You're right in both cases; both guides are wrong. (Here's one that seems to get it right.)
Note that the first manual you mentioned (the one you linked to in a comment) makes the same mistake again two lines further down, where it says that the limit satisfies $s^2-s-2=0$, which should be $s^2-\sqrt s-2=0$.