Show that $s_n<2$ 
If $s_1=\sqrt{2}$ and $s_{n+1}=\sqrt {2+s_n}$ for $n\geq 1$ show that
  $s_n<2$ $\forall, n\geq1$ and $s_n$ is convergent.

The second part I supposed that is immediate from Cauchy sequence definition that $s_n$ is convergent.
The first part I think to use induction
For $n=1$ we have that $s_1=\sqrt{2}<2$
Suppose that is true for $n=k$ then we need to show for $n=k+1$. I wrote a few terms of this recursive sequence
$$s_1=\sqrt{2},\quad s_2=\sqrt{2+\sqrt{2}},\quad s_3=\sqrt{2+s_2}=\sqrt{2+\sqrt{2+\sqrt{2}}}$$
Then I think that $s_{n+1}$ can written as
$$s_{n+1}=\sum_{i=1}^n2^{\frac{1}{2n}}$$
I do not know how to argue that the above term is less than $2$. 
 A: As peterwhy mentions,
$$s_n<2\implies s_{n+1}=\sqrt{2+s_n}<\sqrt{2+2}=2\tag{$\color{green}\checkmark$}$$
To show that it is convergent, you want to show that it is monotone increasing, which combined with the fact that it is bounded above will mean it converges:
$$s_n>s_{n-1}\implies s_{n+1}=\sqrt{2+s_n}>\sqrt{2+s_{n-1}}=s_n\tag{$\color{green}\checkmark$}$$
and thus you are done.
A: By the induction assumption, for some $k\in\mathbb N$,
$$s_k < 2$$
Consider $n = k+1$,
$$s_{k+1} = \sqrt{2+s_k} < \sqrt{2 + 2} = 2$$
Together with the fact that $s_1 = \sqrt 2 < \sqrt 4 = 2$, by induction, all $s_n < 2$.

For $0<x<2$, $x^2 < 2x$ and $2x < 2 + x$.
To prove that the sequence $s_n$ is increasing,
$$\begin{align*}s_{n+1} &= \sqrt{2+s_n}\\
&> \sqrt{s_n + s_n}\\
&= \sqrt{2s_n}\\
&> \sqrt{s_n^2}\\
&= s_n
\end{align*}$$
Since the sequence $s_n$ is increasing and bounded above, limit exists.
A: Hint: $\,s_{n+1}-2=(\sqrt {2+s_n}-2) \cdot \cfrac{\sqrt {2+s_n}+2}{\sqrt {2+s_n}+2} = \cfrac{2+s_n-4}{\sqrt {2+s_n}+2} = \cfrac{s_n-2}{\sqrt {2+s_n}+2}\,$, therefore $s_{n+1}-2$ has the same sign as $s_n -2\,$ and, by induction/telescoping, the same sign as $s_1-2\,$.
A: Using the fact that the square root function is increasing, and the induction hypothesis that $s_k<2$, we have:
$$s_{k+1}=\sqrt{2+s_k}<\sqrt{2+2}=2$$
A: Assuming that the sequence does converge (you prove that later) write the limit as "S".  Letting n go to infinity in $s_{n+1}= \sqrt{2+ s_n}$ we get $S= \sqrt{2+ S}$.  Squaring both sides, $S^2= 2+ S$ so S must satisfy $S^2- S- 2= (S- 2)(S+ 1)= 0$ so S is either 2 or -1.  Since all terms are positive, the limit (again, if there is a limit) must be 2.  
Two show that this is convergent (so that the above is correct) you can show this sequence is increasing and bounded above.  Since we got 2 as the limit above, it makes sense to show that $S_n< 2$ for all n.  Yes, I would do that using induction on n.  When n= 1, $S_n= \sqrt{2}< 2$.  Assume that, for some k, $S_k< 2$.  Then $S_{k+1}= \sqrt{2+ S_k}< \sqrt{2+ 2}= \sqrt{4}= 2$.  
