<s>[INST] Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$ I have the following recursive relation (sequence):
\begin{align}
a_1 = \sqrt{2}, \quad a_{n+1} = \sqrt{2 + a_n}
\end{align}
My Try:
I'm a little skeptical of my manipulations near the end but it looks like it works out. 
Base Case:
Let $n=1$ then
\begin{align}
&a_2 = \sqrt{2 + a_1} \\
&a_2 < 2 \\
&\sqrt{2 + \sqrt{2}} < 2
\end{align}
The base case holds. 
Induction hypothesis: 
Let $n=k$
$$a_1 = \sqrt{2} \quad a_{k+1} = \sqrt{2 + a_k} \quad a_{k+1} = \sqrt{2}$$
Induction Step: 
Now we have to prove that $a_{k+2} < 2$. Let $n=k+1$.
\begin{align}
a_{k+2} &= \sqrt{2 + a_{k+1}} \\
\implies a_{k+2} &= \sqrt{2 + \sqrt{2 + a_k}} \\
\end{align}
Now we have to show that $a_{k+2} < 2$. 
\begin{align}
a_{k+2} &< 2\\
\sqrt{2 + \sqrt{2 + a_k}} &< 2\\
2 + \sqrt{2 + a_k} &< 4 \\
\sqrt{2 + a_k} &< 2 \\
\end{align}
Q.E.D 
Are my steps correct? 
Thanks for your time!
 [/INST] $b_i:= a_i^2$ Then $$ b_1=2,\ b_{n+1}=2+\sqrt{b_{n}}$$
Show that $b_i$ is an increasing sequence bounded by $4$ :
(1) $b_1< 4$ If $b_n<4$ then $b_{n+1}=2+\sqrt{b_n}< 2+2=4$.
(2) $b_2=2+\sqrt{2} > b_1$ If $b_n>b_{n-1}$ then $$
b_{n+1}=2+\sqrt{b_n} > 2+\sqrt{b_{n-1}}=b_n $$
</s>