<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] Your base case and inductive step are both overly complicated. 
Base case: We have $a_1 = \sqrt{2}$, which is less than $2$.
Inductive step: Assume we have $a_k < 2$
Now $a_{k+1} = \sqrt{2 + a_k} < \sqrt{2 + \sqrt{2}} < \sqrt{2 + 2} = 2$.
It may also be useful to remark that the sequence is strictly positive, so it is also bounded below.

The error you are making is that your base case proves $a_2$ but it should prove $a_1$ and your inductive step proves $a_{k+2}$ when you should be proving $a_{k+1}$.
Also, I'm a bit confused by your inductive step. There's no need to write the definition of the sequence in the inductive step and I'm confused why you have $a_{k+1} = \sqrt{2}$. Did you mean $a_{k+1} < \sqrt{2}$? Because if so, your inductive step (if you are assuming it for $n = k$) should say $a_k < \sqrt{2}$. Indeed, the statement you are trying to prove is $\{P_k : \text{each } a_k \text{ is bounded by } 2\}$. So in the inductive step you assume $P_n$ which is "$a_n$ is bounded by $2$"
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