<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] I think you're making a bit of confusion. On one hand you are never proving that the sequence is increasing, on the other hand your argument is a bit too complicated. Here is how I would go about proving it:


*

*$\mathbf{n = 1}$: First note that $a_1 < a_2$ if and only if
$$
a_1^2 = 2 < 2 + \sqrt{2} = a_2^2
$$
which is clearly true. Further, $a_2 < 2$ because $2 + \sqrt{2} < 4$.

*Assume that $a_{n-1} < a_n < 2$. Then we need to prove that $a_n < a_{n+1} < 2$. For the first part note that $a_n < a_{n+1}$ is equivalent to $a_n^2 < 2 + a_n$, i.e.
$$
a_n (a_n - 1) < 2
$$
which is true because $a_n < 2$ implies $a_n - 1 < 1$. For the second part, note that $a_{n+1} < 2$ is equivalent to
$$
2 + a_n < 4
$$
which is true because $a_n < 2$ by hypothesis.
</s>