Prove that sequence $a_1 = \sqrt{2}$, $a_{n+1} = \sqrt{2 + a_n}$ is bounded above by 3 I need a little bit of help (just a hint, please) with an induction proof on this sequence, which I need to prove is bounded above by 3.
$$
a_1 = \sqrt{2}
$$
$$
a_{n+1} = \sqrt{2 + a_n}
$$
My attempt:
$$
a_k < 3
$$
$$
a_k + 2 < 5
$$
$$
\sqrt{a_k + 2} < \sqrt{5}
$$
$$
a_{k+1} < \sqrt{5}
$$
... and I don't know where to go from here.
If I were to find a limit of this sequence, which way would I have to go? Should I try to rewrite the sequence into a formula?

How to find a limit of recursive sequence $$a_{n+1} = \dfrac{a_n^2 + 1}{2}$$? @Anurag A
