limits - 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?

Answer

Once you have $a_{k+1} < \sqrt{5}$, you can use that $\sqrt{5} < 3$ to prove that $a_{k+1} < 3$. Hence by induction all terms of this sequence are bounded by $3$.

Now for the limit part, your sequence is bounded above, if you can show that it is an increasing sequence then it follows (see a theorem about monotone convergence) that the sequence should have a limit. Once that is established you can assume that $\lim_{n \to \infty}a_n=a$. Now you have
$$\lim_{n \to \infty}a_{n+1}=\sqrt{2+\lim_{n \to \infty}a_{n}}.$$
Solve for $a$.