calculus - How to prove that a recursive sequence converges?

For this question, I'm not sure how the fact that $|a_n|<6$ effects the result of this question. I'm not really sure how to complete the proof. Here is what I have so far. Can anyone please help me out?

Consider the sequence defined by $a_{n+1} =

\sqrt{2+a_n}$ if $n ≥ 1$ and $a_1 = 1$.
Suppose that $|a_n| < 6$ for all n ∈ N. Does $\{a_n\}$ converge or diverge? Make sure to
fully prove your claim and cite appropriate theorem(s) and hypothesis conditions.

$a_2 = \sqrt{2+1} = \sqrt{3}$

$a_3 = \sqrt{2+\sqrt{3}}$

Wts there exists an M st $a_n \le M$

Choose M = 2 since the sequence eventually apporaches 2.

Base Case: $a_1 = 1 < 2$

Induction Hypothesis: Let k ∈ N be arbitrary.

Assume $a_k \le 2$

Induction Step: $a_{k+1} \to a_k$

$a_{k+1} = \sqrt{2+a_k} < \sqrt{2+2} = 2$

Therefore by induction, the sequence is bounded above

$a_n^2 - a_{n+1}^2 = a_n^2 - \sqrt{{2+a_n}}^2 = a_n^2 - a_n - 2 = (a_n-2)(a_n+1)$

$a_n^2 - a_{n+1}^2 <0$

$a_n^2 < a_{n+1}^2$

$a_n < a_{n+1}$

Therefore by the difference test, $\{a_n\}$ is strictly increasing.

Therefore by the bounded monotone convergence theorem,$\{a_n\}$ converges.