How do you prove that the following sequence is well defined and convergent How do you prove that the following sequence is well defined and convergent?

$a_{n+1}=\sqrt{2+\sqrt{a_n}}$ with $a_1=0$.
Hints:

$$a_1=0\;,\;\;a_2=\sqrt{2+0}=\sqrt2\;,\;\;a_3=\sqrt{2+\sqrt2}\;,\ldots$$

Using induction :

$$\forall\,n\;,\;\;a_n\ge 0 :\;\;a_n:=\sqrt{2+\sqrt{a_{n-1}}}\stackrel{\text{ind. Hyp.}}\ge\sqrt{2+0}=\sqrt2\ge0$$

$$\forall\,n\;,\;\;a_{n+1}\ge a_n:\;\;a_{n+1}:=\sqrt{2+\sqrt{a_n}}\stackrel{\text{Ind. Hyp.}}\ge\sqrt{2+\sqrt{a_{n-1}}}=:a_n$$

$$\forall\,n\;,\;\;a_n\le 2:\;\;a_n:=\sqrt{2+\sqrt{a_{n-1}}}\stackrel{\text{Ind. Hyp.}}\le\sqrt{2+\sqrt2}\le\sqrt{2+2}=2$$

Conclude the sequence converges .
You can also solve a simple quadratic equation.  It should be clear that the sequence converges to solution the positive solution to:
$$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}},$$
giving
$$x=\sqrt{2+x},$$
or
$$x^2=2+x,$$
whose positive solution is $x=2.$