real analysis - Show that $sqrt{2+sqrt{2+sqrt{2...}}}$ converges to 2

Consider the sequence defined by
$a_1 = \sqrt{2}$, $a_2 = \sqrt{2 + \sqrt{2}}$, so that in general, $a_n = \sqrt{2 + a_{n - 1}}$ for $n > 1$.
I know 2 is an upper bound of this sequence (I proved this by induction). Is there a way to show that this sequence converges to 2? What I think is that the key step is to prove 2 is the least upper bound of this sequence. But how?

Answer

Let $x = \sqrt {2 + \sqrt {2 + \sqrt {2 + \cdots}}}$. Then, note that $$x^2 = 2 + \sqrt {2 + \sqrt {2 + \cdots}} = 2 + x \implies x^2 - x - 2 = 0.$$ Note that the two solutions to this equation are $x=2$ and $x=-1$, but since this square root cannot be negative, it must be $2$.