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?
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$\begingroup$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 $2$.
$\endgroup$ 12 $\begingroup$Set $a_{n+1}=\sqrt{2+a_n}$ and $a_0= \sqrt{2}$. We want to calculate the limit of $a_n$.
First, $0<a_n<2$ implies that $0<a_{n+1} <\sqrt{2+2} =2$ and $0<a_0<2$ then $a_n<2$ for any integer $n\geq 0$.
Second, we have $a_{n+1}^2 -a_n^2= 2+a_n-a_n^2 =(2-a_n)(1+a_n)>0$ then $a_n$ is increasing.
Since $a_n$ is increasing and bounded from above it converges to $a\in[0,2]$ and then $a=\sqrt{2+a}$ and clearly we have $a=2$.
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