calculate the limit for $\sqrt{1 + \sqrt{2 + \sqrt{1 + \sqrt{2 + ......}}}}$

Question

What I got right now is: $a_0 = \sqrt{1 + \sqrt{2}}$, $a_1 = \sqrt{1 + \sqrt{2 + a_0}}$, and $a_n = \sqrt{1 + \sqrt{2 + a_{n-1}}}$. I can say $a = \sqrt{1+\sqrt{2 + a}}$, right? If yes, how to show that $a = \sqrt{1+\sqrt{2 + a}}$?

Answer 1

The sequence is bounded above by $2$: certainly $a_0<2$ since $\sqrt{2}<3$

Once you know $a_{n-1}<2$, $a_n<\sqrt{1+\sqrt{2+2}}=\sqrt{3}<2$

increasing: $a_1>a_0$ clearly, and use induction again

So bdd above, incerasing -> convergent

Answer 2

Consider for $x\geq 0$ the map $$ f(x)=\sqrt{1+\sqrt{2+x}}.$$ The derivative of the map is $$ 0\leq f'(x)=\frac{1}{4\sqrt{1+\sqrt{2+x}}\; \sqrt{2+x}}\leq \frac{1}{4}$$ By the MVT $|f(x)-f(y)|\leq \frac{1}{4}|x-y|$ is a Lipschitz contraction of ${\Bbb R}_+$ which is a complete metric space. Whence the sequence $a_{n+1}=f(a_n)$, $a_0=0$ converges to a unique fixed point of $f$.