- Define the sequence $(x_n)$ recursively by setting
$$ \begin{align} x_1 &= \sqrt{2} \\ x_{n+1} &= \sqrt{2 + x_n} \text{ for all $n \in 1,2,3,\ldots$} \end{align} $$ (a) Show that the sequence $(x_n)$ converges.
(b) Let $\lambda = \lim_{n \to \infty} x_n$. Show $\lambda^2 - \lambda - 2 = 0$.
Hi, I was wondering if the proof to this question for part(a) is correct. We can prove this by the monotone convergence theorem. We can show that the sequence is monotone increasing and bounded above. We know that $x_1 = \sqrt{2}$ and $x_2 = \sqrt{2+\sqrt{2}}$. Hence, $x_1 \leq x_2 \leq 2$, which will serve as our base case for our induction step. Suppose $x_n \leq x_{n+1} \leq 2$. $x_n \leq x_{n+1} \leq 2 \iff x_n + 2 \leq x_{n+1} + 2 \leq 4 \iff \sqrt{x_n + 2} \leq \sqrt{x_{n+1} + 2} \leq 2$. Therefore, $x_n \leq x_{n+1} \leq 2 \implies x_{n+1} \leq x_{n+2} \leq 2$. By induction, we know that the sequence is increasing. We also have shown that it's bounded. Therefore, by the monotone convergence theorem, $\lim x_n$ converges.
Also, I am pretty sure the sequence converges to 2, but I don't know how to prove that. It would make part (b) so much easier. I was thinking of doing something with $\sup \{x_1, x_2, x_3, ...\}$
For Part $(b)$, note that since $x_n$ converges, say to $\lambda$, then
$$\begin{align} \lim_{n\to \infty}x_{n+1}&=\lambda\\\\ &=\lim_{n\to\infty}\sqrt{x_n+2}\\\\ &=\sqrt{\lambda+2} \end{align}$$
Hence $\lambda =\sqrt{\lambda+2}$ and you can finish now.