Define $x_1:=\sqrt{2}$ and recursively $x_n=\sqrt{x_{n-1}+2}$ for every $n\geq 2$.
How can I show $(x_n)_{n\in\mathbb N}$ is bounded? What is the limit $\displaystyle\lim_{n\to \infty} x_n$?
Indeed, I was able to show $(x_n)_{n\in\mathbb N}$ is bounded showing it's a Cauchy sequence. For this I first showed the inequality: $$|x_{n+2}-x_{n+1}|<\frac{1}{2}|x_{n+1}-x_n|.$$ However, that was too much work. Can I show the sequence is bounded directly?
Thanks