Let the sequence $\{a_n\}$ be defined as $a_1 = \sqrt 2$ and $a_{n+1} = \sqrt {2+a_n}$.
Show that $a_n \le$ 2 for all $n$, $a_n$ is monotone increasing, and find the limit of $a_n$.
I’ve been working on this problem all night and every approach I’ve tried has ended in failure. I keep trying to start by saying that $a_n>2$ and showing a contradiction but have yet to conclude my proof.
Hints:
1) Given $a_n$ is monotone increasing and bounded above, what can we say about the convergence of the sequence?
2) Assume the limit is $L$, then it must follow that $L = \sqrt{2 + L}$ (why?). Now you can solve for a positive root.
Complementary to Macavity’s answer, the following hints:
For boundedness: $\sqrt x$ is monotone increasing: If $x_1 < x_2$, then $\sqrt{x_1} < \sqrt{x_2}$.
For monotonicity: Since $a_n \le 2$ for all $n$, we have:
$$2+a_n \ge a_n+a_n = 2a_n \ge a_n\cdot a_n = a_n^2$$