# Show $a_{n+1}=\sqrt{2+a_n},a_1=\sqrt2$ is monotone increasing and bounded by $2$; what is the limit?

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.

#### Solutions Collecting From Web of "Show $a_{n+1}=\sqrt{2+a_n},a_1=\sqrt2$ is monotone increasing and bounded by $2$; what is the limit?"

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:

1. For boundedness: $\sqrt x$ is monotone increasing: If $x_1 < x_2$, then $\sqrt{x_1} < \sqrt{x_2}$.

2. For monotonicity: Since $a_n \le 2$ for all $n$, we have:

$$2+a_n \ge a_n+a_n = 2a_n \ge a_n\cdot a_n = a_n^2$$