What's The Limit of The Recursion $a_{n+1} = \sqrt{2+\sqrt{a_n}}$ with $a_0 = \sqrt{2}$?

Question

I'm following a study guide for an exam and the very first question is:

Show that the sequence defined by $a_{n+1} = \sqrt{2+\sqrt{a_n}}$ with $a_0 = \sqrt{2}$ is increasing, bounded and calculate it's limit as $n \to \infty$.

The first parts are easy enough, and the proof for boundedness shows that, since $a_n < 2$ for all $n$, then: $$a_k < 2$$ $$2 + \sqrt{a_k} < 2 + \sqrt{2}$$ $$a_{k+1} = \sqrt{2+\sqrt{a_n}} < \sqrt{2+\sqrt{2}}$$

This appears to be a pretty good bound for the sequence, but the limit should also be a fixed point for the sequence. Using this and performing some polynomial division, we get: $$ P(x) = (x^2 - 2)^2 - x = (x-1)(x^3 + x^2 - 3x - 4)$$

Performing the recursion on the computer ( and checking with Wolfram) we get a limit $L \approx 1.831177207208337$. However, $\sqrt{2+\sqrt{2}} \approx 1.8477590650225735$. How am I supposed to get a closed form solution for this thing during a timed exam? Am I missing something?

Answer 1

I agree with what you have done here - if the limit is $L$, you want to solve

$$L = \sqrt{2+\sqrt{L}}$$

which gives $L \approx 1.8312$ as in your answer.

I suspect there is a mistake in the question, and it meant to say $a_{n+1} = \sqrt{2+a_n}$. Which you can work out has a limit of 2.

The first equation isn't easy to solve at all - you have $L^2 - \sqrt{L} - 2 = 0$, which is possible to solve by taking this as a quartic in $\sqrt{L}$ and applying the quartic formula, but is very unlikely what was intended on the exam! I think more likely it was a typo unfortunately.