Mathematical Reflections 2014, Issue 2 - Problem U297

Problem:
Let $a_0 = 0, a_1 = 2$, and $a_{n+1}=\sqrt{2-\dfrac{a_{n-1}}{a_n}}$ for $n \geq 0$. Find $\displaystyle \lim_{n \to \infty} 2^n a_n$.

Proposed by Titu Andreescu.

Solution:
We will show by induction on $n$ that $a_n=2 \sin \dfrac{\pi}{2^n}$ for all $n \in \mathbb{N}$. If $n=0$, it's obvious. Suppose that $a_n=2 \sin \dfrac{\pi}{2^n}$ for some $n \in \mathbb{N}$. Then, $$a_{n+1}=\sqrt{2-\dfrac{2\sin (\pi/2^{n-1})}{2\sin (\pi/2^n)}}=\sqrt{2-\dfrac{4\sin (\pi/2^n)\cos (\pi/2^n)}{2\sin (\pi/2^n)}}=\sqrt{2-2\cos \dfrac{\pi}{2^n}}=2\sin \dfrac{\pi}{2^{n+1}}.$$
Therefore, $$2^n a_n=2^{n+1}\sin\dfrac{\pi}{2^n}=2^{n+1}\left(\dfrac{\pi}{2^n}+o((\pi/2^n)^2)\right),$$ which implies $$\lim_{n \to \infty} 2^n a_n=2\pi.$$