IMC 2001 · Day 2 · P8

Obviously $a_2 = \sqrt{2 - \sqrt{2}} < \sqrt{2}$.

Since the function $f(x) = \sqrt{2 - \sqrt{4 - x^2}}$ is increasing on the interval $[0, 2]$ the inequality $a_1 > a_2$ implies that $a_2 > a_3$. Simple induction ends the proof of monotonicity of $(a_n)$. In the same way we prove that $(b_n)$ decreases (just notice that $g(x) = \dfrac{2x}{2 + \sqrt{4 + x^2}} = 2 \Bigl/ \left( 2/x + \sqrt{1 + 4/x^2} \right)$). It is a matter of simple manipulation to prove that $2 f(x) > x$ for all $x \in (0,2)$, this implies that the sequence $(2^n a_n)$ is strictly increasing. The inequality $2 g(x) < x$ for $x \in (0,2)$ implies that the sequence $(2^n b_n)$ strictly decreases. By an easy induction one can show that $a_n^2 = \dfrac{4 b_n^2}{4 + b_n^2}$ for positive integers $n$. Since the limit of the decreasing sequence $(2^n b_n)$ of positive numbers is finite we have $$\lim 4^n a_n^2 = \lim \frac{4 \cdot 4^n b_n^2}{4 + b_n^2} = \lim 4^n b_n^2.$$ We know already that the limits $\lim 2^n a_n$ and $\lim 2^n b_n$ are equal. The first of the two is positive because the sequence $(2^n a_n)$ is strictly increasing. The existence of a number $C$ follows easily from the equalities $$2^n b_n - 2^n a_n = \left( 4^n b_n^2 - \frac{4^{n+1} b_n^2}{4 + b_n^2} \right) \Bigl/ \left( 2^n b_n + 2^n a_n \right) = \frac{(2^n b_n)^4}{4 + b_n^2} \cdot \frac{1}{4^n} \cdot \frac{1}{2^n (b_n + a_n)}$$ and from the existence of positive limits $\lim 2^n b_n$ and $\lim 2^n a_n$.

Remark. The last problem may be solved in a much simpler way by someone who is able to make use of sine and cosine. It is enough to notice that $$a_n = 2 \sin \frac{\pi}{2^{n+1}} \quad \text{and} \quad b_n = 2 \tan \frac{\pi}{2^{n+1}}.$$