KAIST POW 2012-19 : A limit of a sequence involving a square root

A limit of a sequence involving a square root

 Let $a_0 = 3$ and $a_n = a_{n-1} + \sqrt{{a_{n-1}}^2 + 3}$ for all $n \geq 1$. Determine $\lim_{n\to\infty} \frac{a_n}{2^n}$.

 

 Proof. Note that $\cot \theta = \cot 2\theta + \sqrt{\cot^2 2\theta + 1}$ for $0 < \theta < \pi / 2$. Since $a_n $ is determined by $a_0 = 3 = \sqrt{3} \cot {\pi / 6}$, we get $$a_n = \sqrt{3} \cot (\frac{\pi}{6\times 2^n})$$ Therefore, $$\lim_{n\to\infty} \frac{a_n}{2^n} = \lim_{n\to\infty} \frac{6\sqrt{3}}{\pi} \frac{\pi / (6\times 2^n)}{\tan(\pi / (6\times 2^n))} = \frac{6\sqrt{3}}{\pi}$$