This is a Quickie A1001 from Mathematics Magazine, v 83, n 3, June 2010.

Find the limit

\lim_{n\rightarrow\infty}\frac{2^n}{a_n}

where a_1 = 0 and a_{n+1}=\sqrt{{a_n}^2+1}.

Solution

Note that 1/a_2 = 1 = \tan(\pi/2^2). By induction, if 1/a_n = \tan(\pi/2^n), then, for positive angles less than \pi/2, the Tangent Half-Angle Formula gives

\begin{align} \tan(\pi/2^{n+1}) &= \frac{-1+\sqrt{1+\tan^2(\pi/2^n)}}{\tan(\pi/2^n)} = \frac{-1+\sqrt{1+a_{n}^{-2}}}{a_{n}^{-1}} \\&=-a_{n}+\sqrt{a_{n}^{2}+1}=\frac{1}{\sqrt{a_{n}^2+1}+a_n} \\&= \frac{1}{a_{n+1}}\end{align}

Therefore,

\lim_{n\rightarrow\infty}\frac{2^n}{a_n} = \lim_{n\rightarrow\infty}2^{n}\tan(\frac{\pi}{2^n}) = \pi\lim_{n\rightarrow\infty}(\frac{2^n}{\pi}\tan(\frac{\pi}{2^n}))=\pi.