2
$\begingroup$

This question already has an answer here:

If $a,b$ are positive quantities such that $(a<b)$ and if

\begin{align} a_1 &= \frac{a+b}2 & b_1 & = \sqrt{a_1 b} \\ a_2 &= \frac{a_1+b_1}2 & b_2 &= \sqrt{a_2 b_1} \\ &\phantom{36pt} \vdots & & \\ a_n &= \frac{a_{n-1}+b_{n-1}}2 & b_n & = \sqrt{a_n b_{n-1}} \\ &\phantom{36pt} \vdots & & \end{align} then show that $\displaystyle\lim_\limits{{n\to \infty}} b_n=\frac{\sqrt{b^2-a^2}}{\arccos(\frac{a}{b})}$

$\endgroup$

marked as duplicate by Arnaud D., Lee David Chung Lin, Song, YiFan, Lord Shark the Unknown Mar 15 at 5:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2
$\begingroup$

If we assume that $b=1$ and $a=\cos\theta$ with $\theta\in\left(0,\frac{\pi}{2}\right)$ we get:

$$ a_1 = \frac{1+\cos\theta}{2} = \cos^2\frac{\theta}{2},\qquad b_1=\cos\frac{\theta}{2}$$

$$ a_2 = \cos\frac{\theta}{2}\left(\frac{1+\cos\frac{\theta}{2}}{2}\right),\qquad b_2=\cos\frac{\theta}{2}\cos\frac{\theta}{4}$$ so by induction it follows that $$ b_n = \prod_{k=1}^{n}\cos\frac{\theta}{2^k} = \frac{\sin\theta}{2^n\sin\frac{\theta}{2^n}} $$ and $\lim_{n\to +\infty}b_n = \frac{\sin\theta}{\theta}. $ The claim easily follows by rescaling, since $\theta=\arccos\frac{a}{b}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.