Arithmetic-Geometric mean sequences limit 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 $\displaystyle\lim_\limits{{n\to \infty}} b_n=\frac{\sqrt{b^2-a^2}}{\arccos(\frac{a}{b})}$

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 $\theta=\arccos\frac{a}{b}$.
