If $a,\ b$ are positive quantities such that $a<b$ and $a_1 = \frac{a+b}{2},\ b_1 = \sqrt{a_1b},\ a_2 = \frac{a_1+b_1}{2},\ b_2 =\sqrt{a_2b_1},\ ...,\ a_n = \frac{a_{n-1}+b_{n-1}}{2},\ b_n = \sqrt{a_nb_{n-1}},\ ...$ then show that $$\lim_{n\to\infty} b_n = \frac{\sqrt{b^2-a^2}}{\arccos(\frac{a}{b})}$$
I know how to obtain limit in case a sequence is defined recursively. But,here I don't know how to proceed.
A modification of Gauss's proof I've linked in the comments:$$f(x, y) = \frac{\sqrt{y^2-x^2}}{\arccos(\frac{x}{y})},\ x< y $$ Let's try and see if $$f(a_n,\ b_n) = f\left(\frac{a_{n-1}+b_{n-1}}{2},\ \sqrt{\frac{a_{n-1}b_{n-1}+b_{n-1}^2}{2}}\right) = f(a_{n-1}, \ b_{n-1}) $$ really holds. $$\frac{a_n}{b_n} = \frac{\sqrt{a_{n-1}+b_{n-1}}}{\sqrt{2b_{n-1}}} $$ $$\sqrt{b_n^2-a_n^2} = \frac{\sqrt{b_{n-1}^2-a_{n-1}^2}}{2} $$ $$2\arccos\left(\frac{a_n}{b_n}\right) = \arccos\left(2\left(\frac{\sqrt{a_{n-1}+b_{n-1}}}{\sqrt{2b_{n-1}}} \right)^2-1 \right) = \arccos\left(\frac{a_{n-1}}{b_{n-1}}\right) $$ And so the equality really does hold, and we can see that, by denoting $L$ the limit: $$f(a,\ b) = \lim_{x<y,\ (x, \ y)\to (L,\ L)} f(x,\ y) = L $$ I know that this 'proof' (I didn't prove that the limit exists) doesn't give any sort intuition of why $f$ is how it is, but at least it works.