# Tag Info

Here is a detailed outline, not entirely made rigorous. Setting $b_n \stackrel{\rm def}{=} \frac{1}{6}-Y_n$, with $b_0 = 1/6$, we have $$b_{n+1} = \frac{4}{3}\left( \sqrt{1+\frac{3}{2} b_n - \frac{9}{2} b_n^2} - 1\right)\tag{1}$$ (I like to set things near zero.) By induction, $b_n \geq 0$ for every $n$, and a simple computation shows that \$b_{n+1} - b_n \...