Let $a,b \in [0,1]$. Set $a_1 = a$, $b_1 = b$, and define inductively $a_{n+1} = a_n(1-b_n)$ and $b_{n+1} = b_n(1-a_n)$. Then
$$\sum_n^\infty a_nb_n = \text{min}(a,b)$$
We have proven this result for $\omega$-complete effect monoids (of which $[0,1]$ is an example) in this paper, but I was wondering if it was already a known result for $a,b\in[0,1]$.
Note that the special case $a=b$ gives $a_n=b_n$ and hence $a_{n+1} = a_n(1-a_n)$ and $\sum_n^\infty a_n^2 = a$. In particular, for $a=\frac12$ the first few terms are $a_1=\frac12$, $a_2=\frac14$, $a_3 = \frac{3}{16}$, $a_4=\frac{39}{256}$. Does this ring a bell for anyone?
