TITLE: Simultaneous recurrence relations: $a_n = a_{n-1} (1 - b_n), b_n = b_{n-1} (1-a_n)$
QUESTION [2 upvotes]: I want to solve the following simultaneous recurrence relations:
\begin{align*}
a_n &= a_{n-1} (1-b_n) \\
b_n &= b_{n-1} (1-a_n).
\end{align*}
Initial conditions are $a_1 = b_1 = 1$ and $a_2 + b_2 = 1$.
The symmetric case, $a_2 = b_2 = 1/2$, has a simple solution: $a_n = b_n = 1/n$. 
I have combined the equations through $a_n + b_n$, $a_n b_n$, and $a_n/b_n$ without getting closer to a solution. 
If my algebra hasn't failed me, the two equations imply the following:
$$a_n a_{n-2} + a_n a_{n-1} a_{n-2} = a_{n-1}^2 + a_n a_{n-1}^2.$$
If it's of any help, it seems that $a_n$ approaches $2 - 1/a_2$ when $n$ tends to infinity for $a_2 \geq 1/2$.

REPLY [2 votes]: From the self referential recursions,
$$
a_n=a_{n-1}(1-b_n)\tag1
$$
and
$$
b_n=b_{n-1}(1-a_n)\tag2
$$
Plugging $(2)$ into $(1)$ and $(1)$ into $(2)$ and solving, we get the proper recursions
$$
a_n=\frac{a_{n-1}(1-b_{n-1})}{1-a_{n-1}b_{n-1}}\tag3
$$
$$
b_n=\frac{b_{n-1}(1-a_{n-1})}{1-a_{n-1}b_{n-1}}\tag4
$$
From which it follows that
$$
\frac{1-a_n}{1-b_n}=\frac{1-a_{n-1}}{1-b_{n-1}}\tag5
$$
and therefore,
$$
\frac{1-a_n}{1-b_n}=\frac{1-a_2}{1-b_2}\tag6
$$
Furthermore, $(1)$ and $(2)$ imply that
$$
\frac{a_n}{b_n}=\frac{a_{n-1}}{b_{n-1}}\frac{1-b_{n-1}}{1-a_{n-1}}\tag7
$$
Therefore, $(6)$ and $(7)$ yield
$$
\frac{a_n}{b_n}=\frac{a_2}{b_2}\left(\frac{1-b_2}{1-a_2}\right)^{n-2}\tag8
$$
Solving $(6)$ and $(8)$ simultaneously, we get
$$
a_n=\frac{\frac{1-a_2}{1-b_2}-1}{\frac{b_2}{a_2}\left(\frac{1-a_2}{1-b_2}\right)^{n-1}-1}\tag9
$$
and
$$
b_n=\frac{\frac{1-b_2}{1-a_2}-1}{\frac{a_2}{b_2}\left(\frac{1-b_2}{1-a_2}\right)^{n-1}-1}\tag{10}
$$

If $a_2\gt b_2$, then $(9)$ says
$$
\lim_{n\to\infty}a_n=\frac{a_2-b_2}{1-b_2}\tag{11}
$$
and $(10)$ says
$$
\lim\limits_{n\to\infty}b_n=0\tag{12}
$$

If $a_2\lt b_2$, then $(9)$ says
$$
\lim_{n\to\infty}a_n=0\tag{13}
$$
and $(10)$ says
$$
\lim\limits_{n\to\infty}b_n=\frac{b_2-a_2}{1-a_2}\tag{14}
$$
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