<s>[INST] Simultaneous recurrence relations: $a_n = a_{n-1} (1 - b_n), b_n = b_{n-1} (1-a_n)$ 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 $a_2 \geq 1/2$.
 [/INST] Since$$b_n=1-\frac{a_n}{a_{n-1}}$$
we get
$$1-\frac{a_n}{a_{n-1}}=\left(1-\frac{a_{n-1}}{a_{n-2}}\right)(1-a_n)$$
Multiplying the both sides by $a_{n-1}a_{n-2}$ gives
$$a_{n-1}a_{n-2}-a_na_{n-2}=a_{n-1}a_{n-2}-a_na_{n-1}a_{n-2}-a_{n-1}^2+a_na_{n-1}^2,$$
i.e.
$$a_{n-1}^2(a_n-1)=a_na_{n-2}(a_{n-1}-1)$$
Dividing the both sides by $a_{n}a_{n-1}$ gives
$$\frac{a_{n-1}(a_n-1)}{a_{n}}=\frac{a_{n-2}(a_{n-1}-1)}{a_{n-1}}$$
So, if we define $$c_n=\frac{a_{n-1}(a_n-1)}{a_n}$$
then we have
$$c_n=c_{n-1}=\cdots =c_2=\frac{a_1(a_2-1)}{a_2}=-\frac{b_2}{a_2}:=a$$
So,
$$\frac{a_{n-1}(a_n-1)}{a_n}=a\implies \frac{1}{a_n}+\frac{a}{a_{n-1}}=1\implies d_n-d_{n-1}=\frac{1}{(-a)^n}$$
where
$$d_n=\frac{1}{(-a)^n}\cdot\frac{1}{a_n}$$
So, for $n\ge 2$, we have
$$d_n=d_1+\sum_{k=2}^{n}\frac{1}{(-a)^k}=-\frac{1}{a}+\frac{a}{a+1}\left(\frac{1}{a^2}-\frac{1}{(-a)^{n+1}}\right)$$
which holds for $n=1$.
So, we get
$$a_n=\frac{a+1}{1-(-a)^n},\qquad b_n=\frac{a+1}{1-(-a)^n}(-a)^{n-1},$$
i.e.
$$\color{red}{a_n=\frac{1}{\displaystyle\sum_{k=0}^{n-1}\left(\dfrac{b_2}{a_2}\right)^k},\qquad b_n=\frac{\left(\dfrac{b_2}{a_2}\right)^{n-1}}{\displaystyle\sum_{k=0}^{n-1}\left(\dfrac{b_2}{a_2}\right)^k}}$$
</s>