Convergence of two recursive sequence how could I find out whether those recursive sequences are convergent?
Let $a_{n+1} = a_1(1-a_n-b_n) + a_n \\ b_{n+1} = b_1(1-a_n-b_n) + b_n$, where
$a_1 = a \\ b_1 = b$ ($a, b \in (0, 1)$)

$$
a_{n+1}=a(1-a_n-b_n)+a_n\text{ ...(1)}\\
b_{n+1}=b(1-a_n-b_n)+b_n\text{ ...(2)}\\$$
$(1)+\alpha\times(2):$
$$a_{n+1}+\alpha b_{n+1}=(1-a-\alpha b)a_n+(-a+\alpha(1-b))b_n+a+\alpha b$$
$$\frac{-a+\alpha(1-b)}{1-a-\alpha b}=\alpha$$
$$b\alpha^2+(a-b)\alpha-a=0$$
$$(\alpha-1)(b\alpha+a)=0$$
i) $(1)+(2):$
$$a_{n+1}+b_{n+1}=(1-a-b)(a_n+b_n)+(a+b)$$
$$a_{n+1}+b_{n+1}-1=(1-a-b)(a_n+b_n-1)$$
$$\therefore a_n+b_n=(1-a-b)^{n-1}(a+b)\text{ ...(3)}$$
ii) $b\times(1)-a\times(2):$
$$b\cdot a_{n+1}-a\cdot b_{n+1}=b\cdot a_n-a\cdot b_n$$
$$b\cdot a_n-a\cdot b_n=...=b\cdot a_1-a\cdot b_1=ba-ab=0$$
$$\therefore b_n=\frac{b}{a}a_n\text{ ...(4)}$$
$(3)$ & $(4)$:
$$a_n=a\cdot (1-a-b)^{n-1}$$
$$b_n=b\cdot (1-a-b)^{n-1}$$
Therefore the sequence converges if
$$\mid 1-a-b\mid < 1$$
