Double recurrence of first order The sequences $(a_n)_{n\geq 0},(b_n)_{n\geq 0}$ are given by $a_0>0,b_0>0$ and
$$a_{n+1}=3a_nb_n(a_n+b_n),~~~b_{n+1}=a_n^3+b_n^3,~~~ \forall n\geq 0.$$
The problem is to prove that the following limits exist and to compute them:
$$\lim_{n\to\infty}\frac{a_n}{a_n+b_n};~~~\lim_{n\to\infty}\frac{a_0^3+a_1^3+\dots +a_{n-1}^3}{a_n}.$$
I have noticed that $a_{n+1}+b_{n+1}=(a_n+b_n)^3$ and thus
$$a_n+b_n=(a_0+b_0)^{3^n},~~~\forall n\geq 0.$$
I found no way to decide the monotony or boundedness of the sequences and to compute the limits.
 A: Here's how to do the first one. Note that
$$ \frac{b_{n+1}}{a_{n+1}} = \frac{a_n^2-a_nb_n +b_n^2}{3a_n b_n} = \frac{1}{3}\bigg(\frac{a_n}{b_n} - 1 + \frac{b_n}{a_n}\bigg).$$
The obvious substitution $w_n = \frac{a_n}{b_n}$ now yields
$$ \frac{1}{w_{n+1}} = \frac{1}{3}\bigg(w_n- 1 + \frac{1}{w_n}\bigg).$$
The limit $w = \lim_{n\to\infty}w_n$, if it exists, should then satisfy
$$ 2w^{-1} = w-1; \qquad w = 2. $$
Maybe you can prove by induction that $w_n$ is monotonous and bounded. This allows you to compute the first limit.
For the second one: using the above and the relation for $a_{n+1}$, you can see that $$\lim_{n\to\infty} \frac{a_{n-1}^3}{a_{n}} = 4/9. $$
But then the second-to-last term in the sum is $$\lim_{n\to\infty} \frac{a_{n-2}^3}{a_{n}} = \lim_{n\to\infty} \frac{a_{n-2}^3}{a_{n-1}^3}\frac{a_{n-1}^3}{a_{n}}$$
Using $a_n(1+\frac{b_n}{a_n}) = (a_0 + b_0)^{3^n}$ we get
$$\lim_{n\to\infty}\frac{a_{n-2}^3}{a_{n-1}^3} = \lim_{n \to \infty} (a_0 + b_0)^{(3^{n-1}-3^{n})} =0.$$
Similarly, all other terms in the sum also approach zero. So the second limit would be $4/9$.
