limit of two entirely different sequences Let be two sequences $a_n$ and $b_n$ with $a_0 $ and $b_0$ positive real numbers such that
$$a_{n+1}=3a_nb_n(a_n+b_n)$$
and
$$b_{n+1}=a_n^3+b_n^3$$
Find the limit of
$$\lim \frac{a_0^3+a_1^3+...+a_{n-1}^3}{a_n}$$
I obtained that
$a_n+b_n=(a_0+b_0)^{3^n}$
by induction.
If it helps
I do not know how to obtain the sum $a_0^3+a_1^3+...+a_{n-1}^3$ what should I do. Any idea is welcomed.
 A: The sum $a_n+b_n$ behaves nicely. We want to know which part of it is $a_n$. In other words, let $x={a\over a+b}$; how will that change with each iteration?
$$x_{n+1}={a_{n+1}\over a_{n+1}+b_{n+1}} = {3a_nb_n(a_n+b_n)\over(a_n+b_n)^3}=3x_n(1-x_n)$$
That's the logistic map with the greatest parameter value that still lets it converge to a single limit. In other words, $\lim\limits_{n\to\infty}x_n={2\over3}$.
With that in mind, the rest is simple:

*

*$a_0+b_0<1$: the numerator is bounded from below, the denominator tends to 0, so the answer is $\infty$.

*$a_0+b_0=1$: the numerator grows indefinitely, the denominator is bounded from above, so the answer is still $\infty$.

*$a_0+b_0>1$: the numerator is basically $a_{n-1}^3$ (the rest does not matter), and $a_{n-1}^3\approx\Big({2\over3}(a_{n-1}+b_{n-1})\Big)^3=\left({2\over3}\right)^3(a_n+b_n)$, while the denominator is $a_n\approx{2\over3}(a_n+b_n)$ , so the answer is $\left({2\over3}\right)^2=\mathbf{\color{red}{4\over9}}$.

So it goes.
