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.

  • $\begingroup$ Please anybody, I need some help $\endgroup$
    – shangq_tou
    Apr 8, 2021 at 14:21

1 Answer 1


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.

  • 1
    $\begingroup$ I checked with Mathematica, for $a_0+b_0 >1$, we have $$\lim \frac{a_0^3+a_1^3+...+a_{n-1}^3}{a_n} = +\infty$$ $\endgroup$
    – NN2
    Apr 8, 2021 at 16:07
  • 1
    $\begingroup$ Mathematica is dumb. Sure, it often comes in handy, much like a stick or a stone. $\endgroup$ Apr 8, 2021 at 16:08
  • $\begingroup$ Lol :)))) _________ $\endgroup$
    – NN2
    Apr 8, 2021 at 16:10
  • $\begingroup$ Perhaps for large numbers, Mathematicas sucks. In deed, the numerator is bounded by $$(n-2)a_0^3 + a_{n-1}^3< N <(n-2)a_{n-2}^3 + a_{n-1}^3 $$ and $$\frac{(n-2)a_{0}^3}{a_n} \xrightarrow{n \to \infty +\infty} 0$$ $$\frac{(n-2)a_{n-2}^3}{a_n} \xrightarrow{n \to \infty +\infty} 0$$ and then the limit is equal to $$ \frac{a_{n-1}^3}{a_n} \xrightarrow{n \to \infty +\infty} \frac{4}{9}$$ $\endgroup$
    – NN2
    Apr 8, 2021 at 16:49
  • 1
    $\begingroup$ Well, in that case $a_n$ tends to $2\over3$, so each summand is about $8\over27$, and there is a lot of them. $\endgroup$ Apr 8, 2021 at 19:40

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