# 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.

• Please anybody, I need some help Apr 8, 2021 at 14:21

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.

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