7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Please anybody, I need some help $\endgroup$
    – shangq_tou
    Apr 8, 2021 at 14:21

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
6
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.