3

Okay, so I have this problem:

Given the sequence $(x_n)_{n\in\mathbb{N}}$ defined by $x_{n+1}=\dfrac{3x_n^2}{(1+x_n)^3-1}$, with $x_1>0$, find $\displaystyle\lim_{n\to\infty}x_n$ and $\displaystyle\lim_{n\to\infty}nx_n$.

I've proved that the first limit equals $0$, by proving the sequence is bounded and monotonic, but I have no idea what to do about the second one. Could you give me a hint?

ryagami
  • 848
  • 3
    Bounded and monotonic gives you that the sequence converges, and to something that is no MORE/LESS than the bound, not that the sequence converges to 0 – Alan Jan 11 '15 at 23:26
  • @Alan Who is good understanding doesn't need many words. The sequence is always positive and the limit satisfies $x(x+3)=0$. – Pp.. Jan 11 '15 at 23:32
  • 2
    @pp.. I would argue that it's important to not put down incorrect words, as it would lead to trying to use similar reasoning in future problems where you don't have the additional information that makes it true – Alan Jan 11 '15 at 23:35
  • 1
    @Alan don't worry, I know how the first one is done. I've also shown that if it converges, it must converge to $0$, so that's not a problem. The problem is I have no idea how to do the second one. – ryagami Jan 11 '15 at 23:37
  • 1
    @ryagami An idea, as in undertermined limit on the reals, is to use L'Hospital. Well, for sequences it is called Stolz's theorem (check the spelling, I never know how to spell his name). So, instead of computing the limit of $nx_n=\frac{n}{1/x_n}$, we compute the limit of $\frac{(n+1)-n}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}$. So, essentially one needs to deal with $\frac{1}{x_{n+1}}-\frac{1}{x_n}$. – Pp.. Jan 11 '15 at 23:37
  • Notice now, that from the given recurrence you can write the recurrence for $1/x_n$ and with it study its increments. – Pp.. Jan 11 '15 at 23:38
  • @Pp.. Okay, I'm so stupid. Stolz does it fine. Thanks! – ryagami Jan 11 '15 at 23:40
  • @Alan "proved A by B" doesn't mean that B is the whole proof. So the statement is not incorrect. As I said "A bon entendeur ..." – Pp.. Jan 11 '15 at 23:44
  • @Pp.. I think that mathematical statements/proofs must be crystal clearly written and explained, otherwise things can go wrong pretty easily. Alan was right in remarking that monotone + bounded sequence does not give, at all, that the limit is zero. That the limit fulfills $;x(x+3)=0;$ is something you had to work out, not something that stems from the OP's question, in particular when he addresses Alan's comment he says that "I proved that if it converges then it converges to zero", and this begs the question: did you prove the seq. is mon. and bounded or not? – Timbuc Jan 11 '15 at 23:55
  • @Timbuc That is what you think. What I think is that way too often those who don't know how to solve a problem like to pick on the irrelevant and obvious. I enjoy calling upon that type of people. – Pp.. Jan 11 '15 at 23:59
  • I can't see how the comment by Alan is irrelevant: it is not only relevant but also true. What you enjoy or not is irrelevant for the mathematical side of this problem. – Timbuc Jan 12 '15 at 00:00
  • See also: http://math.stackexchange.com/a/3220/1102 – Aryabhata Jan 12 '15 at 02:22

1 Answers1

3

So that the question doesn't appear as unanswered.

Following the hint I wrote in the comment we use Stolz's theorem with $\frac{n}{1/x_n}$.

The given recurrence is equivalent to $$x_{n+1}=\frac{3x_n}{x_n^2+3x_n+3}.$$ If we write $y_n:=1/x_n$ then we get $$y_{n+1}=y_n+1+\frac{1}{3y_n}$$ from where $$y_{n+1}-y_n=1+\frac{1}{3y_n}\to1$$ Therefore $\frac{(n+1)-n}{y_{n+1}-y_n}\to1$. By Stolz's theorem we get that $nx_n=\frac{n}{1/x_n}=\frac{n}{y_n}\to 1$.

Pp..
  • 5,973