2

So, I'm trying to establish the convergence of the sequence $\{a_n\}$ defined as:

$$a_{n+1} = \frac{1}{p}( (p-1)a_n+\frac{a}{a_n^{p-1}}) \text{ , } n\in \mathbb{N} $$

where $a>0$ and $a_1>0$.

So, before you vote to close this question because it has been probably asked numerous times earlier, I want to say that I've already established the convergence on my own, but I have reached two conclusions that can contradict each other and I can't understand why this happens. Maybe I'm just too tired now, but I've been thinking over this for almost an hour and I have no more energy to solve the issue on my own:

The first result is obtained just by applying AM-GM:

$$a_{n+1} = \frac{1}{p}( (p-1)a_n+\frac{a}{a_n^{(p-1)}}) \geq \sqrt[p]{a_n^{p-1} \cdot \frac{a}{a_n^{p-1}}} = \sqrt[p]{a}$$

So, for $n \in \mathbb{N}$, $a_{n+1} \geq \sqrt[p]{a}$

So, this makes sense to me, but one can easily see that we also have:

$$a_{n+1}-a_{n}= \frac{1}{p}( (p-1)a_n+\frac{a}{a_n^{p-1}}) - \frac{1}{p}( (p-1)a_{n-1}+\frac{a}{a_{n-1}^{p-1}})$$ $$\implies p(a_{n+1}-a_{n})= (p-1)(a_{n}-a_{n-1}) + a(\frac{1}{a_n^{p-1}} + \frac{1}{a_{n-1}^{p-1}})$$

So, this is where my confusion starts:

Let's prove by induction that if $a_1 < \sqrt[p]{a}$ then $\{a_n\}$ is increasing and if $a_1 > \sqrt[p]{a}$ then $\{a_n\}$ is decreasing. If $a_1 = \sqrt[p]{a}$ then $\{a_n\}=\{a_1\}$ becomes a constant sequence.

So, to establish the base of induction, if we manipulate the equations we get:

$$a_2 - a_1 = \frac{a-a_1^p}{pa_1^{p-1}}$$

This proves the base of induction in all the three cases. The rest follows immediately from $\displaystyle p(a_{n+1}-a_{n})= (p-1)(a_{n}-a_{n-1}) + a(\frac{1}{a_n^{p-1}} + \frac{1}{a_{n-1}^{p-1}})$

Now, here is the contradiction:

If $a_1 < \sqrt[p]{a}$ then $\{a_n\}$ must be an increasing sequence as I just proved by induction, while, the AM-GM inequality says that all the terms of the sequence beyond $a_2$ must be greater than or equal to $\sqrt[p]{a}$! This implies that the sequence must diverge to $+\infty$ I guess. Right?

But I'm starting to think, by doing numerical examples, that in a real situation, if we start with $a_1 < \sqrt[p]{a}$, the sequence first jumps to some number above $\sqrt[p]{a}$ and then it starts to decrease until it converges to $\sqrt[p]{a}$ at the end.

Can someone tell me what's going on?

EDIT

OK, I had a sign error as gammatester pointed out. I modified my proof and now I think I have solved the problem correctly:

$$a_{n+1}-a_{n} = \frac{a-a_{n+1}^p}{pa_{n+1}^{p-1}}$$

Since $a_{n+1} \geq \sqrt[p]{a}$ we conclude $a_{n+1} \leq a_{n}$. So, the sequence is decreasing and bounded below, hence, it's convergent.

user66733
  • 7,379
  • IMO there is no contradiction: You have shown that $a_2 \ge a_1$ but you have not shown that $a_2^p \le a$ – gammatester Feb 28 '14 at 10:42
  • @gammatester: Yes, so, do you think that the sequence diverges to $\infty$? Because that's the only thing which makes sense to me, but isn't it true that the sequence converges to $\sqrt[p]{a}$ even if $a_1<\sqrt[p]{a}$? I remember we had something like that in numerical analysis or I'm wrong? – user66733 Feb 28 '14 at 10:51
  • No, it simply is not true that your sequence is strictly increasing. Once you have $a_n^p > a$ then you get $a_{n+1} < a_n$. – gammatester Feb 28 '14 at 10:57
  • @gammatester: I'm talking about the case where $a_1 < \sqrt[p]{a}$. Check the induction part again. – user66733 Feb 28 '14 at 11:01
  • Yes, I know. But as already said, you have not shown that $a_2^p < a$ in general (you even noticed that in your numerical experiments!). – gammatester Feb 28 '14 at 11:08
  • I just noticed a sign error and maybe this is a source of your confusion: In the second line of your formula $$a_{n+1}-a_{n}= \frac{1}{p}( (p-1)a_n+\frac{a}{a_n^{p-1}}) - \frac{1}{p}( (p-1)a_{n-1}+\frac{a}{a_{n-1}^{p-1}})$$ $$\implies p(a_{n+1}-a_{n})= (p-1)(a_{n}-a_{n-1}) + a(\frac{1}{a_n^{p-1}} + \frac{1}{a_{n-1}^{p-1}})$$ the last term shoud be $$a\left(\frac{1}{a_n^{p-1}} - \frac{1}{a_{n-1}^{p-1}}\right)$$ – gammatester Feb 28 '14 at 11:55
  • @gammatester: That solves the issue, but now I have to look for a proof that shows the sequence converges. Thanks anyway. – user66733 Feb 28 '14 at 12:11
  • The fallacy is NOT in the sign eror. – Did Feb 28 '14 at 14:25

1 Answers1

1

What you showed is exactly this:

If $a_n\lt\sqrt[p]{a}$ then $a_{n+1}\gt a_n$.

To use this at rank $n+1$ to deduce that $a_{n+2}\gt a_{n+1}$, one would need to know that $a_{n+1}\lt\sqrt[p]{a}$, which is not so. Actually, $(p-1)x+a/x^{p-1}\geqslant p\sqrt[p]{a}$ for every $x\gt0$ hence $a_n\geqslant\sqrt[p]{a}$ for every $n\geqslant2$. So, no contradiction of ZF here...

Did
  • 279,727
  • Yeah... so, the sequence diverges to $+\infty$? Is that right? – user66733 Feb 28 '14 at 10:54
  • Not at all. You really ought to draw the graph of the function $x\mapsto((p-1)x+a/x^{p-1})/p$, you know... Then everything becomes clear. – Did Feb 28 '14 at 10:58
  • So, there's something wrong with my induction? Please check the induction part to see what I mean. I'm sure there must be something wrong with my induction but I can't see why... I've proved the base case for $n=1$, then I've assumed that it is true for each $a_{n}-a_{n-1}$ and I've proved that the same holds for $a_{n+1}-a_{n}$. – user66733 Feb 28 '14 at 11:06
  • I explained exactly what is wrong in your induction (did you read my answer?). The induction step assumes that $a_n\leqslant\sqrt[p]{a}$ and deduce from this that $a_{n+1}\geqslant a_n$ (and this part is correct). So, if $a_1\leqslant\sqrt[p]{a}$ then $a_{2}\geqslant a_1$. But, to show that $a_3\geqslant a_2$, you need to apply a second time the induction step, hence you need that $a_2\leqslant\sqrt[p]{a}$--and this ain't so. – Did Feb 28 '14 at 11:10
  • What you're saying makes sense. Except one thing, isn't this how inductions are done? I mean isn't it enough to show that the statement holds for $n=1$ to make sure about the base case? Why do I need to show that it also holds for $n=2$? I mean if so, then induction is useless because before we can prove something by induction we must make sure that it makes sense for every $n < k$.. I guess I have a misunderstanding about math induction. Is the mistake I'm committing now the same kind of fallacy that Polya describes when he proves all horses are the same color or all girls are blonde? – user66733 Feb 28 '14 at 11:30
  • The induction is flawed. In an induction, one shows that $H_n\implies H_{n+1}$ for every $n\geqslant1$, with $H_n=$ some statement depending on $n$. What would be $H_n$ in this case? – Did Feb 28 '14 at 14:24
  • Well, for the case where $a_1 < \sqrt[p]{a}$ set $H={n \in \mathbb{N}: a_{n+1}-a_{n}>0 }$. I proved that $1 \in H$, hence $H \neq \emptyset$. I've also shown that if $n \in H$ then $n+1 \in H$. So, $H=\mathbb{N}$ by the use of the well-ordering principle. To answer your question, $H_n$ would be defined as the statement $a_{n+1}-a_n>0$. No? I don't need induction to solve this problem though, my modified proof doesn't use induction, but it would be good if I could understand why my induction was wrong. – user66733 Feb 28 '14 at 15:42
  • With this definition of $H$, you never proved that if $n\in H$ then $n+1\in H$. The hypothesis you used to show that $a_{n+1}-a_n\gt0$ (i.e. $n+1\in H$) is not $a_n-a_{n-1}\gt0$ (i.e. $n\in H$) but the completely unrelated condition $a_n\leqslant\sqrt[p]{a}$. – Did Feb 28 '14 at 15:57
  • I'd used this equation: $\displaystyle p(a_{n+1}-a_{n})= (p-1)(a_{n}-a_{n-1}) + a(\frac{1}{a_n^{p-1}} + \frac{1}{a_{n-1}^{p-1}})$. If $a_n-a_{n-1}>0$ (assuming $H_n$ is true) then since $a(\frac{1}{a_n^{p-1}} + \frac{1}{a_{n-1}^{p-1}})$ is always positive it was implied that $a_{n+1}-a_{n}>0$. I think you misunderstood the induction part in my post. Then gammamaster realized that I had a sign error in my calculations... – user66733 Feb 28 '14 at 16:05
  • I am aware that you had a sign error but, as I already mentioned in a comment, this is not the reason why the induction is wrong. The reason is the one I described in a comment. Sorry but, once again, do you read and meditate and ponder and mull over answers and comments as much as you should? – Did Feb 28 '14 at 16:13
  • Yes, I do read answers and comments attentively. In your last comment you said that "you never proved that if $n \in H$ then $n+1 \in H$. And you also said that I hadn't used $a_n-a_{n-1}>0$ but instead I had used the completely unrelated condition $a_n \leq \sqrt[p]{a}$. In my defence, I showed you what equation I had used for the inductive step to establish $n \in H \implies n+1 \in H$. Since $a>0$ and $a_n>0$ if you use the wrong equation I mentioned, you'll see that it implies $a_{n+1}-a_{n}>0$ (i.e. $H_{n+1}$ is true). Doesn't it? That's where I started to get confused. – user66733 Feb 28 '14 at 16:20
  • I see; But then, how were you planning to treat the case $a_1\gt\sqrt[p]{a}$? – Did Feb 28 '14 at 16:58
  • I can't treat it with that equation, but to be honest, I hadn't checked that one on paper. I had assumed from intuition that the sequence would decrease if $a_1 > \sqrt[p]{a}$. But now, if you check my solution after the edit, you'll see that the new solution makes sense and doesn't require mathematical induction. Anyway, thanks for your help Did, I appreciate it. – user66733 Feb 28 '14 at 22:07