convergent sequence

**Julius** · Feb 27th 2010, 06:00 AM

Let $\alpha \in (0, 1]$ be a given real number and let a real sequence $\{a_n\}^{\infty}_{n=1}$ satisfy the inequality
$a_{n+1}\leq \alpha a_n + (1-\alpha)a_{n-1}$ for n>1
Prove that if $\{a_n\}$ is bounded, then it must be convergent.
my idea was to show that the sequence is monotonic, and then it would converge, but that inequality didnt really help me showing it

**chisigma** · Feb 27th 2010, 07:35 AM

Setting $\Delta_{n} = a_{n+1} - a_{n}$ the sequence of the $\Delta_{n}$ is the solution of the difference equation...

$\Delta_{n} = - (1-\alpha)\cdot \Delta_{n-1}$ (1)

... with the 'initial condition' $\Delta_{0} = a_{1} - a_{0}$ . Now if is $|1-\alpha|<1$ the sequence is a geometric sum and it converges to...

$a_{\infty} = a_{0}+ \frac{\Delta_{0}}{2-\alpha}$ (2)

If $|1-\alpha| \ge 1$ the sequence doesn't converge...

Kind regards

$\chi$ $\sigma$

**Laurent** · Feb 27th 2010, 07:52 AM

Originally Posted by chisigma

Setting $\Delta_{n} = a_{n+1} - a_{n}$ the sequence of the $\Delta_{n}$ is the solution of the difference equation...

$\Delta_{n} = - (1-\alpha)\cdot \Delta_{n-1}$ (1)

Not really: note that there is an inequality in the original question.

However, I don't have a correct proof yet...

**chisigma** · Feb 27th 2010, 08:10 AM

Originally Posted by Laurent

Not really: note that there is an inequality in the original question.

However, I don't have a correct proof yet...

The fact is that in the 'original question' there is not anly an inequality, but also the 'variable parameter' $\alpha$ and that can only produce confusion. For this reason, waiting that Julius better explains his problem, I preferred to assume equality and to find the condition of covergence as function of $\alpha$ ...

Regarding the fact that You don't have yet a 'correct proof'... I'm very sorry fot that and do hope You have better luck in the future

! ...

Kind regards

$\chi$ $\sigma$

**Julius** · Feb 27th 2010, 08:32 AM

well, i can only show the original text of the problem, there is nothing more that i know

i hope i didnt miss anything.. also, im not sure if assuming equality is right here

**chisigma** · Feb 27th 2010, 09:03 AM

In my opinion this probelm is presented in a form a little 'confused'... anyway some effort to solve it has to be made...

In effect we can write the 'difference equation' as...

$\Delta_{n} \le - (1-\alpha)\cdot \Delta_{n-1} \rightarrow \Delta_{n} = \beta\cdot \Delta_{n-1}$ (1)

... where $\beta\le 0$ and apply the result previously found. If $-1<\beta\le 0$ the sequence converges, in other cases it doesn't...

Kind regards

$\chi$ $\sigma$

**Laurent** · Feb 27th 2010, 11:59 AM

To ChiSigma: My point in noting that you misread the question was certainly not to offend you; this is a common mistake for all of us and doesn't need justifying. On an unrelated side, I don't understand your last proof at all: what does the arrow mean and what is $\beta$ ?

Here is the proof I finally came up with.

Studying monotonicity was a good idea, even though the sequence $(a_n)_n$ has no reason for being monotonous itself. From the inequality $a_{n+1}\leq \alpha a_n+(1-\alpha)a_{n-1}$ , we get $a_{n+1}\leq \max(a_n,a_{n-1})$ , hence also $\max(a_{n+1},a_n)\leq \max(a_n,a_{n-1})$ , which is to say that the sequence $u_n=\max(a_n,a_{n+1})$ is decreasing. Since $(a_n)_n$ is bounded, so is $(u_n)_n$ , hence it converges: $u_n\to u\in\mathbb{R}$ .

At this point, it seems that we are almost done. However a further argument is needed to conclude. Here is the simplest I could find (there must be others). Let $n\geq 1$ . On one hand, $a_n\leq u_n$ (hence $\limsup_n a_n\leq u$ ), but we need a lower bound for $a_n$ . If $u_n\neq a_n$ , then by definition $a_n\leq a_{n+1}$ , and thus $a_{n+2}\leq\alpha a_{n+1}+(1-\alpha)a_n\leq a_{n+1}$ , so that $u_{n+1}=a_{n+1}$ , and by the same argument $u_{n-1}=a_{n-1}$ . Hence, either $u_n=a_n$ or $u_{n+1}\leq \alpha a_n +(1-\alpha)u_{n-1}$ , i.e. $a_n\geq \frac{1}{\alpha}(u_{n+1}-(1-\alpha)u_n)$ . Since $\frac{1}{\alpha}(u_{n+1}-(1-\alpha)u_n)\to u$ , we have the lower bound we need (in both cases, the lower bound has same limit $u$ as the upper bound). This finishes the proof.

**chisigma** · Feb 27th 2010, 11:59 AM

Finally we have to do an observation regarding : 'prove that if $\{a_{n}\}$ is bounded, then it must be convergent'...

In fact the difference equation...

$\Delta_{n} = \beta \cdot \Delta_{n-1}$ (1)

... for $\beta= -1$ produces a bounded but not converging sequence... what I suggest to Julius at this point is to consult some other textbooks

...

Kind regards

$\chi$ $\sigma$

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