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Math Help - convergent sequence

  1. #1
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    convergent sequence

    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
    Last edited by Julius; February 27th 2010 at 08:06 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    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
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  3. #3
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    Quote Originally Posted by chisigma View Post
    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...
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Laurent View Post
    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
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  5. #5
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    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
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  6. #6
    MHF Contributor chisigma's Avatar
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    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
    Last edited by chisigma; February 27th 2010 at 01:04 PM. Reason: purely orthografic modifications
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  7. #7
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    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.
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  8. #8
    MHF Contributor chisigma's Avatar
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    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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