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Math Help - Bounded sequences.

  1. #1
    Super Member Showcase_22's Avatar
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    Bounded sequences.

    This is an analysis question I got stuck on:

    Prove that if a sequence is eventually bounded then it is bounded.
    I decided to make a_N my sequence. I need to show that this is bounded if a_{N+n} is bounded.

    So my first step was:

    |a_{n+N}|<\epsilon

    I was not sure where to go from here. Is there a way to separate a_{n+N}? I was also thinking that changing my epsilon to an expression would help but I can't see what to change it to.

    Help would be very much appreciated.
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    You have correctly said that "eventually" means "from some point onwards", in other words for n>N. But you haven't used the definition of "bounded" (which doesn't use epsilon at all). Write down the definition of what it means for the sequence a_{n+N} to be bounded, and then think how that would be affected if you include the finite number of terms a_1,\,a_2,\ldots,\,a_N at the beginning of the sequence.
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    Super Member Showcase_22's Avatar
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    You have correctly said that "eventually" means "from some point onwards", in other words for n>N. But you haven't used the definition of "bounded" (which doesn't use epsilon at all). Write down the definition of what it means for the sequence to be bounded, and then think how that would be affected if you include the finite number of terms at the beginning of the sequence.
    I was trying to use epsilon as my starting point but now I see that it was a red herring!

    A sequence (a_n) of real numbers is called a bounded sequence if |a_n| \leq M for all n and some positive real number M. I understand what this means and i'm fine with it.

    The finite number of terms at the beginning of the sequence are bounded between the maximum of a_n and the minimum of a_n for n \leq N. Therefore these finite number of terms are already bounded and it is only necessary to prove that the infinite number of terms that follow ( |a_n| \leq M \ \forall \ n>N) are bounded.

    This is where I think the question essentially answers itself. Form the above argument, the finite number of terms at the start are bounded and the question states that the sequence is "eventually bounded" so all the infinite number of terms for n>N are also bounded. Hence the entire sequence is bounded.

    Is this right?
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    Quote Originally Posted by Showcase_22 View Post
    A sequence (a_n) of real numbers is called a bounded sequence if |a_n| \leq M for all n and some positive real number M. I understand what this means and i'm fine with it.
    To finish I suggest using B = M + \sum\limits_{n = 1}^{N - 1} {\left| {a_n } \right|} as your bound.
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  5. #5
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Showcase_22 View Post

    The finite number of terms at the beginning of the sequence are bounded between the maximum of a_n and the minimum of a_n for n \leq N. Therefore these finite number of terms are already bounded and it is only necessary to prove that the infinite number of terms that follow ( |a_n| \leq M \ \forall \ n>N) are bounded.

    This is where I think the question essentially answers itself. From the above argument, the finite number of terms at the start are bounded and the question states that the sequence is "eventually bounded" so all the infinite number of terms for n>N are also bounded. Hence the entire sequence is bounded.
    So you're saying this is correct? =O

    To finish I suggest using <br />
B = M + \sum\limits_{n = 1}^{N - 1} {\left| {a_n } \right|}<br />
as your bound.
    Could I also use B=M+ max (a_n) ? It's not that I don't like your answer, i'm just trying to test my understanding my offering alternative solutions.
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    Quote Originally Posted by Showcase_22 View Post
    So you're saying this is correct? =O

    Could I also use B=M+ \max (a_n) ? It's not that I don't like your answer, i'm just trying to test my understanding my offering alternative solutions.
    Yes, the method is correct, and the bound B=M+ \max (a_n) is partially correct. You would need to replace \max (a_n) by \max (|a_n|) , and make it clear that the "max" is meant to be taken over n=1,2,...,N. You could also use the bound \max\{|a_1|,|a_2|,\ldots,|a_N|,M\}.
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  7. #7
    Super Member Showcase_22's Avatar
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    Why is it necessary to include the modulus sign? In my previous post I put the modulus sign in and then changed it because I thought it wasn't needed. =S

    Also, what's the lower bound? Is it just <br />
B=-M- \max |a_n|<br />
?
    Last edited by Showcase_22; October 14th 2008 at 09:29 AM.
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  8. #8
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    Quote Originally Posted by Showcase_22 View Post
    Why is it necessary to include the modulus sign?
    If you don't use the modulus, then you have to provide both an upper bound and a lower bound. Taking the modulus is a device for dealing with both at once, because |a_n|<M means -M<a_n<M.
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