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Math Help - asking if this squeeze theorem proof is correct

  1. #1
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    Smile asking if this squeeze theorem proof is correct

    Hello,

    I would like to ask if the proof below, concerning the squeeze theorem is correct. I guess someone else must have come up with it but I didn't find anything in inte rnet...

    Gn =< An =< Bn ... where sequences (Gn), (Bn) -> a, then a is a limit for (An) as well.
    [=< less or equal, -> converges to, <=> equivalent]

    Proof
    -Consider sequences (Xn), (Yn) of positive or zero terms! (1)

    Gn=<An <=> Gn=An-Xn <=> |Gn-a|=|An-Xn-a| <=> |An-Xn-a|<e <=> |An-(a+Xn)|<e <=> An->(a+Xn)

    Bn>=An <=> Bn=An+Yn <=> |Bn-a|=|An+Yn-a| <=> |An+Yn-a|<e <=>
    |An -(a-Yn)|<e <=> An->(a-Yn)

    This is:
    {An->(a+Xn) , An->(a-Yn)} => a+Xn=a-Yn <=> Xn=Yn=0 because follows from (1) they are positive or zero.

    Please, if anyone finds that there is a mistake, or it is correct, I would be very glad to know.

    Thank you all, in advance!
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  2. #2
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    Quote Originally Posted by Melsi View Post
    Hello,

    I would like to ask if the proof below, concerning the squeeze theorem is correct. I guess someone else must have come up with it but I didn't find anything in inte rnet...

    Gn =< An =< Bn ... where sequences (Gn), (Bn) -> a, then a is a limit for (An) as well.
    [=< less or equal, -> converges to, <=> equivalent]

    Proof
    -Consider sequences (Xn), (Yn) of positive or zero terms! (1)

    Gn=<An <=> Gn=An-Xn <=> |Gn-a|=|An-Xn-a| <=> |An-Xn-a|<e <=> |An-(a+Xn)|<e <=> An->(a+Xn)


    This makes no sense: in order to obtain the limit of A_n you must make n\to\infty , so
    how

    come n still appears in the expression A_n\xrightarrow [n\to\infty]{} a+X_n ??!

    The same can be said about the following step, too.

    Tonio



    Bn>=An <=> Bn=An+Yn <=> |Bn-a|=|An+Yn-a| <=> |An+Yn-a|<e <=>
    |An -(a-Yn)|<e <=> An->(a-Yn)

    This is:
    {An->(a+Xn) , An->(a-Yn)} => a+Xn=a-Yn <=> Xn=Yn=0 because follows from (1) they are positive or zero.

    Please, if anyone finds that there is a mistake, or it is correct, I would be very glad to know.

    Thank you all, in advance!
    .
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  3. #3
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    Thank you very much for your answer. It is great to have your thought checked, I appreciate it.

    Having in mind not to waste yours or others time I would like to make a final try and chang it a bit:

    Gn=< An <=> Gn=An-Xn <=>|(An-Xn)-a|<e after suitable index <=>(An-Xn)->a

    Bn>= An <=> Bn=An+Yn <=>|(An+Yn)-a|<e after suitable index <=>(An+Yn)->a

    1. are the above two lines correct enough in order to make the consumption that Xn=Yn=0 then An->a?
    2. isn't the random positive e enough to make the index n to run infinitely.

    Thank you very much for your help!
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  4. #4
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    I think that your approach will not lead you to a proof.
    Here is an outline which is very short on specifics.

    Suppose that \epsilon>0 .
    By carefully selecting indices we can get the following.
    |B_n-G_n|\le |B_n-a|+|a-G_n|<\frac{\epsilon }{4}+\frac{\epsilon }{4} =\frac{\epsilon }{2} .

    Now 0<|A_n-G_n|\le |B_n-G_n|<\frac{\epsilon }{2}

    So |A_n-a|\le |A_n-G_n|+|G_n-a|<\epsilon.
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  5. #5
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    Very good solution!

    Well, I think this one is excellent, I have already analyzed it in my notebook and it is very satisfying.

    I am very pleased with the answer and so grateful,
    Thank you very much!
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