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Math Help - Sequence Proof (Hard)

  1. #1
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    Sequence Proof (Hard)

    Hi! I need some help with this question:



    (a)
    Statement reads: For every \epsilon there exists a point in the sequence N \in N after which all terms lie between the lines y= L-\epsilon and y= L+ \epsilon.

    Now if we interchange the quantifiers it reads: There exists N \in N for every \epsilon > 0 such that for every point greater than or equal to N, all terms are between the lines y= L-\epsilon and y= L+ \epsilon.

    (sigh), it's the same thing! What's the difference if we interchange the two quantifiers? And how do we go about proving it? I tried everything, even drew a diagram. But man it’s hard...

    So any help here is appreciated.
    Last edited by Roam; October 5th 2009 at 12:35 AM.
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  2. #2
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    Hello Roam
    Quote Originally Posted by Roam View Post
    Hi! I need some help with this question:



    (a)
    Statement reads: For every \epsilon there exists a point in the sequence N \in N after which all terms lie between the lines y= L-\epsilon and y= L+ \epsilon.

    Now if we interchange the quantifiers it reads: There exists N \in N for every \epsilon > 0 such that for every point greater than or equal to N, all terms are between the lines y= L-\epsilon and y= L+ \epsilon.

    (sigh), it's the same thing! What's the difference if we interchange the two quantifiers? And how do we go about proving it? I tried everything, even drew a diagram. But man itís hard...

    So any help here is appreciated.
    No, it isn't the same thing at all. In the first statement, we choose \epsilon first, and then find the N for which |a_n -L|<\epsilon whenever n\ge N. In other words, the value of N depends upon the value of \epsilon.

    In the second statement, there is a fixed value of N, such that whenever n\ge N, |a_n-L| can be made arbitrarily small.

    Grandad
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    Thanks Grandad, I see the difference now.

    But I still don't know what method to use in order to prove the second statement. Could you give me some clues?
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    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Roam View Post
    Thanks Grandad, I see the difference now.

    But I still don't know what method to use in order to prove the second statement. Could you give me some clues?
    Notice: |a_{n}|=|(a_{n}-a)+a|\leq\\|a_{n}-a|+|a|. If we subtract |a| form both sides, we have that, |a_{n}|-|a|\leq\\|a_{n}-a|. Thus, |a_{n}-a|\geq\\||a_{n}|-|a||.

    Now for the proof...

    Pick and arbitrary \epsilon>0. Since (a_{n}) converges to some limit a, there exists a natrual number N such that, whenever n\geq\\N implies |a_{n}-a|<\epsilon. However, ||a_{n}|-|a||\leq\\|a_{n}-a|<\epsilon also. Therefore, ||a_{n}|-|a||<\epsilon.
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    Quote Originally Posted by Danneedshelp View Post
    Notice: |a_{n}|=|(a_{n}-a)+a|\leq\\|a_{n}-a|+|a|. If we subtract |a| form both sides, we have that, |a_{n}|-|a|\leq\\|a_{n}-a|. Thus, \color{red}|a_{n}-a|\geq\\||a_{n}|-|a||.
    You have skipped a vital step there
    See this proof.
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  6. #6
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Plato View Post
    You have skipped a vital step there
    See this proof.
    Thanks,

    I just skipped to the result needed for the proof, but thanks for pointing that out.
    Last edited by Danneedshelp; October 5th 2009 at 07:58 PM.
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  7. #7
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    Quote Originally Posted by Danneedshelp View Post
    Notice: |a_{n}|=|(a_{n}-a)+a|\leq\\|a_{n}-a|+|a|. If we subtract |a| form both sides, we have that, |a_{n}|-|a|\leq\\|a_{n}-a|. Thus, |a_{n}-a|\geq\\||a_{n}|-|a||.

    Now for the proof...

    Pick and arbitrary \epsilon>0. Since (a_{n}) converges to some limit a, there exists a natrual number N such that, whenever n\geq\\N implies |a_{n}-a|<\epsilon. However, ||a_{n}|-|a||\leq\\|a_{n}-a|<\epsilon also. Therefore, ||a_{n}|-|a||<\epsilon.
    Hi!

    I was thinking of using a proof by contradiction. I'm not sure how to do it but something like:

    Suppose a_n \neq L for any n \geq N. And then somehow try to prove that this assumption violates the condition:

     <br />
\exists N \in \mathbb{N}  \forall \epsilon > 0 \forall n \geq N |a_n -L|< \epsilon<br />

    Is this a good strategy?
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