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Math Help - Sequences and Limits

  1. #1
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    Sequences and Limits

    Question:
    Give an example of a divergent sequence (a_n)_{n\in\mathbb{N}} that satisfies the following conditions:

    (i) For every  \epsilon >0, there exists an X such that for infinitely many x>X,

     \left| a_n - 3 \right| < \epsilon


    (ii) There exists an  \epsilon >0 and a  X \in\mathbb{N} such that for all  x>X ,

     \left| a_n - 3 \right| < \epsilon


    ================================================== =======

    I have absolutely no idea how to do these For the infinitely many x>X one, I'm thinking this could involve a trig function since it doesn't converge and repeats every 2\pi . Please help.
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  2. #2
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    Quote Originally Posted by WWTL@WHL View Post
    Question:
    Give an example of a divergent sequence (a_n)_{n\in\mathbb{N}} that satisfies the following conditions:

    (i) For every  \epsilon >0, there exists an X such that for infinitely many x>X,

     \left| a_n - 3 \right| < \epsilon


    (ii) There exists an  \epsilon >0 and a  X \in\mathbb{N} such that for all  x>X ,

     \left| a_n - 3 \right| < \epsilon
    Let a_n = 0,3,0,3,0,3....
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Let a_n = 0,3,0,3,0,3....
    Thanks for the reply.

    But if I had a sequence like that:  a_n = 0 for n is odd, and  a_n = 3 for n is even....

    ...how does this satisfy the criteria of being less than \epsilon? I mean, if a_n = 0, then you have |-3| = 3 < \epsilon and what if epsilon was 0.6, or 2 etc.?

    Sorry if I'm being stupid.
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  4. #4
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    Hey, after spending the best part of an hour jotting stuff down on paper, I've come up with this for part (i)

     a_n = \frac{\epsilon}{2}\cos x + 3

    is that a divergent sequence? I think it is. And it satisfies the conditions, right? I mean if you let X = -1, then are infinitely many x>X.

    no idea about (ii) though Any ideas?
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  5. #5
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    Quote Originally Posted by WWTL@WHL View Post
    But if I had a sequence like that:  a_n = 0 for n is odd, and  a_n = 3 for n is even....
    ...how does this satisfy the criteria of being less than \epsilon? I mean, if a_n = 0, then you have |-3| = 3 < \epsilon and what if epsilon was 0.6, or 2 etc.?
    You seem to have misread your own question.
    There are infinitely many even integers. So if j is even we have:
    \left| {a_j  - 3} \right| = \left| {3 - 3} \right| = 0 < \varepsilon .

    Moreover for the second condition, letting \varepsilon  = 4 fulfills the condition.
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  6. #6
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    Quote Originally Posted by Plato
    Moreover for the second condition, letting \varepsilon  = 4 fulfills the condition.
    Indeed it does, thanks a lot mate!


    Quote Originally Posted by Plato View Post
    You seem to have misread your own question.
    There are infinitely many even integers. So if j is even we have:
    \left| {a_j  - 3} \right| = \left| {3 - 3} \right| = 0 < \varepsilon .
    Yep, I did. Sorry about that.

    But if j is odd we have:

    \left| {a_j  - 3} \right| = \left| {0 - 3} \right| = 3 < \varepsilon - but what if epsilon was 2? That's the question I was trying to ask. There doesn't seem to be a way around it.

    EDIT: Nevermind I understand it now. I've still been misreading my question. I think I get it now. Thanks Plato and ThePerfectHacker
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