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Math Help - proove series question..

  1. #1
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    proove series question..

    the question in this link:
    http://img372.imageshack.us/img372/7929/76106583tn6.gif

    this question is obvious
    its common sense

    if a(n) converges to a

    then if we take the function who takes the biggest member
    of course it will pick the closest to "a"

    it take the largest member from a1 to an

    for example:
    a1=6 a2=5 a3=6 a4=1 a5=8 a6=7

    b1=6 b2=6 b3=6 b4=6 b5=8 b6=8

    i dont know how to transform these word into math

    ??

    the definition of convergens states that
    if a sequence is bounded and monotonic then it converges

    An->a
    Bn=max{till An }
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  2. #2
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    anyone?
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  3. #3
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    i could say that if An s monotonic and converges then An+1 also converges to "a"

    so Bn ->a
    but here i cant say that An is monotonic

    ??
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Since no one else is trying this I will give a suggestion. This is not a full solution but merely a possible building block for you to work off of. Because of the neccessary montonicity of a_n why not consider three cases: a_n<a_{n+1}\cdots, a_n>a_{n+1},\cdots, or a_n=a_{n+1}=\cdots

    Case Three needs no explanation.

    If case two is true then obviously \max\left\{a_n,a_{n+1},\cdots\right\}=a_n\to{a}

    And if case one is the case then we would have that \max\left\{a_n,a_{n+1},\cdots{a_{n^2}}\right\}=a_{  n^2}

    So now we just need to use the fact that a_n\to{a} to show that a_{n^2}\to{a}

    By the definition of a_n\to{a} we have that for every \varepsilon>0 there exists a N such that N\leqslant{n} implies d\left(a_n,a\right)<\varepsilon

    So now because of the above we have that d\left(a_{n^2},a\right)<\varepsilon whenever N\leqslant{n^2}\implies{\lceil{\sqrt{N}\rceil\leqs  lant{n}}} so a_{n^2}\to{a}
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  5. #5
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    its seems like a fool proof what else do i need to prove here?

    case1: An<An+1
    case2: An>An+1
    case3: An=An+1

    if case1 is true:
    then MAX will pick the member with the biggest n which is An^2
    and you showed how to prove that if An->0 then An^2->0

    if case 2 is true:
    it will pick An and by definition An goes to 0

    if case3:
    then no matter what it will pick because its the same.
    and all the members are equaled to An
    so they all go to 0.
    so it lookes pretty much complete
    what do i miss??
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    its seems like a fool proof what else do i need to prove here?

    case1: An<An+1
    case2: An>An+1
    case3: An=An+1

    if case1 is true:
    then MAX will pick the member with the biggest n which is An^2
    and you showed how to prove that if An->0 then An^2->0

    if case 2 is true:
    it will pick An and by definition An goes to 0

    if case3:
    then no matter what it will pick because its the same.
    and all the members are equaled to An
    so they all go to 0.
    so it lookes pretty much complete
    what do i miss??
    If my proof is sufficient enough for you then I suppose you just need to rephrase it in your own words.
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