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Math Help - Explanation needed on a sequence proof

  1. #1
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    Explanation needed on a sequence proof

    The Theorem is every convergent sequence is bounded:

    (a_n)\rightarrow\alpha Then, by Theorem 2.1 (in my book), (|a_n|)\rightarrow |\alpha|.

    Quite arbitrary, let us choose \epsilon =1 (so I could pick any number not just 1?), and then use the definition of a limit to conclude there exists N such that \left| |a_n|-|\alpha|\right |< 1, \ \forall n>N.

    That is for n > N,

    |\alpha|-1<|a_n|<|\alpha|+1.

    (I don't understand what is going with the max part)
    Hence, \forall n\geq 1

    |a_n| \leq \text{max}\{|a_1|,|a_2|,\cdots , |a_N|, |\alpha|+1\}

    and so (a_n) is bounded.
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    The Theorem is every convergent sequence is bounded: (a_n)\rightarrow\alpha Then, by Theorem 2.1 (in my book), (|a_n|)\rightarrow |\alpha|.
    choose \epsilon =1 , and then use the definition of a limit to conclude there exists N such that \left| |a_n|-|\alpha|\right |< 1, \ \forall n>N.
    That is for n > N,
    |\alpha|-1<|a_n|<|\alpha|+1.
    (I don't understand what is going with the max part)
    Hence, \forall n\geq 1
    |a_n| \leq \text{max}\{|a_1|,|a_2|,\cdots , |a_N|, |\alpha|+1\} and so (a_n) is bounded.
    Actually it is n\ge N then |a_n|<|\alpha|+1.
    That is, from N on |\alpha|+1 bounds the sequence.
    But we do not know what happens to |a_j| if 1\le j<N.
    That is where the max comes in.
    Here is the way I taught it: let M = \sum\limits_{k = 1}^N {\left| {a_k } \right|} .
    Now \left( {\forall n} \right) we have |a_n|<|\alpha|+1+M
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  3. #3
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    Quote Originally Posted by Plato View Post
    Actually it is n\ge N then |a_n|<|\alpha|+1.
    That is, from N on |\alpha|+1 bounds the sequence.
    But we do not know what happens to |a_j| if 1\le j<N.
    That is where the max comes in.
    Here is the way I taught it: let M = \sum\limits_{k = 1}^N {\left| {a_k } \right|} .
    Now \left( {\forall n} \right) we have |a_n|<|\alpha|+1+M
    Why can you just add M at the end? It is understandable abs(a) + 1 + M is greater than a_n since a_n \leq M so anything added to M would be greater than a_n.
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  4. #4
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    Quote Originally Posted by dwsmith View Post
    Why can you just add M at the end? It is understandable abs(a) + 1 + M is greater than a_n since a_n \leq M so anything added to M would be greater than a_n.
    That is a simple to answer.
    If 1\le n\le N then |a_n|\le M\le M+|\alpha|+1

    If n\ge N then |a_n|\le |\alpha|+1\le M+|\alpha|+1
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