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Thread: 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:

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

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

    That is for n > N,

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

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

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

    and so $\displaystyle (a_n)$ 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: $\displaystyle (a_n)\rightarrow\alpha$ Then, by Theorem 2.1 (in my book), $\displaystyle (|a_n|)\rightarrow |\alpha|$.
    choose $\displaystyle \epsilon =1$ , and then use the definition of a limit to conclude there exists N such that $\displaystyle \left| |a_n|-|\alpha|\right |< 1, \ \forall n>N$.
    That is for n > N,
    $\displaystyle |\alpha|-1<|a_n|<|\alpha|+1$.
    (I don't understand what is going with the max part)
    Hence, $\displaystyle \forall n\geq 1$
    $\displaystyle |a_n| \leq \text{max}\{|a_1|,|a_2|,\cdots , |a_N|, |\alpha|+1\}$ and so $\displaystyle (a_n)$ is bounded.
    Actually it is $\displaystyle n\ge N$ then $\displaystyle |a_n|<|\alpha|+1$.
    That is, from $\displaystyle N$ on $\displaystyle |\alpha|+1$ bounds the sequence.
    But we do not know what happens to $\displaystyle |a_j|$ if $\displaystyle 1\le j<N$.
    That is where the max comes in.
    Here is the way I taught it: let $\displaystyle M = \sum\limits_{k = 1}^N {\left| {a_k } \right|} $.
    Now $\displaystyle \left( {\forall n} \right)$ we have $\displaystyle |a_n|<|\alpha|+1+M$
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    Quote Originally Posted by Plato View Post
    Actually it is $\displaystyle n\ge N$ then $\displaystyle |a_n|<|\alpha|+1$.
    That is, from $\displaystyle N$ on $\displaystyle |\alpha|+1$ bounds the sequence.
    But we do not know what happens to $\displaystyle |a_j|$ if $\displaystyle 1\le j<N$.
    That is where the max comes in.
    Here is the way I taught it: let $\displaystyle M = \sum\limits_{k = 1}^N {\left| {a_k } \right|} $.
    Now $\displaystyle \left( {\forall n} \right)$ we have $\displaystyle |a_n|<|\alpha|+1+M$
    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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    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 $\displaystyle 1\le n\le N$ then $\displaystyle |a_n|\le M\le M+|\alpha|+1$

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