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Math Help - Cauchy Sequence proof

  1. #1
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    Cauchy Sequence proof

    I don't understand a part of this proof.

    A Cauchy Sequence is bounded.

    pf) Let {a_n} be a cauchy sequence

    then there is N  \in N such that Absolute value ( a_n - a_m) \leq \epsilon

    Since [tex] abs(  a_n) \leq abs(a_N) +1

    Let M = max {abs (a1), abs(a2) , ... abs(aN), abs(an) +1} <-------------- I don't understand this part.

    abs(an)  \leq M \forall n \in N

    Thank you.
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  2. #2
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    Re: Cauchy Sequence proof

    Because this is a Cauchy sequence, you know that for some N, if n,m> N, then |a_n- a_m|<  1. In particular, if a_n> a_{N+1}, we have |a_n- a_{N+1}|= a_n- a_{N+1}< 1 so that a_n< a_{N+1}+ 1. So a_{N+1}+ 1 is an upper bound for all terms with n> N.

    The part you are asking about, where they look at |a_1|, up to |a_N| is the easy part- those together with the bound a_{N+1}+ 1 form a finite set. And a finite set of numbers always contains a "largest member" which must be an upper bound on the entire sequence.
    Thanks from EliteAndoy
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