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Math Help - Cauchy Sequence

  1. #1
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    Cauchy Sequence

    I need to prove that every Cauchy sequence is bounded. Here is my work thus far:

    {a(sub n)}n=1 to infinity is a Cauchy sequence. A sequence is said to be Cauchy if for all E>0 there exists N in the natural numbers such that if m,n>=N, then |a(sub n)-a (sub m)|<E.
    Proof: Let E=1. Then there exists N in the natural numbers such that m,n>=N implies that |a(sub n)-a(sub m)|<1. By the reverse triangle inequality, ||a(sub n)|-|a(sub m)||<=|a(sub n)-a(sub m)|<1 for m,n>=N, so |a(sub n)|<=|a(sub N+1)|+1 for n>=N. (Fix m=N+1). Let M=max{|a(sub 1)|, |a(sub 2)|, ..., |a(sub N+1)|+1} then |a(sub n)|<=M.

    Can you tell me if I am missing any information?
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  2. #2
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    Re: Cauchy Sequence

    Quote Originally Posted by lovesmath View Post
    I need to prove that every Cauchy sequence is bounded.
    Because this may well be a most important theorem in sequences, I think its proof needs to done correctly.

    1>0 so \exists N such that if n\ge N then |a_n-a_N|<1
    Of course that means |a_n|\le 1+|a_N| if n\ge N

    Let B = 1 + \sum\limits_{k = 1}^N {|a_k |} .

    Thus |a_n|\le B for \forall n~.
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