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Math Help - Cauchy sequence

  1. #1
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    Cauchy sequence

    I know that (an) and (bn) are cauchy sequences. I need to prove or disprove that (an+bn) is also cauchy.

    Thanks for any suggestions.
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  2. #2
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    Quote Originally Posted by JaysFan31 View Post
    I know that (an) and (bn) are cauchy sequences. I need to prove or disprove that (an+bn) is also cauchy.

    Thanks for any suggestions.
    The sequence, \{ a_n \} is a Cauchy sequence. Thus, for n,m\geq N_1 we have,
    |a_n-a_m| < \epsilon for any \epsilon>0

    The sequence, \{ b_n\} is a Cauchy sequence. Thus, for n,m \geq N_2 we have, |b_n-b_m|< \epsilon for any \epsilon>0

    Consider the sequence, \{a_n+b_n\}. For any \epsilon /2>0 we can find N_1,N_2 satisfy those two conditions on top. Let N=\max\{N_1,N_2\} then,
    |a_n-a_m|<\epsilon/2 and,
    |b_n-b_m|<\epsilon/2.
    Addition yields,
    |a_n-a_m|+|b_n-b_m|<\epsilon
    Triangular inequality (Thanks to CaptainBlank for teaching this move to me).
    We have,
    |a_n+b_n-(a_m+b_m)|<\epsilon for n,m\geq N.
    Q.E.D.
    Last edited by ThePerfectHacker; October 23rd 2006 at 06:17 PM.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    The sequence, \{ a_n \} is a Cauchy sequence. Thus, for n,m\geq N_1 we have,
    |a_n-a_m| < \epsilon for any \epsilon>0

    The sequence, \{ b_n\} is a Cauchy sequence. Thus, for n,m \geq N_2 we have, |b_n-b_m|< \epsilon for any \epsilon>0

    Consider the sequence, \{a_n+b_n\}. For any \epsilon /2>0 we can find N_1,N_2 satisfy those two conditions on top. Let N=\max\{N_1,N_2\} then,
    |a_n-a_m|<\epsilon/2 and,
    |b_n-b_m|<\epsilon/2.
    Addition yields,
    |a_n-a_m|+|b_n-b_m|<\epsilon
    Triangular inequality (Thanks to CaptainBlank for teaching this move to me).
    We have,
    |a_n+b_n-(a_m+b_m)|<\epsilon for n,m\geq N.
    Q.E.D.
    This needs to start something like:

    As \{a_n\} is a Cauchy sequence, for all \epsilon>0 there exists a N_1 such that for all n,m>N_1 |a_n-a_m|<\epsilon.

    With something similar for \{b_n\}, \epsilon and N_2.

    Then the rest follows as ImPerfectHacker has shown.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    This needs to start something like:

    As \{a_n\} is a Cauchy sequence, for all \epsilon>0 there exists a N_1 such that for all n,m>N_1 |a_n-a_m|<\epsilon.

    With something similar for \{b_n\}, \epsilon and N_2.

    Then the rest follows as ImPerfectHacker has shown.

    RonL
    I know, I was too lazy to write out: "For all ... there exits ... such that... then..."
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  5. #5
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    Using the same from above, is (an*bn) Cauchy?
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  6. #6
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    Quote Originally Posted by JaysFan31 View Post
    Using the same from above, is (an*bn) Cauchy?
    Actually I have a much simpler proof that I should have stated for both the problems.

    A sequence is a Cauchy sequence if and only if it is convergent.

    Using this fact we can easily show that an*bn and an+bn is a cauchy sequence.

    If an and bn are Cauchy sequences then they are convergent sequences. Then an*bn and an+bn are convergent sequences. That means then an*bn and an+bn are Cauchy sequences.
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