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Math Help - Cauchy sequences

  1. #1
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    Cauchy sequences

    Let {x sub n} be a sequence and suppose that the sequence {x sub (n+1) - x sub n} converges to 0. Give an example to show that the sequence {x sub n} may not converge. Hence, the condition that abs(x sub n-x sub m) is less than sigma for all m, n greater than or equal to N is crucial in the definition of a Cauchy sequence.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by friday616 View Post
    Let {x sub n} be a sequence and suppose that the sequence {x sub (n+1) - x sub n} converges to 0. Give an example to show that the sequence {x sub n} may not converge. Hence, the condition that abs(x sub n-x sub m) is less than sigma for all m, n greater than or equal to N is crucial in the definition of a Cauchy sequence.
    How about the sequence in \mathbb{Q}: 1, 1.4, 1.41, 1.414, 1.4142...

    The idea of this sequence is that if it were in \mathbb{R}, it would converge to \sqrt{2}; however, since \sqrt{2}\notin\mathbb{Q}, this sequence doesn't converge to anything, though the points do get arbitrarily close together. (You know the deal: \forall~\epsilon>0, \exists~N such that  n,m>N implies that |x_n-x_m|<\epsilon)
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  3. #3
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    Quote Originally Posted by friday616 View Post
    Let {x sub n} be a sequence and suppose that the sequence {x sub (n+1) - x sub n} converges to 0. Give an example to show that the sequence {x sub n} may not converge. Hence, the condition that abs(x sub n-x sub m) is less than sigma for all m, n greater than or equal to N is crucial in the definition of a Cauchy sequence.
    Let x_n:= \sum_{j=1} ^{n} \ \frac{1}{j} then x_{n+1} - x_n = \frac{1}{n+1} \rightarrow 0 but x_n doesn't converge
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  4. #4
    Super Member redsoxfan325's Avatar
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    That's a good one.
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  5. #5
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    How about  x_n = log(n) .

    We have  x_{n+1} - x_n = log(n+1) - log(n) = log(\frac{n+1}{n}) \rightarrow 0 .
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