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Math Help - Cauchy Sequence

  1. #1
    Member roshanhero's Avatar
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    Cauchy Sequence

    How can we show that the sequence (-1)^{n+1}is bounded but isnot a Cauchy Sequence?
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    Is it true that (\forall n)[|x_n|\le 1]~?

    Cauchy sequences converge in \mathbb{R}.
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  3. #3
    Member roshanhero's Avatar
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    Yeap the given sequence is bounded, but how can we show that it is not a Cauchy by using the definition of Cauchy?
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    Try taking an arbitrary n \in \mathbb{N} and then take m = n+1 and use them in the definition of a cauchy sequence (and think why this works)
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    A sequence of real numbers converges if and only if it is not Cauchy. So one way to show that this sequence is not Cauchy is to show that it does not converge.
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    Quote Originally Posted by Defunkt View Post
    Try taking an arbitrary n \in \mathbb{N} and then take m = n+1 and use them in the definition of a cauchy sequence (and think why this works)
    You had me confused for a moment! The definition of Cauchy sequence is that \lim_{m,n\to 0} |a_n- a_m|= 0 with m and n going to infinity independently. You could not use m= n+1 to prove a sequence is Cauchy but it certainly can be used to give a counter example.
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  7. #7
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    That's a very good point, and a good exercise for the thread creator would be to conjure up a counter example for the case \lim_{n\to \infty} |a_{n+1}- a_n|= 0 - ie. a sequence for which the last limit is zero but the sequence does not converge.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Defunkt View Post
    That's a very good point, and a good exercise for the thread creator would be to conjure up a counter example for the case \lim_{n\to \infty} |a_{n+1}- a_n|= 0 - ie. a sequence for which the last limit is zero but the sequence does not converge.
    What about a_n=H_n!
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