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Math Help - Limit of two sequences

  1. #1
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    Limit of two sequences

    Let an and bn be two convergent sequences. Prove: if for every even n: an<=bn, and for every odd n: an>=bn, then lim an = lim bn.

    Any ideas? I'm looking for a formal proof.
    Thanks!
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  2. #2
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    Re: Limit of two sequences

    This can be proved using the following facts.

    (1) The limit of a subsequence of a converging sequence equals the limit of the sequence.

    (2) If an and bn are converging sequences and an ≤ bn for all n, then lim an ≤ lim bn.
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  3. #3
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    Re: Limit of two sequences

    The said exercise is presented in the book before defining subsequences and fact (1), so I'm not supposed to use it. That's why I was looking for a more basic proof, almost only by the definition of limit of sequence. However, I can use the following rules, in case they're helpful:
    "If a sequence converges, then its limit is unique."
    "Every convergent sequence is bounded."
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  4. #4
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    Re: Limit of two sequences

    Let A = lim aₙ and B = lim bₙ. For a given ε > 0, choose N such that for all n > N we have |aₙ - A| < ε and |bₙ - B| < ε. Pick any even n > N; then inequalities in the previous sentence imply that A < aₙ + ε and bₙ < B + ε. Therefore, A < aₙ + ε ≤ bₙ + ε < B + 2ε. Thus, for the given ε we showed that A < B + 2ε. Since this holds for any positive ε, it must be that A ≤ B. Similarly, you can show that A ≥ B.
    Thanks from loui1410
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  5. #5
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    Re: Limit of two sequences

    Quote Originally Posted by loui1410 View Post
    The said exercise is presented in the book before defining subsequences and fact (1), so I'm not supposed to use it. That's why I was looking for a more basic proof, almost only by the definition of limit of sequence. However, I can use the following rules, in case they're helpful:
    "If a sequence converges, then its limit is unique."
    "Every convergent sequence is bounded."
    Suppose that {\lim _{n \to \infty }}{a_n} = A\;\& \;{\lim _{n \to \infty }}{b_n} = B. Going for a contradiction, suppose that A<B. Now define \varepsilon  = \frac{{B - A}}{2} > 0.
    So we have \left( {A - \varepsilon ,A + \varepsilon } \right) \cap \left( {B - \varepsilon ,B + \varepsilon } \right) = \emptyset.

    Because of convergence, almost all the a_n's are in \left( {A - \varepsilon ,A + \varepsilon } \right) and almost all the b_n's are in \left( {B - \varepsilon ,B + \varepsilon } \right).
    But that is impossible because of the alteration of odd and even terms.
    Thanks from loui1410
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  6. #6
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    Re: Limit of two sequences

    emakarov, thank you - that's a great solution, I wouldn't have thought about it.
    Plato's solution, however, seems more natural to me as it's similar to some other examples I've already seen.
    Thank you both.
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