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Math Help - Real analysis; sequences and limits

  1. #1
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    Real analysis; sequences and limits

    I think I need to do this by contradiction, but I'm stuck.

    Prove:

    If the sequence <an> has the limit A, the sequence <bn> has the limit B, and an<bn for all n, show A< or = B.
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    Re: Real analysis; sequences and limits

    Nah, no need for contradiction. If \displaystyle \begin{align*} a_n < b_n \end{align*}, then \displaystyle \begin{align*} b_n = a_n + c_n \end{align*}, where \displaystyle \begin{align*} c_n > 0  \end{align*} for all n, which must go to some positive value (C) since B is finite. Then

    \displaystyle \begin{align*} \lim_{n \to \infty} b_n &= \lim_{n \to \infty} \left( a_n + c_n \right) \\ &= \lim_{n \to \infty} a_n + \lim_{n \to \infty} c_n \\ &= A + C \end{align*}

    But we also know this limit is B, therefore B = A + C, and thus \displaystyle \begin{align*} B \geq A \end{align*}.
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    Re: Real analysis; sequences and limits

    How do you know B is finite? Also, how do you know the c sub n has limit C?

    In class, we have done these type of proof using the definition of converge. I can follow this, mostly, but I don't think I'm allowed to do the proof this way.

    Thanks,
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  4. #4
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    Re: Real analysis; sequences and limits

    If a sequence has a limiting value of B, then B is a number and is clearly finite.

    In order for all the b terms to be greater than the a terms, then it's the same as adding some value to the a terms. Since the a terms go to some value A, in order for the b terms to also go to some value, then the remaining c terms also have to go to some value.
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    Re: Real analysis; sequences and limits

    Hi,

    In direct answer to your question about proof by contradiction, I offer the following:

    Real analysis; sequences and limits-mhfsequencelimita.png
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    Re: Real analysis; sequences and limits

    This is helpful, but did you let epsilon = -L/2? which is positive because L is less than zero. But where did you get -L/2? I'm guessing it's from some scratch work that should be obvious to me, but isn't.

    Also, we haven't proved or used that the limit of the difference two convergent sequences is the the difference of the limits. I know it's true, but since we haven't proved it I can't use it.
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    Re: Real analysis; sequences and limits

    Quote Originally Posted by amyw View Post
    This is helpful, but did you let epsilon = -L/2? which is positive because L is less than zero. But where did you get -L/2? I'm guessing it's from some scratch work that should be obvious to me, but isn't.
    Also, we haven't proved or used that the limit of the difference two convergent sequences is the the difference of the limits. I know it's true, but since we haven't proved it I can't use it.
    This is a well-known problem used to teach a particular way of proof.
    Suppose that b<a then b<\frac{a+b}{2}<a

    So let \varepsilon  = \frac{{a - b}}{2} > 0. Now note that a - \varepsilon  = \frac{{b + a}}{2}\;\& \;b + \varepsilon  = \frac{{b + a}}{2}.
    That means \left( {b - \varepsilon ,b + \varepsilon } \right) \cap \left( {a - \varepsilon ,a + \varepsilon } \right) = \emptyset .

    But because (b_n)\to b almost all of the terms are in \left( {b - \varepsilon ,b + \varepsilon } \right).

    AND because (a_n)\to a almost all of the terms are in \left( {a - \varepsilon ,a + \varepsilon } \right).

    Surely you can see a contradiction there?
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    Re: Real analysis; sequences and limits

    Hi Amy,

    I hope the following is yet more "insight" into your problem:

    Real analysis; sequences and limits-mhfsequencelimit2.png
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  9. #9
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    Re: Real analysis; sequences and limits

    Thanks, this makes perfect sense to me now that I've had some sleep.
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