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Thread: Analysis: show xn is a null sequence

  1. #1
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    Analysis: show xn is a null sequence

    Prove that x_{n+1}-1/2x_n\to 0 \Rightarrow x_n\to 0.

    My approach:
    Let \varepsilon>0. Then there exists an N\in \Bbb N s.t. \forall n\geq N, |x_{n+1}-1/2x_n|<\varepsilon. It follows that -\varepsilon\leq -\varepsilon+1/2x_n<x_{n+1}<\varepsilon \leq \varepsilon+1/2x_n. Then for any n\geq N, we have |x_{n+1}|<\varepsilon.
    But I have assumed that \forall n\in \Bbb N, x_n\geq 0 which may not be the case. Can I argue that when x_n<0, we have -\varepsilon+1/2x_n<-\varepsilon<x_{n+1}<\varepsilon+1/2x_n<\varepsilon? Is my proof correct?

    Thank you!
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  2. #2
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    Re: Analysis: show xn is a null sequence

    Hey ach4124.

    You have "sandwiched" your epsilon so that the absolute value of your x_n is less than it (which is the thing about epsilons often going to zero even if they don't actually approach it "technically") which is important.

    The argument of the epsilon has to be correct in terms of xn satisfying that inequality so from that point of view it looks good.

    You could also use things like the triangle inequality but I'm not sure what your teacher has instructed (in analysis, you can use lots of existing identities to prove these sorts of things).
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    Re: Analysis: show xn is a null sequence

    Are you actually required to use \delta-\epsilon? I would write that x_{n+1}=\tfrac12x_n and use that to write a closed form expression for x_n in terms of n.
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    Re: Analysis: show xn is a null sequence

    Thank you for your response. At second glance, I think my proof is flawed! In my proof, 1/2x_n does not 'hold any place', so I guess we can extend the statement to x_{n+1}-bx_n\to 0\Rightarrow x_n\to 0 for b\in \Bbb Q, but when b=1, x_n=\sum 1/n shows that the statement does not hold.

    I couldn't prove it using triangle inequality |x_{n+1}-1/2x_n|\leq |x_{n+1}|+1/2|x_n|. Yet, |x_{n+1}-1/2x_n|<\varepsilon does not imply |x_{n+1}|+1/2|x_n|<\varepsilon?

    I should use the reverse one. ||x_{n+1}|-1/2|x_n||\leq |x_{n+1}-1/2x_n|<\varepsilon, so -\varepsilon\leq -\varepsilon+1/2|x_n|<|x_{n+1}|<-\varepsilon\leq \varepsilon+1/2|x_n|. This is a bit nicer than my proof because we don't need to care about the sign of x_n.
    Last edited by ach4124; Aug 13th 2016 at 09:53 PM.
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    Re: Analysis: show xn is a null sequence

    Thanks, Archie. That's a nice method!
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    Re: Analysis: show xn is a null sequence

    As an alternative:
    x_{n+1}-\tfrac12x_n=0 \implies x_{n+1}-x_n=-\tfrac12x_n, so the sequence \{x_n\} is decreasing when x_n is positive and increasing when x_n is negative.

    Also x_{n+1}=\tfrac12x_n tells us that x_{n+1} has the same sign as x_n.

    Thus, the sequence \{x_n\} is either increasing and bounded above (by zero), or it is decreasing and bounded below (by zero), or (trivially) every x_n is equal to zero. This tells us that the series converges.

    When a series converges, \lim_{n \to \infty} x_{n+1}-x_{n} =0. But since x_{n+1}=\tfrac12x_n, we have \lim_{n \to \infty} \tfrac12x_n-x_{n} =\lim_{n \to \infty} -\tfrac12x_n =0 and so \lim_{n \to \infty} x_{n} =0 as required.
    Thanks from ach4124
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    Re: Analysis: show xn is a null sequence

    Was the original question supposed to be x_{n+1}-\tfrac12x_n=0 as I read to be, or x_{n+1}-\tfrac12x_n \to 0 as you wrote?
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    Re: Analysis: show xn is a null sequence

    It is x_{n+1}-\frac{1}{2}x_n\to 0.
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    Re: Analysis: show xn is a null sequence

    That's unusual. You might want to check that with your teacher. I'll have to think about what can be done with it though.
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  10. #10
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    Re: Analysis: show xn is a null sequence

    I found this exercise in Selected Problems in Real Analysis, an AMS book. The solution given involves sums: "Let \tilde{x}_n=x_n-x_{n-1}/2. Then x_n-\frac{x_1}{2^{n-1}}=\sum_{1<k\leq n} \frac{\tilde{x}_k}{2^{n-k}}=\sum_{1<k\leq m}+\sum_{m<k\leq n}. Since \tilde{x}_k\to 0, for large m the second sum is small for all n\geq m. For fixed m the first sum is arbitrary small if n is sufficiently large."
    How does that prove x_n\to 0? Does my argument using reverse triangle inequality suffice to prove the statement? I don't understand why we need to consider sums.
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