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Two subsequential limit questions

  1. Feb 21, 2012 #1
    I have two questions involving subsequential limits. One I have started and understand what I need to do, the second I really don't know how to start.

    First question

    Let 0 ≤ a < b < +∞. Define the sequence a[itex]_{n}[/itex] recursively by setting a[itex]_{1}[/itex] = a, a[itex]_{2}[/itex] = b, and a[itex]_{n+2}[/itex]=(a[itex]_{n}[/itex]+a[itex]_{n+1}[/itex])/2 [itex]\forall[/itex]n.

    Show that the sequences a[itex]_{2n}[/itex] and a[itex]_{2n-1}[/itex] are monotonic and convergent. Does a[itex]_{n}[/itex] converge? To what?

    Solution attempt
    This is the second part of the question. The first part I already proved, and shows that if a[itex]_{2n}[/itex] and a[itex]_{2n-1}[/itex] both converge to the same limit then a[itex]_{n}[/itex] converges to that limit.

    I can show that both subsequences are bounded. I can also show that if their limits exists, they are the same (by plugging in a[itex]_{2n}[/itex]→L a[itex]_{2n-1}[/itex]→L and using the recurrence relations). The limit is obviously a+(2/3)b (though I need to prove it).

    So all I need to do is show that a[itex]_{2n-1}[/itex] is nondecreasing and a[itex]_{2n}[/itex] is nonincreasing. Unfortunately I can't find the right induction proof to show this. Then I need to find one of the limits (either the full sequence or the even or odd subsequence).

    Second question
    Let λ[itex]\in[/itex][0,1]. Show that there exists a sequence r[itex]_{n}[/itex] such that r[itex]_{n}[/itex]→λ. r[itex]_{n}[/itex][itex]\in[/itex]{0,1/2[itex]^n[/itex],2/2[itex]^n[/itex],...,(2[itex]^n[/itex]-1)/2[itex]^n[/itex],1}.

    Solution attempt
    The correct theorem is presumably:

    Let S denote the set of subsequential limits of a sequence s[itex]_{n}[/itex]. Suppose t[itex]_{n}[/itex] is a sequence in S[itex]\cap[/itex]ℝ and that t[itex]_{n}[/itex]→t. Then t belongs to S.

    However I can't figure out how to apply it.

    Thanks for your help.
  2. jcsd
  3. Feb 21, 2012 #2
    For the first question, the limit should be a+2(b-a)/3.

    Can you write the subsequences in a closed form?

    For the second question, let e > 0. What about N such that 1 / 2N < e?
  4. Feb 21, 2012 #3
    The numerators look like a Fibonacci type sequence (a Horadam Sequence?). The bottom is just 2^(n-2). You're correct about the limit, I mistyped. Unfortunately I can't figure out if this helps. Is the pattern more obvious if you split off the odd/even sequences? I can't seem to find a pattern in each individual subsequence.

    I'm starting to see the proof for the second problem. As I recall if [itex]\exists[/itex]N such that [itex]\forall[/itex]n≥N |a[itex]_{n+1}[/itex]-a[itex]_{n}[/itex]|<1/2[itex]^{n}[/itex] then the sequence is Cauchy. Because the sequence would be bounded between 0 and 1 the limit exists and must lie in [0,1]. However, that doesn't prove that any real number in that range can be made the limit of the sequence. Can you help with that bit?
  5. Feb 22, 2012 #4
    The Fibonacci sequence itself can be lifted into a closed form exponential expression, and the subsequences are exponential, or geometric sums. There might be a way to write the whole sequence in a closed form as an alternating geometric sum, now that I think about it. In any case, you want to have a form such that F(1) = a, F(2) = b, F(3) = (a + b) / 2, F(4) = ((a + b) / 2 + b) / 2...

    For #2: You want to have for all e > 0 there is an N such that for all n ≥ N, |an+1-an|< 1/2N < e.

    Anyway, given real r, how many dyadic rationals of the form i / 2n are you guaranteed to find in the interval (r - 1/2n, r + 1/2n)?

    Edit: Also, it may be easier to think about #1 in the interval [0, 1]. Generalizing to [a, b] is fairly trivial.
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