Two subsequential limit questions

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In summary, the conversation involves two questions about subsequential limits. The first question asks to define a sequence using recursion and show that its subsequences are monotonic and convergent. The second question asks to show the existence of a sequence that converges to a given limit in a specified range. The conversation also includes solution attempts for both questions and requests for further explanations and help.
  • #1
hiro
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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.
 
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  • #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?
 
  • #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?
 
  • #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.
 

1. What is a subsequential limit?

A subsequential limit is a type of limit that occurs when a sequence has multiple accumulation points, meaning there are multiple values that the sequence approaches as the index approaches infinity. Unlike a regular limit, a subsequential limit does not require the sequence to approach a single specific value.

2. How is a subsequential limit different from a regular limit?

A regular limit occurs when a sequence approaches a single specific value as the index approaches infinity. A subsequential limit, on the other hand, occurs when a sequence has multiple accumulation points and does not necessarily approach a single specific value.

3. How do you find the subsequential limit of a sequence?

To find the subsequential limit of a sequence, you must first identify the accumulation points of the sequence. This can be done by finding the limit of any subsequence of the original sequence. The subsequential limit is then the set of all accumulation points of the sequence.

4. Can a sequence have more than one subsequential limit?

Yes, a sequence can have multiple subsequential limits if it has multiple accumulation points. These subsequential limits may or may not be equal, depending on the sequence.

5. Why is it important to study subsequential limits?

Subsequential limits are important in the study of limits and convergence of sequences. They allow for a more comprehensive understanding of the behavior of a sequence and can help identify important properties of the sequence, such as boundedness or convergence.

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