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Unbounded sequence

  1. Jan 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Either give an example or show that no example exists.

    An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound.

    2. Relevant equations


    3. The attempt at a solution

    example: {k}3k=-∞

    and by this notation i mean that k starts at -∞ and ends at 3, and k∈ℤ .

    I chose -∞ as the lower bound because I wanted this sequence to be unbounded. Then I chose 3 as the upper bound because that is what the original statement asked for.


    My questions:
    Does this example fulfill the original statement because I am unsure of what the question means by "and no β < 3 is an upper bound." The book uses β as the upper bound in previous pages.

    Also does the way I wrote the sequence mean what I want it to mean? Because I am also confused on the notation.
     
  2. jcsd
  3. Jan 23, 2017 #2

    Mark44

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    How can an unbounded sequence have an upper bound?
    Can you list the first, say, five terms in this sequence?
     
  4. Jan 23, 2017 #3

    Ray Vickson

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    Perhaps an "unbounded sequence" means what everybody else calls an "infinite sequence". An infinite sequence can be bounded or unbounded.
     
  5. Jan 23, 2017 #4

    LCKurtz

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    If you mean what I think you do, a better way to write it would be ##a_k = 4-k,~k = 1 ..\infty##.
     
  6. Jan 23, 2017 #5
    This is the definition of an unbounded sequence I've been using from online,

    a sequence is bounded if it is bounded above and below <=> if ∃k∈ℝ such that | xn | ≤ k ∀n∈ℕ.


    The first five terms in this sequence would be
    -∞, -∞+1, -∞+2, -∞+3, -∞+4


    I was thinking that an unbounded sequence can have an upper bound if it goes to infinity in some direction but converges to a number as well.
     
  7. Jan 23, 2017 #6
    I checked the answer in the back of the book, and the answer is fn=3-n which seems similar to ak=4-k, k=1...∞.

    I think I would have gotten this same answer had the original statement been " An unbounded sequence for which 3 is an upper bound ".

    I'm still confused by the bold part: "An unbounded sequence for which 3 is an upper bound, and no β < 3 is an upper bound. "
    I think that it means no number less than 3 can be an upper bound, but isn't this redundant since we said 3 is an upper bound in the first part of the statement?
     
  8. Jan 23, 2017 #7

    Mark44

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    No, these aren't numbers.
     
  9. Jan 23, 2017 #8

    LCKurtz

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    They are the same, assuming ##n## starts at ##0## in ##3-n##.
    No, it isn't redundant. Since ##3## is a term of the sequence, no number ##x## less than ##3## can be an upper bound. How could it be if ##x < 3##?
     
  10. Jan 23, 2017 #9

    Mark44

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    This sequence is bounded since all of the terms are less than or equal to 3.
    I think the problem is poorly worded.
     
  11. Jan 23, 2017 #10

    LCKurtz

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    Aren't you confusing "bounded above" with "bounded"?
     
  12. Jan 23, 2017 #11

    Mark44

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    I don't think so. The definition of bounded sequence I am using is that it is a sequence that is bounded above and bounded below. That is, for some M > 0, |an| < M for all n in Z+ (or similar restriction on n).
    See https://proofwiki.org/wiki/Definition:Bounded_Sequence
     
  13. Jan 23, 2017 #12
    Ok I think I get it. So we're told 3 is an upper bound. That means that the greatest number in this sequence is less than or equal to 3.

    we're also told that no number less than 3 is an upper bound.

    Therefore 3 must be included in this sequence.

    If we were not told that no number less than 3 is an upper bound, then our sequence could have been something like xn=-10-n, n=0 . . ∞

    Am i understanding this correctly?
     
  14. Jan 23, 2017 #13

    LCKurtz

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    Then why do you say in post #9 that ##a_k = 4-k,~k=1..\infty## is bounded?
     
    Last edited: Jan 23, 2017
  15. Jan 23, 2017 #14

    LCKurtz

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    Yes, almost. You could have it true if the sequence just got arbitrarily close to ##3## but less than ##3##.
     
  16. Jan 23, 2017 #15
    Ok, thank you both for your help, helped me a lot on this question .
     
  17. Jan 23, 2017 #16

    Mark44

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    I should have said "bounded above."
     
  18. Jan 23, 2017 #17

    LCKurtz

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    Right. That's what I pointed out in post #10 in the first place.
     
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