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Prove the limit theorem: a_n < b_n -> A-B

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Given the sequence {a_n} converges to A and {b_n} converges to B, and a_n <_ b_n for all n>_ n*, prove A <_ B


    2. Relevant equations

    x+ epsilon < y for every positive real epsilon, implies x <_ y
    A - B = (A - a_n) + (b_n - B) + (a_n - b_n)

    3. The attempt at a solution

    I want to show A - B is not positive, I know a_n - b_n is not positive by hypothesis, and A - a_n and b_n - B are bounded by (-e/2, e/2) where e is any postive number, hence their sum is bounded by (-e,e). But since their sum is not bounded by zero, I can't figure out how to apply (x + e) - y negative implies (x - y) non positive.
     
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  3. Sep 30, 2009 #2

    LCKurtz

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    If you suppose A > B and take epsilon (1/3)(A-B) can you find a contradiction?
     
  4. Sep 30, 2009 #3
    i should have mentioned earlier, the problem statement explicitly states do not prove by contradiction.
     
  5. Sep 30, 2009 #4

    LCKurtz

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    If you see how to do the indirect argument I am hinting at, you can easily modify it to be a direct argument. (Shhh... I won't tell anybody if you look at the indirect argument first).:wink:
     
  6. Sep 30, 2009 #5
    Well I was asked to prove the this by contradiction before and i was able to do it with (a-b)/2 and i still dont see how to do it directly :(

    EDIT:
    hah i think i got it!
    EDIT:
    just kidding...
     
    Last edited: Sep 30, 2009
  7. Sep 30, 2009 #6

    LCKurtz

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    Would a proof that showed for any ε > 0 that A ≤ B + ε be considered a direct proof?
     
  8. Sep 30, 2009 #7
    so if I use
    |a_n - A| < f(A,B)
    |b_n - A| < g(A,B)

    I can split up each inequality and ultimately end up with something like
    a_n > -f(A,B) + A
    b_n < G(A,B) + B

    so a_n - b_n > -f(A,B) - g(A,B) + A - B

    so if i want to show a_n - b_n > A - B, the function -f + -g must be positive. But this is impossible since both of these functions must be strictly positive, hence negative f plus negative g is also negative.
    So i'm still stuck
     
  9. Oct 1, 2009 #8
    If you know liminf, limsup, then this is easy.

    Since [tex]a_n \leq b_n[/tex], or [tex]0 \leq b_n - a_n[/tex]. Then it follows that [tex]0 \leq \liminf (b_n - a_n) \leq \limsup (b_n - a_n)[/tex]. But since [tex]\lim a_n = A[/tex] and [tex]\lim b_n = B[/tex], it follows that [tex]\lim (b_n - a_n) = B - A[/tex], which implies the limit inferior and limit superior must equal. That is, [tex]\limsup (b_n - a_n) = \liminf (b_n - a_n) = B - A[/tex]. Thus, we have that [tex]0 \leq B - A[/tex], rearrange then you're done.
     
  10. Oct 1, 2009 #9

    Landau

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    For every e>0 there is N such that n>N implies |a_n-A|<e and |b_n-B|<e.
    In other words: A-e<a_n<A+e and B-e<b_n<B+e. Try to put these two together.

    \\edit: as LCKurtz remarked, I gave away too much, so I removed the last part.
     
    Last edited: Oct 2, 2009
  11. Oct 1, 2009 #10

    LCKurtz

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    And here I've been under the impression that we aren't supposed to actually work homework problems for them.
     
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