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

  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.
  2. jcsd
  3. LCKurtz

    LCKurtz 8,404
    Homework Helper
    Gold Member

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

    LCKurtz 8,404
    Homework Helper
    Gold Member

    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. 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 :(

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

    LCKurtz 8,404
    Homework Helper
    Gold Member

    Would a proof that showed for any ε > 0 that A ≤ B + ε be considered a direct proof?
  8. 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. 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. Landau

    Landau 905
    Science Advisor

    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. LCKurtz

    LCKurtz 8,404
    Homework Helper
    Gold Member

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