# prove the limit theorem: a_n < b_n -> A-B

by shoescreen
Tags: limit, prove, theorem
 P: 15 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.
 HW Helper Thanks PF Gold P: 7,000 If you suppose A > B and take epsilon (1/3)(A-B) can you find a contradiction?
P: 15
 Quote by LCKurtz If you suppose A > B and take epsilon (1/3)(A-B) can you find a contradiction?
i should have mentioned earlier, the problem statement explicitly states do not prove by contradiction.

HW Helper
Thanks
PF Gold
P: 7,000

## prove the limit theorem: a_n < b_n -> A-B

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).
P: 15
 Quote by LCKurtz 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).
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...
 HW Helper Thanks PF Gold P: 7,000 Would a proof that showed for any ε > 0 that A ≤ B + ε be considered a direct proof?
 P: 15 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
 P: 84 If you know liminf, limsup, then this is easy. Since $$a_n \leq b_n$$, or $$0 \leq b_n - a_n$$. Then it follows that $$0 \leq \liminf (b_n - a_n) \leq \limsup (b_n - a_n)$$. But since $$\lim a_n = A$$ and $$\lim b_n = B$$, it follows that $$\lim (b_n - a_n) = B - A$$, which implies the limit inferior and limit superior must equal. That is, $$\limsup (b_n - a_n) = \liminf (b_n - a_n) = B - A$$. Thus, we have that $$0 \leq B - A$$, rearrange then you're done.
 Sci Advisor P: 905 For every e>0 there is N such that n>N implies |a_n-A|
 HW Helper Thanks PF Gold P: 7,000 And here I've been under the impression that we aren't supposed to actually work homework problems for them.

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