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

by shoescreen
Tags: limit, prove, theorem
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shoescreen
#1
Sep30-09, 12:29 PM
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.
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LCKurtz
#2
Sep30-09, 01:00 PM
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If you suppose A > B and take epsilon (1/3)(A-B) can you find a contradiction?
shoescreen
#3
Sep30-09, 01:03 PM
P: 15
Quote Quote by LCKurtz View Post
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.

LCKurtz
#4
Sep30-09, 01:21 PM
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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).
shoescreen
#5
Sep30-09, 01:33 PM
P: 15
Quote Quote by LCKurtz View Post
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...
LCKurtz
#6
Sep30-09, 02:16 PM
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Would a proof that showed for any ε > 0 that A ≤ B + ε be considered a direct proof?
shoescreen
#7
Sep30-09, 02:22 PM
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
fmam3
#8
Oct1-09, 03:15 AM
P: 84
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.
Landau
#9
Oct1-09, 06:51 PM
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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.
LCKurtz
#10
Oct1-09, 06:55 PM
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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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