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

In summary, the problem statement asks for a proof that A-B is not positive, but the indirect argument can be modified to be a direct argument.
  • #1
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Homework Statement


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


Homework Equations



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

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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  • #2
If you suppose A > B and take epsilon (1/3)(A-B) can you find a contradiction?
 
  • #3
LCKurtz said:
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.
 
  • #4
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:
 
  • #5
LCKurtz said:
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:

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 don't see how to do it directly :(

EDIT:
hah i think i got it!
EDIT:
just kidding...
 
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  • #6
Would a proof that showed for any ε > 0 that A ≤ B + ε be considered a direct proof?
 
  • #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
 
  • #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.
 
  • #9
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.
 
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  • #10
And here I've been under the impression that we aren't supposed to actually work homework problems for them.
 

1. How do I prove the limit theorem "a_n < b_n -> A-B"?

To prove the limit theorem, you must first understand the concept of limits in mathematics. A limit is the value that a function or sequence approaches as its input or index approaches a certain value. In this particular theorem, we are comparing two sequences, a_n and b_n, and stating that if a_n is always less than b_n, then the difference between their limits, A and B respectively, will also follow this relationship. To prove this, you would need to use the definition of limits and the properties of inequalities.

2. What is the significance of the "a_n < b_n" condition in the limit theorem?

The "a_n < b_n" condition is essential in the limit theorem because it establishes the relationship between the two sequences. It tells us that the terms of a_n will always be less than the terms of b_n, which is a crucial factor in determining the relationship between their limits. Without this condition, we cannot prove the limit theorem.

3. Can you give an example of how to use this limit theorem in a real-life situation?

One example of using this limit theorem in a real-life situation is in the stock market. Let's say we have two stocks, A and B, whose values are represented by the sequences a_n and b_n. If it is known that the value of stock A is always less than the value of stock B, then we can use this limit theorem to predict that the difference between their values, A-B, will also follow this relationship. This information can be useful for investors in making decisions about which stock to invest in.

4. Are there any exceptions to this limit theorem?

Like any other mathematical theorem, there are certain conditions and assumptions that must be met for this limit theorem to hold. One exception is when the limits A and B do not exist. In this case, the theorem would not hold as we cannot determine the relationship between the limits. Additionally, if the condition "a_n < b_n" is not satisfied for some terms of the sequences, the theorem may not hold.

5. Is this limit theorem used in other theorems or proofs?

Yes, this limit theorem is often used in conjunction with other theorems and proofs in mathematics. It is a fundamental concept that is used to prove other limit theorems, such as the squeeze theorem and the comparison test for series. It is also applicable in various fields of science, such as physics and engineering, where limits play a significant role in understanding and predicting phenomena.

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