If a_n < b_n for all n, then lim a_n < lim b_n

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In summary, we are trying to prove that a ≤ b, given that a_n and b_n are convergent sequences with limits a and b respectively, and that a_n ≤ b_n for all n. To do this, we use proof by contradiction and assume that a > b. By choosing a specific ε, we can show that a_n > b_n, which contradicts our initial assumption. Therefore, our assumption that a > b is impossible, and thus a ≤ b.
  • #1
demonelite123
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If [itex] a_n [/itex] and [itex] b_n [/itex] are convergent sequences and [itex] a_n ≤ b_n [/itex] for all n, show that a ≤ b where a and b are the limits of [itex] a_n [/itex] and [itex] b_n [/itex] respectively.

since a_n and b_n are convergent, there exists an N1 such that [itex] |a_n - a| < ε [/itex] for all n > N1 and an N2 such that [itex] |bn - b| < ε [/itex] for all n > N2. I then choose N = max(N1, N2) so for all n > N, the 2 inequalities are satisfied. Since i want to show that a ≤ b, i take [itex] a < a_n + ε ≤ b_n + ε [/itex], but i am stuck here since b_n + ε is not less than b. Since this leads to a dead end, can someone give me a hint on how to approach this problem?
 
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  • #2
Prove by contradiction that a > b is impossible.
 
  • #3
Hint: if [itex]a > b[/itex], then you can find an [itex]\epsilon > 0[/itex] such that [itex]a - \epsilon > b + \epsilon[/itex].
 
  • #4
demonelite123 said:
If [itex] a_n [/itex] and [itex] b_n [/itex] are convergent sequences and [itex] a_n ≤ b_n [/itex] for all n, show that a ≤ b where a and b are the limits of [itex] a_n [/itex] and [itex] b_n [/itex] respectively.

since a_n and b_n are convergent, there exists an N1 such that [itex] |a_n - a| < ε [/itex] for all n > N1 and an N2 such that [itex] |bn - b| < ε [/itex] for all n > N2. I then choose N = max(N1, N2) so for all n > N, the 2 inequalities are satisfied. Since i want to show that a ≤ b, i take [itex] a < a_n + ε ≤ b_n + ε [/itex], but i am stuck here since b_n + ε is not less than b. Since this leads to a dead end, can someone give me a hint on how to approach this problem?

Do a proof by contradiction. Assume that a>b. Pick ε=(a-b)/4. Doesn't that lead to a contradiction?
 
  • #5
thanks for your replies. i am working on it but still having trouble finding a contradiction. i know that a > b and a_n ≤ b_n along with [itex] |a_n - a| < ε [/itex] and [itex] |b_n - b| ≤ ε [/itex] will lead me to a contradiction, but i have been manipulating these all together and still can't find the contradiction.

i've also tried setting ε = (a-b)/4 like Dick said, but I'm not sure of how to get a contradiction from that either. i get that a_n < (5a-b)/4 and b_n < (a + 3b)/4 which doesn't tell me much. can you explain a little more about how the contradiction arises in this case?
 
  • #6
Assuming ## a > b ##, and choosing ## \epsilon ## so that ## a - \epsilon > b + \epsilon > b_n##, ## a_n > a - \epsilon > b + \epsilon > b_n ##, which contradicts ##a_n \le b_n ##.
 
  • #7
thanks for your reply voko!

i know that you told me to do that earlier but i wasn't sure if that in itself had to be proved or not. Is that step considered elementary or does it actually need to be rigorously proved as well?

i was trying to do this problem using only the inequalities given and using only operations/properties proved true for inequalities such as adding/subtracting things on both sides. but like i said earlier, i wasn't able to get very far.
 
  • #8
demonelite123 said:
i know that you told me to do that earlier but i wasn't sure if that in itself had to be proved or not. Is that step considered elementary or does it actually need to be rigorously proved as well?

Since you are requested to prove that, it is best not to make any assumptions as to the rigor. The sketch I gave above is still not completely formal, since one needs to prove that it is possible to choose ## \epsilon ## with the required property (even though it might seem obvious). Dick suggested one particular way, but it is still up to you to prove that such an ## \epsilon ## satisfies ## a - \epsilon > b + \epsilon ##.
 

1. What does the statement "a_n < b_n for all n" mean?

This statement means that for every term in the sequence a, the value of a is less than the corresponding term in the sequence b. In other words, a is always smaller than b.

2. How does this statement relate to the concept of limits?

This statement is related to the concept of limits because it is comparing the behavior of two sequences as n approaches infinity. By stating that a_n < b_n for all n, it is implying that the limit of a_n must also be less than the limit of b_n.

3. Is it possible for lim a_n to be equal to lim b_n if a_n < b_n for all n?

Yes, it is possible for lim a_n to be equal to lim b_n in this scenario. This would occur when both sequences have the same limit, regardless of the fact that a_n is always less than b_n.

4. Can the statement "a_n < b_n for all n" be reversed to "lim a_n < lim b_n"?

No, this statement cannot be reversed. Just because a_n < b_n for all n does not necessarily mean that lim a_n < lim b_n. This is because the limit is a limit of the entire sequence and not just a particular term.

5. How does this statement apply to real-world situations?

This statement can be applied to real-world situations where two variables are being compared. For example, if the speed of a car is always less than the speed limit, then the limit of the car's speed must also be less than the limit of the speed limit. This can also be applied to other scenarios, such as comparing the growth rates of two populations or the temperatures of two substances over time.

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