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

  1. Sep 25, 2012 #1
    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?
     
  2. jcsd
  3. Sep 25, 2012 #2
    Prove by contradiction that a > b is impossible.
     
  4. Sep 25, 2012 #3

    jbunniii

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    Hint: if [itex]a > b[/itex], then you can find an [itex]\epsilon > 0[/itex] such that [itex]a - \epsilon > b + \epsilon[/itex].
     
  5. Sep 25, 2012 #4

    Dick

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    Do a proof by contradiction. Assume that a>b. Pick ε=(a-b)/4. Doesn't that lead to a contradiction?
     
  6. Sep 26, 2012 #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?
     
  7. Sep 26, 2012 #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 ##.
     
  8. Sep 26, 2012 #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.
     
  9. Sep 26, 2012 #8
    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 ##.
     
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