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Calculus Proof

  1. Oct 27, 2005 #1

    kreil

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    I have this problem on my HW assignment and I just need a little help getting started...


    Let {an} be a convergent sequence and let [itex]l=\lim_{n{\rightarrow}\infty}a_n[/itex]

    Prove that if l < p then:


    [tex]\exists N{\in}N \forall n \ge N : a_n < p[/tex]


    Thanks!
     
    Last edited: Oct 28, 2005
  2. jcsd
  3. Oct 27, 2005 #2

    Physics Monkey

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    You have something backwards I think. If L > P then intuitively you expect all the [tex] a_n [/tex] to eventually be close to L and thus greater (not less) than P, right? Can you make progress with this intuitive picture in mind?
     
  4. Oct 27, 2005 #3
    Counterexample: Let an = 1 - (1/(n+1)). Then [itex]\lim_{n\rightarrow\infty}a_n = 1[/itex]. Choose p = 1/10 < 1. Then [itex]\forall N \in \mathbb{N} \exists n\ge N : a_n > p[/itex].
     
  5. Oct 28, 2005 #4
    Not necessarily. Let an = 1/(n+1). Then [itex]\lim_{n\rightarrow\infty} a_n = 0[/itex]. If you choose p=9/10 > 0, all the an's near l are less than p. I think a qualifier is missing somewhere.
     
  6. Oct 28, 2005 #5

    Physics Monkey

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    No, hypermorphism, you missed the condition in kreil's original post that L > P. In your case L < P from which it follows, as you have indicated, the [tex] a_n [/tex] will eventually be less than P and near L.
     
    Last edited: Oct 28, 2005
  7. Oct 28, 2005 #6
    Whoops. :rofl: That's probably the right question, then. In that case, kreil should appeal to the definition of convergence.
     
  8. Oct 28, 2005 #7

    kreil

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    sorry its l < p

    kreil
     
  9. Oct 28, 2005 #8

    kreil

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    I did some work on it and I'm pretty sure I got it...

    Show: an<p

    l < p [itex]\implies[/itex] p-l > 0

    [tex]l=\lim_{n{\rightarrow}\infty}a_n < p

    \iff \forall \epsilon > 0 \exists N \in N \forall n \ge N: |a_n-l| < \epsilon[/tex]

    Choose e = p - l > 0

    [tex]\exists N \in N \forall n \ge N: |a_n-l|<\epsilon=p-l[/tex]

    and thus

    [tex]\exists N \in N \forall n \ge N: a_n<l+{\epsilon}=l+p-l=p[/tex]



    Josh
     
    Last edited: Oct 28, 2005
  10. Oct 28, 2005 #9

    Physics Monkey

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    Yeah, great, that makes more sense. Also, your proof looks good.
     
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