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Advanced Calculus Sequence Convergence

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that the sequence {a_n} converges to A if and only if lim n--->∞ (a_n-A)=0.

    2. Relevant equations



    3. The attempt at a solution

    It's an if and only if proof, but I'm not sure how to prove it. Please help!
     
  2. jcsd
  3. Sep 19, 2012 #2

    jbunniii

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    Try writing each statement in epsilon-delta form, and compare.
     
  4. Sep 20, 2012 #3
    I'm not good at writing proofs. So far I have:
    Let {an} converge to A. Given epsilon>0, there exists N>0 s.t. lan-Al<epsilon for all n>N.
    So l((a_n)-A)l<epsilon for all n>n.
    Thus, we can write lim n--->infinity (a_n-A)=0.

    Then, I'm not sure how to prove the statement's converse. Can someone help?
     
  5. Sep 20, 2012 #4

    jbunniii

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    Well, how did you prove it in the forward direction? Can you simply reverse the reasoning?
     
  6. Sep 20, 2012 #5
    I'm not sure how to write the reverse. Would I start with: lim n--->infinity(a_n -A)=0. So given epsilon>0, there exists N>0 s.t. la_n-Al<epsilon for all n>N.

    I don't know if this is correct, and I don't know where to go after that.
     
  7. Sep 20, 2012 #6

    jbunniii

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    Yes, that's correct. But what's the difference between this:

    ([itex]\lim_{n \rightarrow \infty} (a_n - A) = 0)[/itex]: "So given epsilon>0, there exists N>0 s.t. la_n-Al<epsilon for all n>N."

    versus what you wrote earlier:

    ([itex]\lim_{n \rightarrow \infty} a_n = A)[/itex]: "Given epsilon>0, there exists N>0 s.t. lan-Al<epsilon for all n>N."
     
  8. Sep 20, 2012 #7
    I'm not sure. I don't know if I have it written correctly. I feel like I'm working in circles.
     
  9. Sep 20, 2012 #8

    HallsofIvy

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    Look at the definitions! "[itex]a_n[/itex] converges to A" means "Given [itex]\epsilon> 0[/itex], there exist an integer N such that if n> N, [itex]|a_n- A|< \epsilon[/itex]".

    Applying exactly the same definition, "[itex](a_n- A)[/itex] converges to 0" means "Given [itex]\epsilon> 0[/itex], there exist N such that if n> N, [itex]|(a_n-A)- 0|< \epsilon[/itex]".

    But [itex](a_n- A)- 0[/itex] is just [itex]a_n- A[/itex]!

    Remember that definitions in mathematics are "working definitions"- you use the precise words of definitions in problems and proofs.
     
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