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Bounding proof question

  1. Dec 3, 2008 #1
    how to prove that:
    every sequence which convergences has to be bounded
     
  2. jcsd
  3. Dec 3, 2008 #2

    CompuChip

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    First try and explain it in words. Suppose you have a converging sequence. Then what happens to its elements (in particular, to infinitely many of them at the end of the sequence)? What can you do with the finitely many before that?
     
  4. Dec 3, 2008 #3

    HallsofIvy

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    Suppose the sequence converges to L. What can you say about members of the sequence between L+ 1 and L- 1?
     
  5. Dec 3, 2008 #4
    i can say that they
    L-1<=An<=L+1

    what is the next step?
     
  6. Dec 3, 2008 #5

    Mark44

    Staff: Mentor

    You can't say that without knowing something about n. Here is an example:
    {10, -10, 5, -5, 2.5, -2.5, 1.25, -1.25, ...}
    This sequence converges to 0, but it's not true in general that -1 <= a_n <= 1.
     
  7. Dec 3, 2008 #6
    for certain e>0
    |An-L|<e
     
  8. Dec 3, 2008 #7

    Mark44

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    No, for some n that depends on epsilon.

    In the sequence I gave as an example, what does n have to be so that |a_n| < 1?
     
  9. Dec 3, 2008 #8
    n has to be even
    n=2,4,6 etc..
     
  10. Dec 3, 2008 #9

    Mark44

    Staff: Mentor

    Nope.

    Assuming the sequence I gave a few posts back starts with a_0 = 10, a_2 is not within 1 unit of 0, nor are a_4 and a_6.

    I've given you a rather large epsilon (1), and I'm asking for a number N so that if n >= N, then |a_n| < 1. This is not really a hard problem.
     
  11. Dec 4, 2008 #10

    HallsofIvy

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    The problem appears to be that you do not know the definition of convergence.

    What is the definition of "[itex]lim_{n\rightarrow \infty} A_n= L[/itex]"?
     
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