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Cauchy sequences

  1. Nov 5, 2008 #1
    By definition, a sequence a(n) has the Cauchy sequence if for eery E>0 ,there exist a natural number N such that Abs(a(n) - a(m) ) < E for all n, m > N

    Could anyone tell me what is a(m) ? is it a subsequence of a(n) , or could it be any other non related sequence ?
     
  2. jcsd
  3. Nov 5, 2008 #2

    Office_Shredder

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    a(m) is the same sequence as a(n)
     
  4. Nov 5, 2008 #3

    lurflurf

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    a(m) and a(n) are not sequences they are elements of a sequence

    Pehaps the difficulty will be eased by restating the definition differently

    a sequence is Cauchy if for any E>0 there exist a natural number N such that the difference between any two terms beyond N cannot exceed N

    or

    a sequence is Cauchy if for any E>0 there exist a natural number N such that Abs(a(N+n) - a(N+m) ) < E for all n,m that are natural numbers
     
  5. Nov 5, 2008 #4

    HallsofIvy

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    Neither a_m nor a_n in that is a sequence. They are, rather, any two numbers from the original sequence {a_i}, with, of course, m and n larger than N.
     
  6. Nov 5, 2008 #5
    OK thanks

    One more question
    what's the difference between Lim sup a(n) and sup A(n)
    does the limit tells me something else ?
     
  7. Nov 5, 2008 #6

    mathman

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    It is easier to explain by example. Consider the sequence 1, 1/2, 1/3, 1/4,...

    The sup is 1, while the lim sup is 0.
     
  8. Nov 10, 2008 #7
    what basically is the change that lim produced in sup
    why it changes sup=1
    to lim sup=0
    and do this thing hold in all cases that lim sup is not the part of the sequence
     
  9. Nov 11, 2008 #8
    The limit superior of a sequence [itex](a_n)_{n\geq 0}[/itex] is the largest accumulation (or cluster) point of this sequence. An accumulation point is a number c such that in any neighbourhood of c there are infintely many members of the sequence. Analogously, the limit inferior is the least such accumulation point.

    If [itex](a_n)_{n\geq 0}[/itex] is convergent, say with limit a, then
    [tex]
    \lim_{n\to\infty} {a_n} = \limsup_{n\to\infty}{a_n} = \liminf_{n\to\infty}{a_n} = a
    [/tex]
     
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