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Convergence of sequences

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data
    If {cn} is a convergent sequence of real numbers, does there necessarily exist R> 0 such that |cn|≤ R for every n ∈ N? Equivalently, is {cn : n ∈ N} a bounded set of real numbers? Explain why or why not.

    2. Relevant equations

    3. The attempt at a solution
    I would think this is patently obvious given the definition of convergence but I don't really know how to put the proof in words and numbers.

    cn approaches some value c for large n but it can do so in a number of ways so how can i prove that there is always some R that is larger or equal to all elements cn?

    Thanks for the help!
  2. jcsd
  3. Sep 14, 2010 #2
    Write the limit definition of a sequence out. Assume to opposite of your claim (for contradiction).
    So for any N you pick to satisfy an epsilon of the limit, there will be an R as large as you want where R < |Xn|. Where epsilon is fixed already. So your R can be dependant on epsilon and L. Is there some R that would bring about a contradiction?
  4. Sep 14, 2010 #3


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    Science Advisor

    Suppose your sequence converges to L. Then, by definition of "limit", given any [itex]\epsilon> 0[/itex] there must exist some N such that if n> N then [itex]|a_n- L|< \epsilon[/itex].

    Take [itex]\epsilon= 1[/itex], say. Then there exist N such that if n> N, [itex]|a_n- L|< 1[/itex] which is the same as saying [itex]-1< a_n- L< 1[/itex] or [itex]L-1< a_n< L+ 1[/itex]. Now that only restricts [itex]x_n[/itex] for n> N, but the set [itex]\{a_0, a_1, a_2, \cdot\cdot\cdot, a_N\}[/itex] is finite and has a largest member. Every number in the sequence [itex]\{a_n\}[/itex] must be less that the that largest member or L+ 1, whichever is larger.
  5. Sep 14, 2010 #4
    In my case {cn} need not be finite, though. Does that still hold in this case?
  6. Sep 14, 2010 #5


    Staff: Mentor

    Yes. HallsOfIvy is talking about the terms in the sequence c1, c2, ..., cN, the first N terms at the beginning of the sequence.
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