1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook