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Homework Help: Sequence that converges

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data

    I was given this homework problem:

    Show that if ##a_1,a_2, ... ,## is a sequence of real numbers that converges to ##a##, then [tex]lim_{n\to \infty}\frac{\sum^n_{k=1} a_k}{n}=a.[/tex]

    I was provided a solution but my book never went over such examples or the concrete steps to solve such a problem. I am wondering what are the basic first steps to solving these types of problems?

    And if possible, where can I find practice problems like these online? I searched but I couldn't find any.
    Last edited: Sep 29, 2013
  2. jcsd
  3. Sep 29, 2013 #2


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    Dearly Missed

    Remember that for every n, you may rewrite:
  4. Sep 29, 2013 #3


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    There are various techniques for limit problems, but since this problem asks you to start with an arbitrary convergent sequence [itex](a_k)[/itex] the only one which will work is to use what you know about [itex](a_k)[/itex]: for all [itex]\epsilon > 0[/itex] there exists [itex]K \in \mathbb{N}[/itex] such that if [itex]k \geq K[/itex] then [itex]a - \epsilon < a_k < a + \epsilon[/itex].

    That suggests taking an arbitrary [itex]\epsilon > 0[/itex] and its corresponding [itex]K[/itex] and splitting the sum as follows:
    \frac1n \sum_{k=1}^n a_k = \frac1n \sum_{k=1}^{K-1} a_k + \frac1n \sum_{k=K}^n a_k
    (You are interested in the limit [itex]n \to \infty[/itex], so at some stage you will have [itex]n > K[/itex] and you may as well assume that to start with.)

    Your plan is to show that
    a - \epsilon \leq \lim_{n \to \infty} \frac1n \sum_{k=1}^n a_k \leq a + \epsilon
    and since [itex]\epsilon > 0[/itex] was arbitrary it must follow that
    \lim_{n \to \infty} \frac1n \sum_{k=1}^n a_k = a
    as required.
  5. Sep 29, 2013 #4


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    I would use ϵ/2 in one step instead of ϵ, that makes the inequalities easier to show.
  6. Sep 29, 2013 #5
    Thank you very much, pasmith! That cleared some issues I had, thanks for clarifying it for me.
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