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A convergent sequence of reals

  1. Aug 25, 2011 #1
    Call {a1, a2, a3, ...} = {an} a "convergent sequence" if

    [tex]\exists L \in \mathbb{R} : \quad \forall \epsilon > 0 \quad \exists N \in \mathbb{N} : (\forall n > N \quad (n > N \implies |a_n - L| < \epsilon))[/tex]

    in which case we write [itex]\lim_{n \rightarrow \infty} a_n = \lim a_n = L[/itex]. Of course this is the usual definition for sequences of reals. Consider the sequence

    [tex]\{{n \over {n + 1}}\} = \{{1 \over 2}, {2 \over 3}, {3 \over 4}, ... \}[/tex]

    It appears the numbers approach 1, intuitively. We'll try to show by definition that [itex]\lim a_n = 1[/itex]. Given [itex]\epsilon > 0[/itex], we'll try to find the corresponding positive integer N so that we satisfy the above definition. I get this far:

    [tex]|{n \over n + 1} - 1| = |{-1 \over n + 1}| = |{1 \over n + 1}| < \epsilon \implies 1 < {\epsilon}(n + 1)[/tex]

    By the Archimedean property, we can always find (n + 1) so that this is true, i.e. for any [itex]\epsilon > 0[/itex]. But this doesn't help me find the particular fixed integer N which corresponds to our choice of [itex]\epsilon[/itex]. How is this done?
     
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  3. Aug 25, 2011 #2
    In general, the N has to be a function of [itex]\epsilon[/itex]. For example. Take [itex]N = \left\lceil\frac{1}{\epsilon}\right\rceil[/itex] (ceiling function) then [itex]n > N \ge \frac{1}{\epsilon}\Longrightarrow \frac{1}{n+1} < \epsilon[/itex] as required.
     
  4. Aug 25, 2011 #3

    LCKurtz

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    You are given [itex]\epsilon > 0[/itex]. You have
    [tex]|\frac 1 {n+1}| <\epsilon\rightleftarrows \frac 1 \epsilon < n+1 \rightleftarrows n > \frac 1 \epsilon -1[/tex]

    So start your final argument like this. Given [itex]\epsilon > 0[/itex] pick an integer
    [tex]N > \frac 1 \epsilon -1[/tex]
    Then if n > N, ... (work your steps in reverse here).
     
  5. Aug 25, 2011 #4
    Right. I was thinking [itex]N = \lceil{\frac{1}{\epsilon}}\rceil[/itex] would work, but I didn't want to introduce the ceiling function because this is for a teaching exercise. Writing [itex]N > \frac{1}{\epsilon} - 1[/itex] is a good idea too, I was thinking more about writing [itex]N = N(\epsilon)[/itex] explicitly but this works perfectly. In fact, it's a good idea for a lesson example because it shows how limits of sequences are different from limits of functions: we work with integers as arguments, so we need [itex]N > 0[/itex] instead of [itex]\delta > 0[/itex].

    Thanks all.
     
    Last edited: Aug 25, 2011
  6. Aug 25, 2011 #5
    Well if you don't make the requirement that N be a natural number, then the ceiling function is not needed. I don't think that requirement adds anything to the definition.
     
  7. Aug 25, 2011 #6
    ^ Well, the terms in the sequence are indexed by natural numbers, so it wouldn't make sense to use another number as the index. I see that n can be a natural number while the "fixed" N doesn't have to be; but I think it might be better for teaching to think of N as "the point in the sequence past which the sequence is within epsilon of its limit".
     
  8. Aug 25, 2011 #7
    I don't think it is a bad thing to introduce the ceiling function to (I suppose) calculus students. I think it isn't difficult to understand and it is better to teach it when you need it since they would know it is made for something, not just some arbitrary useless definition.
     
  9. Aug 25, 2011 #8
    Yeah. I think I'll do that tomorrow. Thanks.
     
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