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Calculus II i don't understand the proof for the limit comparison test

  1. Jul 10, 2013 #1
    would someone please care to reword this proof for me?


    it talks about ε, which is not even defined and then n0, which is again not defined, what the hell are all these variables... i'm sure someone here could do a better job organizing that crap proof.

  2. jcsd
  3. Jul 10, 2013 #2

    Stephen Tashi

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    You have a minor reason to complain because the proof should say "for each [itex] \varepsilon > 0 [/itex] there exists an integer [itex] n_0 [/itex]..."

    The variable [itex] n_0 [/itex] isn't "undefined" if you can read mathematical statements such as those found in the definition for the limit of a sequence. In that definition there is a statement about "there exists an integer...". Whatever name is used for that integer isn't an "undefined" variable.
  4. Jul 10, 2013 #3
    The proof isn't "crap" at all, but it might be a bit difficult to read if you are not used to it. I will try and clarify the proof given by Wikipedia.

    First, the proof picks a small positive number [itex]\epsilon[/itex] and says that no matter how small it is, there is always a sufficiently big [itex]N_0[/itex] so that for all [itex]n[/itex] greater than it, [itex]\displaystyle \left|\frac{a_n}{b_n}-c\right|[/itex] will be smaller than [itex]\epsilon[/itex]. This is the same thing as saying the ratio of the two sequences converges to c.

    Then, it plays around with this expression until it gets some inequality involving the two sequences where it can apply the direct comparison test and conclude that they are either both convergent or both divergent.
    Last edited: Jul 10, 2013
  5. Jul 10, 2013 #4
    if the limit of a(n)/b(n) = c, then wouldn't {a(n)/b(n)} - c = 0? i don't understand........................
  6. Jul 10, 2013 #5
    There are some problems with your foundations of limits then. The proof simply makes use of the definition of a limit, also called the epsilon-delta definition. Do you know what the statement [itex]\displaystyle \lim_{n\to\infty} a_n[/itex] rigorously means?
  7. Jul 10, 2013 #6
    is it referring to the series "a(n)" as "n" --> infinity?
  8. Jul 10, 2013 #7
    and i never understood the definition of the limit using the epsilon and delta terms, but i know it basically says that the value of a limit approaches essentially that value with separation of epsilon as epsilon decreases and decreases. is this remotely correct? lol
  9. Jul 10, 2013 #8
    I'll take that as a no.
    You seem to have deeper foundational issues than can be solved on a topic like this, or at least I think so. Can you check your private messages? I sent you a link.

    It is very very remotely correct. It is so remotely correct that it is completely useless in proofs like this, but it is still correct. Still, doing calculus with limits without knowing the epsilon-delta definition is, at the very least, ill advised.
  10. Jul 10, 2013 #9

    Stephen Tashi

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    Millenial's diagnosis is correct. You haven't come to grips with the formal definition of limit, so you can't expect to understand proofs that employ that definition.

    I'll go further. Your general approach to learning mathematics is wrong. You are only attempting to understand things in intuitive ways. That is OK as a beginning, but you don't understand mathematics until you can deal with concepts as they are actually defined. You are using the liberal arts approach of "Express the definition in your own words". That won't work for you in mathematics because you don't use precise language.

    It's common for people to express disdain for "legalistic hair-splitting" but that's exactly the way that formal mathematical proofs are conducted. Advocates for math prefer to emphasize that math is creative, useful, fun etc. However, the truth is that mathematics has a very legalistic aspect. You have to read mathematical definitions lke a lawyer reading a contract to find a loophole. If you avoid the legalistic side, you end up with a completely mangled set of ideas about math.
  11. Jul 10, 2013 #10


    Staff: Mentor

    You are confusing the idea of a limit of a ratio with the value of the ratio for a particular n.

    For example, let an = n, and let bn = 2n + 1, where n is a positive integer, n ≥ 1.

    a1/b1 = 1/3
    a2/b2 = 2/5
    a3/b3 = 3/7

    It's fairly clear (I hope) that ##\lim_{n \to \infty}\frac{a_n}{b_n} = 1/2##, but for each specific value of n, an/bn ≠ 1/2, hence an/bn - 1/2 ≠ 0.
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