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Proof of Inequality Between Lower and Upper Bounds

  1. Jul 10, 2014 #1
    Convergence of Divergent Series Whose Sequence Has a Limit

    1. The problem statement, all variables and given/known data
    Suppose [itex]∑a_{n}[/itex] is a series with lim [itex]a_{n}[/itex] = L ≠ 0. Obviously this diverges since L ≠ 0. Suppose we make the new series, [itex]∑(a_{n} - L)[/itex]. My question is this: is there some sufficient condition we could put solely on [itex]a_{n}[/itex] (or maybe its sum) without calculating the limit or referencing it other than the existence of a limit so that the new series converges?

    2. Relevant equations
    I can't think of anything here; I could list all of the convergence tests from Ch. 3 of Baby Rudin but doesn't seem wholly relevant with the way I phrased the question. Feels like this is probably something trivial from some higher level analysis course or analytic number theory course that I haven't had the chance to take yet.

    There's one thing that seems possibly relevant (from Rudin):
    Theorem Suppose [itex]a_{1}≥a_{2}≥a_{3}≥...≥0.[/itex] Then the series [itex]∑a_{n}[/itex] converges if and only if the series [itex]∑2^{k}a_{2^{k}}[/itex] converges.

    3. The attempt at a solution
    I really don't know how to approach this without introducing the L somehow into the proof which isn't what I really wish to explore.

    I thought to make some simple conditions first to try and reduce the nature of the problem, like making the sequence monotonic and non-negative. With the theorem, I was thinking of maybe introducing some sort of scaling maybe with the [itex]2^{k}[/itex] factor? I don't know.

    I think the root test gives convergence for 0 < L < 1, but what about for L > 1? But this sort of references the limit.
     
    Last edited: Jul 10, 2014
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  3. Jul 10, 2014 #2

    micromass

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    That's a pretty vague question. Is this an actual book problem or a question you invented yourself? I don't see any reason why there would be such a simple condition. Well, obviously you can try all the series tests you've seen and adapt it to this case. But I don't really know what kind of answer you want here.
     
  4. Jul 10, 2014 #3
    Invented. Just curious if there exists any such condition to give convergence. I don't know anything about techniques to deal with divergent series, so I'd be satisfied with any specific reference material in lieu of some answer.
     
  5. Jul 10, 2014 #4

    micromass

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  6. Jul 10, 2014 #5
    Alright, thanks. Messed up the title of this thread but think I corrected it. Guess I can't fix the thread title, ah well.

    Criterion of Abel looks promising maybe:
    The criterion of Abel
    Let ∑+∞n=0an be a (real or complex) convergent series . Let (vn)n be a bounded sequence of real numbers which is either nondecreasing or nonincreasing. Then the sequence ∑+∞n=0vnan converges.

    [itex]v_{n}[/itex] could be the [itex]a_{n}[/itex] and would have to figure out some convergent sequence so that the product equals (L - [itex]a_{n}[/itex]).
     
    Last edited: Jul 10, 2014
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