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Convergent sum has possible error that never occurs

  1. Jul 28, 2011 #1
    I have a convergent sum where I use the reciprocal of a_n at each step:

    a_n = a_n / gcd(a_n, b_n) <--- I'm removing common factors.

    This converges as long as I want to run it. Both a_n and b_n are quite dynamic.

    However, if a_n equals b_n then the divide after the gcd would return 1 and the series would diverge.

    Is there a way to prove that a_n and b_n will never be equal?

    I can match this series with another that uses the same method but it has been proven that the divide and gcd will never return a 1.

    Can I use that first series to prove that the second series will never have the divide and gcd return 1?

    Are there any theorems that deal with this condition?

    Any help is appreciated.
     
  2. jcsd
  3. Jul 28, 2011 #2
    As far as I can tell from what you posted, there are no constraints at all on the a's or the b's. They could be anything at all. Obviously, if they could be anything at all, there is no way to prove that a_n and b_n will never be equal. To prove that you would need some additional information about the series that you don't have (or at least haven't mentioned).
     
  4. Jul 28, 2011 #3
    Ok, a_n and b_n will never equal 1, so that takes care of two out of three conditions.

    Now, I seem to remember a mention somewhere about a possible side-effect in a convergent sum. That if the sum converges to some point without the side-effect occurring, that the side-effect would not occur. I have not been able to find that source.

    This is a sieve I'm working on and it works like a champ. I'm just trying to prove that it will always work.

    On the series that has been proven, a_n and b_n are both factorials. On my series, both use the gamma function. Pretty much the same thing.
     
  5. Jul 28, 2011 #4
    It's impossible to answer your question with the information given. Knowing that the a's are a gamma function of something and the b's are a gamma function of something and that neither is 1 is not enough to determine whether the series converges.
     
  6. Jul 28, 2011 #5
    The question I'm trying to answer is: If both series use essentially the same method at each step to identify primes, and one series has been proved to never return a 1 (which indicates that no primes were found at that step), can I use the fact that the second series, which can be kept in lock-step with the first, also never generates a 1. In fact, it returns values that are identical to the first series.

    The differences between the series: for every 2 steps of the first series, there are 2,4,8,16,32... steps for the second series.

    The first series in the original format (single steps) converges. The second series (single steps) can be shown to converge. However, that is incomplete because it is not known whether the 1 will ever occur somewhere down the line.

    So, I need to prove for the second series that the 1 will never occur.
     
    Last edited: Jul 28, 2011
  7. Jul 28, 2011 #6
    In your first post, you said that if a 1 ever occurs, the series will diverge. You just stated that the series "can be shown to converge". A series that diverges does not converge. So, if everything you've stated is correct, you've answered your own question.
     
  8. Jul 28, 2011 #7
    Yes, it would diverge. That is my dilemma. I can take the series up into millions of steps without encountering a 1. The series can be shown to converge because this exception never happens. I need to prove that it will never happen.
     
    Last edited: Jul 28, 2011
  9. Jul 28, 2011 #8
    Once again, you're saying, "I know the answer", then immediately following it by saying "I don't know the answer". You say, "The sequence can be shown to converge because the exception never happens". Then you say, "I need to prove that it will never happen." But if you've already shown that it never happens, what's left to prove?

    My best guess it that there is some distinction in your mind between "showing" that the exception never happens, and "proving" that it never happens. But you are not communicating, or I am not understanding.

    EDIT: OK, while I was typing you changed "never" to "never, ever", in red yet. I'm sorry, but that doesn't help. "Never, ever" may have some extra rhetorical heft, but mathematically I can't see that there is any difference between them. Never is never.
     
  10. Jul 28, 2011 #9
    Sorry about the red never, ever. I need to prove it is impossible the have a missing prime in one of my series' segments. Just showing it to millions of steps is not good enough for the number theorists.

    Incidentally, the other series is based on Betrand's Postulate, where it has been proved that there is one or more primes between n and 2n. The single step series that I crafted steps between powers of 2. The divide/gcd will never return a 1 because there is always at least one prime present. Same for the two-step series which steps between powers of 4.
     
  11. Jul 28, 2011 #10
    OK, so now this is becoming a little clear. When you say you have "shown" something never happens, you just mean that you have never seen it happen. What about your other claims? For instance, when you say that the second series returns values identical to the first series, is that something you have "shown", or something you have "proved"?

    Anyway, I come back the same place. You just haven't given anywhere near enough information to answer your question. As far as anyone could work out from what you've said so far, you have two sequence, a_n and b_n, which we know nothing about except that (1) they arise from gamma functions in some unexplained way, (2) they are never 1, and (3) they match up with the items in some other, also unspecified sequence at predictable intervals (although it's not clear if you know this, or have just never seen a case where it was not true), and (4) the latter unspecified sequence is based in some unspecified way on Bertram's Postulate.

    Isn't it obvious to you that no one could possibly answer your question based on this information?

    And then you do the "I know the answer"/"I don't know the answer" dance once again:

    So, problem solved, right?
     
  12. Jul 28, 2011 #11
    Another reason for the demanding proof is this Riemann identity I discovered by accident:

    [itex]\text{Zeta}[3]==\frac{2 \pi ^4}{315\prod _{n=1}^{\infty } \frac{1}{1+ \frac{1}{\text{Prime}[n]^2-\text{Prime}[n]}} }[/itex]
     
  13. Jul 28, 2011 #12
    I can prove that the mapping of the two-step series mapped to the multiple-step series always returns the same values. However, if my series returned a 1 in one of its steps it will not change this equivalence because each sub-step is part of a product (both series.)
     
  14. Jul 28, 2011 #13
    So, once again, the problem is solved. You have now stated twice in a row, based on two different arguments, that you know your series converges. So, we're finished, right?
     
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