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Sum of a series.

  1. Jul 31, 2005 #1
    find the sum of the series 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/12 + .....,
    where the terms are the reciprocals of the positive integers whose only prime factors are 2's and 3's.

    Here's my work so far:

    1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/12 + ... =
    1 + 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/3 + 1/9 + 1/27 + ... + 1/6 + 1/12 + 1/18 + 1/24 + ... =
    1 + 1/2 + 1/4 + 1/8 + 1/16 + ... + 1 + 1/3 + 1/9 + 1/27 + ... + 1 + 1/6 + 1/12 + 1/18 + 1/24 + ... - 2

    Those first two are geometric series. I don't know what to do with the third one.
     
  2. jcsd
  3. Jul 31, 2005 #2

    Hurkyl

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    Try writing it in sigma notation. (Σ)
     
  4. Jul 31, 2005 #3
    :confused:
    My problem is to find the value of 1 + 1/6 + 1/12 + 1/18 + 1/24 + 1/32 + 1/48 + 1/54...
     
  5. Jul 31, 2005 #4

    Hurkyl

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    That sum is no easier to calculate than the original sum. (In fact, I would be able to figure out the original sum faster than the one in your last post)

    There are other ways to see how to approach this problem, but IMHO, writing it in sigma notation makes the right step much clearer.

    In general, it's good practice to write sums in sigma notation anyways.
     
    Last edited: Jul 31, 2005
  6. Jul 31, 2005 #5
    Well then my problem is "sigma notation". :cry:
     
  7. Jul 31, 2005 #6

    Hurkyl

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    Do you know how to write a simpler sum in sigma notation? Like 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...?
     
  8. Jul 31, 2005 #7
    Yes.
    [tex]\sum_{n=0}^\infty \frac{1}{2^n}[/tex]
     
  9. Jul 31, 2005 #8

    Hurkyl

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    Ok, good. Next question: what's the general form of a number whose prime factors are only 2 and 3?
     
  10. Jul 31, 2005 #9
    That's [tex]2^n3^m[/tex]
     
  11. Jul 31, 2005 #10

    Hurkyl

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    So does that suggest how to write the sum of the reciprocals of every positive integer whose prime factors can only be 2's and 3's in sigma notation?
     
  12. Jul 31, 2005 #11
    Indeed. Is it going to be a double sum?
     
  13. Jul 31, 2005 #12

    Hurkyl

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    Yes, it is!


    Technical note: actually, the double sum is technically different from the original sum. However, the sum turns out to be absolutely convergent (the easiest way to see this fact is that it has no negative terms!), and this sort of "rearrangement" is perfectly legal with absolutely convergent sums, meaning that they'll give the same answer.
     
    Last edited: Jul 31, 2005
  14. Jul 31, 2005 #13
    And is the correct answer 3? :blushing:
     
  15. Jul 31, 2005 #14

    Hurkyl

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    Sounds right. Well done! By the way, don't miss the technical note I edited into my last post.

    By the way, now that you know how the problem is solved, it may be instructive to try and figure out how to do it without sigma notation. Then again, it might not be instructive too, I don't know. :smile:
     
    Last edited: Jul 31, 2005
  16. Jul 31, 2005 #15
    Thanks for the technical note. Had forgot it. o:)
    Without sigma notation? I have no clue at all :tongue:
     
  17. Jul 31, 2005 #16

    shmoe

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    You had it as a double sum right? With a bit of rearranging you should be able to write it as a product of 2 infinite sums, which I suppose you could write without sigma notation if you liked. I'd suggest you try this as the product form you'll get is quite cute and worth the effort.
     
  18. Aug 1, 2005 #17
    Yes, that's exactly what I did.
     
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