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Stupid series I can't seem to find

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    I need a way to write the following series in a sum:
    [tex]\frac{1}{2} + \frac{1}{4} + \frac{1}{32} + \frac{1}{64} + \frac{1}{1024} + \frac{1}{2048} + ...[/tex]

    If you look closely you can see a 2^n pattern in there:
    [tex]\frac{1}{2} + \frac{1}{4} + \frac{0}{8} + \frac{0}{16} + \frac{1}{32} + \frac{1}{64} + \frac{0}{128} + \frac{0}{256} + \frac{1}{512} + \frac{1}{1024} + ...[/tex]

    The denominator obviously follows [tex]2^n[/tex], but the numerator goes 110011001100...
    I can't seem to find a function that will allow me to put this in a single sum:
    [tex]\sum_{n=1}^{\infty} ....[/tex]

    (The sum should converge to 0.8, right?)

    3. The attempt at a solution
    I tried using the mod operator to determine if n was even or odd, something like this:
    [tex]\sum_{n=1}^{\infty} \frac{n \mod 2}{2^n} + \frac{n \mod 2}{2^{n+1}}[/tex]
    This doesn't work, because the terms that are discarded when n is even are not discarded the next 'run' when (n+1) is even... Dunno how to explain this properly, but if you calculate it manually it does this:
    [tex]\frac{1}{2} + \frac{1}{4} + \frac{0}{4} + \frac{0}{8} + \frac{1}{8} + \frac{1}{16} + ...[/tex]
    So if you discard the 0/ ... terms you are just left with the usual 2^(-n) sum...

    I have a feeling I'm close, but I can't figure it out :S
     
    Last edited: Feb 15, 2009
  2. jcsd
  3. Feb 15, 2009 #2

    epenguin

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    If the missing bits were not missing it would add up to 1.

    Now you can put each missing bit into a relation with a bit that is there.

    They are a certain fraction of terms that are there. So their sum is that proportion of 1 IYSWIM.

    Do a diagram and this will look plausible.
     
  4. Feb 15, 2009 #3
    I can't seem to understand what you mean exactly...

    By missing bits you mean the 0/x parts?
    If so, then half of the terms as missing, so to your logic that would mean the sum would converge to 0.5...? (Whatr does IYSWIM mean?)

    But I think it should converge to 0.8 (I said 8 before but that was a mistake, I edited it now).
     
  5. Feb 15, 2009 #4

    Dick

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    If it's the pattern you suggest, (1/2+1/4)*(1/16)=(1/32+1/64). (1/32+1/64)*(1/16)=(1/512+1/1024). It's a geometric series with a common ratio of 1/16. But this doesn't quite fit with the series you quote in the first line (the next term after 1/64 is 1/1024, not 1/512). Is that a typo?
     
  6. Feb 15, 2009 #5

    epenguin

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    IYSWIM means 'if you see what I mean'.
    Just compare each pair of 'absent' terms with the 'present' pair preceding it.
     
  7. Feb 15, 2009 #6
    It was indeed a typo. The correct form is right underneath, where 1/512 is present.

    Alright, that makes a bit more sense. I still fail to see how I can now create a single sum for this series though... If I have a bit more time (maybe later tonight) I'll try to figure it out :)

    Thanks so far.
     
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