1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook