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Simple Sum

  1. Feb 13, 2006 #1
    How can I find the sum of [tex]1/2 - 1/4 + 1/8 - 1/16 + ... - 1/256[/tex]?

    Do I need to group the positive and negative terms?
     
  2. jcsd
  3. Feb 13, 2006 #2

    TD

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    Do you mean the finite sum, as you wrote it, or an infinite sum (the series)?
     
  4. Feb 13, 2006 #3
    Finite sum.
     
  5. Feb 13, 2006 #4

    TD

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    In that case: yes you could add the positive terms and substract the sum of the (absolute values) of terms with a negative sign. It would be rather annoying (bu still doable) work to put them all on the same denominator, can you use a calculator or not?
     
  6. Feb 13, 2006 #5
    No calculator allowed :\

    I only know sigma notation and the formula's for the partial sums of geometric and arithmetic series'.
     
  7. Feb 13, 2006 #6

    TD

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    Oh of course, but that's fine.

    The sum as a whole can be written as:

    [tex]\sum\limits_{n = 1}^8 {\frac{{\left( { - 1} \right)^{n + 1} }}{{2^n }}}[/tex]

    Now, can you split it in two sums and find the general formula for both?
    Or even if you can't find the sigma-notation, can you try to become two geometric series if you look at the positive terms and at the negative terms seperately?
     
  8. Feb 13, 2006 #7
    Oops, I had that answer but I made the mistake of thinking there was 256 terms :\

    How do I figure out the common ratio if it varies from positive to negative?

    I will try and split it up into two sums in the mean time...
     
  9. Feb 13, 2006 #8

    TD

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    Well, if you can find the common ratio by dividing a term by its precessor (is that an English word?) Anyway, if the sequence with terms t_n is geometric, than t_(n+1)/t_n = r with r constant for all n. You can check this and find r this way.
     
  10. Feb 13, 2006 #9
    I get:

    [tex]\sum\limits_{n = 1}^4 {\frac{1}{2 \cdot 4^{n - 1}}} - \sum\limits_{n = 1}^4 {\frac{1}{4 \cdot 4^{n - 1}}}[/tex]
     
  11. Feb 13, 2006 #10

    TD

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    Looks good, but have you checked whether the initial problem wasn't geometric already?
     
  12. Feb 13, 2006 #11
    r = -1/2

    So,

    [tex]\sum\limits_{n = 1}^8 {\left[\frac{1}{2} \cdot \left (-\frac{1}{2}\right)^{n-1}\right]}[/tex]
     
    Last edited: Feb 13, 2006
  13. Feb 13, 2006 #12

    TD

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    Looks good again :smile:
     
  14. Feb 13, 2006 #13
    Thanks for your help!
     
  15. Feb 14, 2006 #14
    I have another problem:

    [tex]S = 1/2 - 1/3 + 1/4 - 1/5 + ...[/tex]

    Find [itex]S_{100}[/itex]

    So, [tex]S_n = \frac{n}{2}(t_1 + t_n) = \frac{100}{2}\left({\frac{1}{2} - \frac{1}{101}\right)[/tex]

    This gives me 24.505 when I think it should be 0.301927

    Also, how can I associate the terms to show that 1 > S > 0? Does grouping positive terms and subtracting the negative terms show this?
     
  16. Feb 14, 2006 #15

    TD

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    The formula you used to find the n-th partial sum is the one for arithmetic series, is that the case here? If you think so, what is the common difference then?
     
  17. Feb 14, 2006 #16
    I guess there is no common difference... but I see no common ratio either?
     
    Last edited: Feb 14, 2006
  18. Feb 14, 2006 #17

    TD

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    Correct, it's not arithmetic nor geometric...
     
  19. Feb 14, 2006 #18
    I see it's a harmonic series, so the recipricals are arithmetic. Does that help me?
     
  20. Feb 14, 2006 #19

    TD

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    I'm not sure, are you supposed to determine this partial sum by some calculation again (no calculator)?
     
  21. Feb 14, 2006 #20
    Yes... never calculator :p

    I only ever got a forumla for the partial sum of a geometric and arithmetic series.
     
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