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Help with describing a sequence

  1. Apr 13, 2008 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data

    Describe the following sequence

    0.3,0.33,0.333,0.3333,0.33333,...


    2. Relevant equations



    3. The attempt at a solution

    I tried to put the sequence as

    [tex]\frac{3}{10},\frac{3}{10}+\frac{3}{100},\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+...[/tex]

    But I didn't see anything happening there. Also I would hope the answer is convergent to [itex]\frac{1}{3}[/itex] since it looks like the number of 3's tend to infinity which would be the number of 3's in [itex]\frac{1}{3}[/itex]
     
  2. jcsd
  3. Apr 13, 2008 #2

    Dick

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    It's 3 times the geometric series (1/10)^n starting at n=1, isn't it? The series sums to 1/9, doesn't it?
     
  4. Apr 14, 2008 #3
    .3,.33,.333,.3333..........
    > 3 [.1+.11+.111+.1111.........]
    Solve in factors of 10 now ...
    go ahead....
     
  5. Apr 14, 2008 #4

    tiny-tim

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    Hi rock.freak667! :smile:

    You're only asked to describe the sequence.

    In other words: find a formula A_n for the nth term.

    For example, if the sequence is 1 8 27 64 …, then you describe it as {n³}.

    Hint: As you point out, they're obviously getting closer to 1/3. So what is the difference between each term and 1/3? :smile:
     
  6. Apr 14, 2008 #5

    rock.freak667

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    well the differences are like this

    1/3-0.3=0.0333...
    1/3-0.33=0.00333...
    1/3-0.333=0.000333...

    So the differences run in a GP of first term a=0.0333... and common ratio r=0.1

    But the sum to infinity = 0.0333../0.9 = 0.333...*1/9 = 1/3 * 1/9 = 1/27..Not 1/3 as I wanted.
     
  7. Apr 14, 2008 #6

    exk

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    rock.freak667 remember that [tex]\sum\frac{1}{10^{n}} = \frac{1}{1-\frac{1}{10}}[/tex] only when n goes from 0 to infinity. Is that the case in your example?
     
  8. Apr 15, 2008 #7

    tiny-tim

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    Hi rock.freak667! :smile:

    (remember, the question asks you to describe the sequence, and you haven't specifically done that yet.)

    So 3An = 1 - what ?

    And so ∑An = … ? :smile:
     
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