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Geometric series question

  1. May 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Determine whether the serie is convergent or divergent , if it is convergent find its sum.

    Ʃn=1 (1 + 2n )/ 3n

    2. Relevant equations

    Ʃa(r)n-1 = a / (1-r) r < 1 is converging or if r > 1 diverging

    3. The attempt at a solution

    Well I can see its a geometric series so i did Ʃ (1/3)^n + (2/3)^n and r < 1 therefore its convergent but I can't figure out the a. I thought it would be one but the answer key says 1/3 and 2/3 as the a for both series
     
  2. jcsd
  3. May 22, 2013 #2

    Dick

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    Find the sum for (1/3)^n and (2/3)^n separately and then add them. They don't have to have a common value of a. The whole series isn't geometric. Only the parts are.
     
  4. May 22, 2013 #3
    Oh I know I mean, the a value for (1/3)^n is said to be 1/3 while the one for (2/3)^n is said to be 2/3. I don't understand how the this came to be. I thought a in each case was 1
     
  5. May 22, 2013 #4

    Dick

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    Your formula in the "Relevant equations" section isn't right or wrong until you put some limits on the summation like you did in the problem statement. What are they?
     
  6. May 22, 2013 #5
    oh sorry its n=1 to infinity. My bad I hadn't notice that!
     
  7. May 22, 2013 #6

    Dick

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    Then that makes the a's what the problem solution suggests, right?
     
  8. May 22, 2013 #7
    I feel so dumb.... I was so use to n-1 I assumed right away r = 1 due to the exponent for the first term. Dear, let's hope rest will make me smarter for my final. Thanks a lot Dick for bearing through this question haha!
     
  9. May 22, 2013 #8

    Dick

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    You are pretty smart if you figured that out so quickly given the hint. You'll do fine. Just pay attention to details. Like limits on summations.
     
    Last edited: May 22, 2013
  10. May 23, 2013 #9

    Ray Vickson

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    Don't forget that an infinite series is defined as the limit of a finite sum, so
    [tex] S = \sum_{n=1}^{\infty} \frac{1+2^n}{3^n} [/tex]
    is defined as
    [tex] S = \lim_{N \to \infty} \sum_{n=1}^{N} \frac{1+2^n}{3^n} [/tex]
    For finite N the sum can be split into two obvious parts, and both parts converge separately as ##N \to \infty.## It follows that the original infinite series is the sum of the two separate series, each of which is a convergent geometric series.
     
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