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Find the Sum of The Geometric Series

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Ʃ(1 to infinity) (2/3)^(3n)


    2. Relevant equations

    For a geometric series, the series converges to a/1-r

    3. The attempt at a solution

    I'm really just confused about how to manipulate this so that it has a form of ar^n, especially since it starts at 1 rather than 0. I know that since the series starts at 1, we ought to have it in the form ar^n-1, but I'm at a loss as to how to go about.

    I'm not really concerned with the answer so much as how one would go about getting it, and really understanding the process. It seems I'm having a hard time figuring out how to manipulate series and get them in the correct forms...

    Any help in the right direction would be great. Thanks!

    This is what I've come up with, but I'm not sure if it's correct...

    We can rewrite the series as ((2/3)^3)^k = (8/27)^(k-1) * (8/27)

    a = 8/27
    r = 8/27

    Computing a/1-r, I get 8/19...
     
    Last edited: May 6, 2012
  2. jcsd
  3. May 6, 2012 #2
    Well, to fix the index problem, try just changing the index from 1 to 0 on your sum, and then subtracting off whatever you added to put the series in the correct form. Then the total sum should remain unchanged.
     
  4. May 6, 2012 #3
    Thanks, I think I got it. I just get totally thrown off when the index isn't in the correct form, or when the series is blatantly obvious, haha.

    Thank you, though!
     
  5. May 6, 2012 #4

    LCKurtz

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    The beginning index doesn't matter. In the sum ##\frac a {1-r}##, ##a## represents the first term, whatever it is with the starting index and ##r## is the common ratio. What did you get for ##r##?
     
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