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Proper form for geometric series with a neg inside

  1. Nov 19, 2013 #1
    1. The problem statement, all variables and given/known data

    Curious about this ...I have to find the sum.


    2. Relevant equations



    3. The attempt at a solution

    Ʃ (1/4)(-1/3)^n from 1 to infinity

    I want to know the proper form and why. Is it (1/4) Ʃ (-1/3)(1/3)^n-1 or (1/4) Ʃ (1/3)(-1/3)^(n-1)

    You get different answers
     
  2. jcsd
  3. Nov 19, 2013 #2

    Mark44

    Staff: Mentor

    What's the point of bringing out a factor of 1/3? You did the same thing in your other thread.

    In the above, your first formula is incorrect. (-1/3)(1/3)n-1 = (-1/3)[1 + 1/3 + 1/9 + ...

    For a geometric series such as ##\sum_{n = 1}^{\infty}r^n##, if |r| < 1, the series converges to r/(1 - r).

    Here's a simplistic explanation:
    Let S = r + r2 + ... + rn + ...
    Then -rS = -r2 - r3 - ... - rn + 1 - ...

    Then S - rS = r, or S(1 - r) = r, so S = r/(1 - r)
     
  4. Nov 19, 2013 #3
    I'm not sure we take out the factor in class because that is how it is written in general in the book formula.

    I thought the sum of a geometric series was S = A / (1 -r) where A is the coefficient in front r. Not r/ (1-r)
     
  5. Nov 19, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Go back and read it again. The exact formula to use depends on whether you have the sum ##\sum_{n=0}^{\infty} A r^n## or ##\sum_{n=1}^{\infty} A r^n##. Do you notice the difference?
     
  6. Nov 19, 2013 #5
    Yeah n = 0 or n = 1. Your first one is OK your second one isn't it needs to be n-1
     
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