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Series Question

  1. Apr 29, 2007 #1
    1. The problem statement, all variables and given/known data
    (infinity)sigma(k = 0) [2(2/6)^k + (-2/10)^k)


    2. Relevant equations
    Geometric Series


    3. The attempt at a solution
    I split these up into two geometric series

    (infinity)sigma(k = 0) [2(1/3)^k]
    2 / (1 - 1/3)
    r = 3

    This diverges.

    (infinity)sigma(k = 0) (-1/5)^k
    1 / (1 + 1/5)
    r = 5/6

    This converges.

    Now, I'm not sure what to do? If it does converge, I need to find its sum o_O.
     
  2. jcsd
  3. Apr 29, 2007 #2

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    The first series doesn't diverge. What makes you think it does?
     
  4. Apr 29, 2007 #3
    Oh oops! I see, the |r| value is the ratio. Not the number you get after. The number you get after is what the series converges to?

    So 3 + 5/6 = 18 / 6 + 5 /6 --> 23 / 6

    Would that be correct?
     
  5. Apr 29, 2007 #4

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    Yea. Although to be more rigorous, you need to show those two series converge absolutely to justify splitting the sum. This is only a problem for the second one, which is an alternating series, but taking the absolute value of each series gives another converging series, so you're ok.

    This might sound like a technicality, and it is, but there are examples where your method would give a wrong answer. For example, it wouldn't work on the series:

    [tex] \sum_{n=1}^\infty ((-1)^n+(-1)^{n+1}) [/tex]

    There are also examples where it would give you a finite answer, but which would be wrong. These are rare, and I can't think of one off hand, but I'll post one if I do.
     
    Last edited: Apr 29, 2007
  6. Apr 29, 2007 #5
    So to prove that, I would use the ratio test to show absolute convergence? Absolute convergence was very confusing reading from the text book.

    I forgot there were other ways to show absolute convergence. Like using harmonic, or comparison or something.
     
  7. Apr 29, 2007 #6

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    A series [itex]\sum a_n[/itex] converges absolutely iff [itex]\sum |a_n|[/itex] converges. If the series is already positive, like the first one, you're done. Since the second isn't, you need to show that by taking the absolute value, which just amounts to changing -1/5 to 1/5, you still get a series that converges.
     
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