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Math Help - Sum 3-1+1/3-1/9+...=9/4

  1. #1
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    Sum 3-1+1/3-1/9+...=9/4

    I have numerically verified that the following sum comes to \frac 94, the same thing the back of the book says. So what'd I do wrong?

    Find the sum of the convergent series
    3-1+\frac 13 - \frac 19 + \mathellipsis.

    I verified that the above sum can be represented as
    \sum_{n \ge 1} \frac{(-1)^{n+1}9}{3^n},

    which can be rewritten as
    \sum_{n \ge 1} \frac{(-1)^{n+1}9}{3^n} = -9\sum_{n \ge 1} \frac{(-1)^{n}}{3^n} = -9\sum_{n \ge 1} \left(\frac{-1}{3}\right)^n.

    I believe I can apply the geometric series to the summation, giving me
    -9\sum_{n \ge 1} \left(\frac{-1}{3}\right)^n = -9 \cdot \frac{1}{1+\frac 13} = -9 \cdot \frac 34 = -\frac{27}{4} \ne \frac 94.

    Ahhh... but now I see that the geometric series equation starts the summation at n=0 (do you find that typing up your answers sometimes helps you realize what's wrong? Why is that?).

    So, I need to subtract -9 \cdot \frac{-1}{3}^0 = -9 from my answer, which (unsurprisingly) gives \frac 94.
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  2. #2
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    Re: Sum 3-1+1/3-1/9+...=9/4

    Hey MSUMathStdnt.

    Given the convergence criteria the sum should be 3 - (1/[1+1/3]) = 3 - 3/4 = 12/4 - 3/4 = 9/4
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  3. #3
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    Re: Sum 3-1+1/3-1/9+...=9/4

    Hello, MSUMathStdnt!

    \text{Find the sum: }\:3-1+\frac 13 - \frac 19 + \hdots

    This is a geometric series with first term a = 3 and common ratio r = \text{-}\tfrac{1}{3}

    The sum is: . S \:=\:\frac{a}{1-r}

    Therefore: . S \;=\;\dfrac{3}{1-\left(\text{-}\frac{1}{3}\right)} \;=\;\frac{3}{\frac{4}{3}} \;=\;\frac{9}{4}

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