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Math Help - Finding the value of infinite sums

  1. #1
    Senior Member chella182's Avatar
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    Finding the value of infinite sums

    Okay, I've got this example in my notes and I don't understand it. My notes say...

    Find \sum_{k=0}^{\infty}3^{-k}.
    Step 1 Find S_{1}, S_{2}...
    Step 2 Find the limit.

    S_{1}=\frac{1}{1\times2}=1-\frac{1}{2}

    But I don't understand that last bit. I know it's true (obviously) but why write it like that?
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    It's a simple geometric series. Other than that, I would not know why you wrote what you did.

    \sum_{k=0}^{\infty}\frac{1}{3^{k}}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}
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  3. #3
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    S_n=\frac{1-3^{-(n+1)}}{1-3^{-1}}=\frac{3}{2}(1-\frac{1}{3^{n+1}}).
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  4. #4
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    Quote Originally Posted by chella182 View Post
    Okay, I've got this example in my notes and I don't understand it. My notes say...

    Find \sum_{k=0}^{\infty}3^{-k}.
    Step 1 Find S_{1}, S_{2}...
    Step 2 Find the limit.

    S_{1}=\frac{1}{1\times2}=1-\frac{1}{2}

    But I don't understand that last bit. I know it's true (obviously) but why write it like that?

    I would do ths problem by looking at the first 3 terms in the series

    From

    \sum_{k=0}^{\infty}3^{-k}

    term 1 t_1 for n=0 is 3^{-0} = 1

    term 2 t_2 for n=1 is 3^{-1} = \frac{1}{3}

    term 3 t_3 for n=2 is 3^{-2} = \frac{1}{9}

    giving the series {1 ,\frac{1}{3},\frac{1}{9}}\cdots

    From here I would find r = \frac{t_3}{t_2} = \frac{t_2}{t_1} = \frac{1}{3}

    and 'a' the first term = 1

    From here

    \lim_{n\to\infty} \frac{1-r^{n+1}}{1-r}

    and for the sum of infinate series

    S_{\infty} \frac{a}{1-r}
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