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Math Help - The sum (to infinity) of a geometric series.

  1. #1
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    The sum (to infinity) of a geometric series.

    I'm trying to calculate the sum of to infinity of the geometric series 99,33,11,...

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

    S_{\infty}  = \frac{99}{1-3} = -\frac{99}{2}

    Is that right?
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  2. #2
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    Hello, Captcha!

    Your r is wrong . . .


    Calculate the sum of the geometric series: . 99,\;33,\;11,\;\hdots
    Each term is divided by 3: . r \,=\,\tfrac{1}{3}

    . . S \;=\; \frac{a}{1-r} \;=\; \frac{99}{1-\frac{1}{3}} \;=\; \frac{99}{\frac{2}{3}} \;=\;\frac{297}{2}

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  3. #3
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    Quote Originally Posted by Captcha View Post
    I'm trying to calculate the sum of to infinity of the geometric series 99,33,11,...

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

    S_{\infty}  = \frac{99}{1-3} = -\frac{99}{2}

    Is that right?
    No, it isn't. For one thing, you should have noticed yourself that you are summing a series of positive numbers but the sum you got is negative. For another, along with that formula for the sum of an infinite geometric series, you should have learned that it is only valid for -1\le r< 1.

    The first term of a geometric series is a and the second is ar. Here, a= 99, alright, but (99)(3) is NOT 33. Your r is wrong. 99r= 33 so what is r?

    Oh, blast! Soroban got in 5 minutes before me. I need to learn to type faster!
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Captcha!

    Your r is wrong . . .


    Each term is divided by 3: . r \,=\,\tfrac{1}{3}

    . . S \;=\; \frac{a}{1-r} \;=\; \frac{99}{1-\frac{1}{3}} \;=\; \frac{99}{\frac{2}{3}} \;=\;\frac{297}{2}

    Ah, I somehow calculated r from 33r = 99. Thank you.
    Quote Originally Posted by HallsofIvy View Post
    No, it isn't. For one thing, you should have noticed yourself that you are summing a series of positive numbers but the sum you got is negative.
    Yeah. I should have.
    For another, along with that formula for the sum of an infinite geometric series, you should have learned that it is only valid for -1\le r< 1
    Yeah. I was told that. Why is that, though? Why does this formula not work for -1\le r< 1? How could we calculate the sum if r > 1 or r \le -1 ?
    The first term of a geometric series is a and the second is ar. Here, a= 99, alright, but (99)(3) is NOT 33. Your r is wrong. 99r= 33 so what is r?
    r = \frac{33}{99} = \frac{1}{3}, of course. Thanks.
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  5. #5
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    Quote Originally Posted by Captcha View Post
    Yeah. I was told that. Why is that, though? Why does this formula not work for -1\le r< 1? How could we calculate the sum if r > 1 or r \le -1 ?
    .
    HI

    The infinite GP formula actually came from this :

    S_n=\frac{a(1-r^n)}{1-r}

    =\frac{a-ar^n}{1-r}

    =\frac{a}{1-r}-\frac{ar^n}{1-r}

    Now , as n approaches infinity

    \lim_{n\rightarrow \infty}S_n=\frac{a}{1-r}-\frac{ar^n}{1-r}

    Consider this : (0.1)^{10} and 2^{10} .. which is approaching 0 ? It would be 0.1^{10} right . That's why in order for \frac{ar^n}{1-r} to be 0 , the ar^n has to first be 0 which would only happen if |r|<1 . If this condition isn't applied , then it would approach infinity instead of 0 .
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