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Math Help - Geometric Series Question

  1. #1
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    Geometric Series Question

    Given that the sum to infinity of a G.P. is denoted by S.

    Show that if a > 0, then S > 4ar except for one value of r.
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  2. #2
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    Re: Geometric Series Question

    Hello, moonchildvulcan!

    \text{Given that the sum to infinity of a G.P. is denoted by }S.

    \text{Show that if }a > 0,\,\text{ then }S > 4ar\,\text{ except for one value of }r.

    The sum of an infinite G.P. is:. \frac{a}{1-r},\,\text{ for }|r| < 1.

    If S is equal to 4ar, we have: . \frac{a}{1-r} \:=\:4ar

    . . \frac{1}{1-r} \:=\:4r \quad\Rightarrow\quad 1 \:=\:4r-4r^2 \quad\Rightarrow\quad 4r^2 - 4r + 1 \:=\:0

    . . (2r-1)^2\:=\:0 \quad\Rightarrow\quad 2r-1 \:=\:0 \quad\Rightarrow\quad \boxed{r \:=\:\tfrac{1}{2}}
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  3. #3
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    Re: Geometric Series Question

    If you start with a(2r-1)^2\ge 0 you can arrive at your conclusion.
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  4. #4
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    Re: Geometric Series Question

    Thanks for that. I still cannot see why S > 4ar in the first place. The rest that follows, I managed to achieve.
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  5. #5
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    Re: Geometric Series Question

    Quote Originally Posted by a tutor View Post
    If you start with a(2r-1)^2\ge 0 you can arrive at your conclusion.
    Sorry but why would you start with this? On what basis? It's establishing why S > 4ar in the first place that is bugging me. I am happy with solving the inequality that follows to show that r = 1/2 is the exception.
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  6. #6
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    Re: Geometric Series Question

    I started by assuming the conclusion to see where it lead.

    \frac{a}{1-r}>4ar

    \Rightarrow a>4ar-4ar^2

    \Rightarrow a(4r^2-4r+1)>0

    \Rightarrow a(2r-1)^2>0

    but this works in reverse too so starting with a(2r-1)^2 > 0 if r \ne \frac{1}{2} you get the result you want.
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