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Math Help - Values of x for a convergent geometric series with common ratio r

  1. #1
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    Values of x for a convergent geometric series with common ratio r

    Hi,

    How do I go about doing this?
    (Sorry, I don't know how to start, therefore I have no attempt.)

    Determine the values of x for a convergent geometric series with common ratio
    \displaystyle r = \frac{2x - 3}{x + 4}.
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    Hi,

    How do I go about doing this?
    (Sorry, I don't know how to start, therefore I have no attempt.)

    Determine the values of x for a convergent geometric series with common ratio
    \displaystyle r = \frac{2x - 3}{x + 4}.
    A geometric series converges if

    \displaystyle |r| < 1 \iff \bigg| \frac{2x-3}{x+4}\bigg| < 1

    Can you finish from here?
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  3. #3
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    \displaystyle \bigg| \frac{2x-3}{x+4}\bigg| < 1

    \displaystyle \left(\frac{2x-3}{x+4}\right)^2 < 1

    (2x + 3)^2 < (x + 4)^2

    4x^2 - 12x + 9 - x^2 - 8x - 16 < 0

    3x^2 - 20x - 7 < 0

    3x^2 - 21x + x - 7 < 0

    (3x + 1)(x - 7) < 0

    \displaystyle -\frac13 < x < 7
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  4. #4
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    Looks good.

    You can also do it without squaring by looking at the 3 cases:

    x<-4

    -4\leq x\leq 3/2

    x>3/2
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