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

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February 6th 2011, 11:28 AM
Hellbent

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
$\displaystyle r = \frac{2x - 3}{x + 4}.$
February 6th 2011, 11:30 AM
TheEmptySet

Quote:

Originally Posted by Hellbent

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?
February 8th 2011, 01:35 AM
Hellbent

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

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

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

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

$3x^2 - 20x - 7 < 0$

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

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

$\displaystyle -\frac13 < x < 7$
February 8th 2011, 02:22 AM
DrSteve

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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