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

**Hellbent** · February 6th 2011, 11:28 AM

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
$\displaystyle r = \frac{2x - 3}{x + 4}.$

**TheEmptySet** · February 6th 2011, 11:30 AM

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?

**Hellbent** · February 8th 2011, 01:35 AM

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

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

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

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

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

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

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

$\displaystyle -\frac13 < x < 7$