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

• Feb 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 $\displaystyle x$ for a convergent geometric series with common ratio
$\displaystyle \displaystyle r = \frac{2x - 3}{x + 4}.$
• Feb 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 $\displaystyle x$ for a convergent geometric series with common ratio
$\displaystyle \displaystyle r = \frac{2x - 3}{x + 4}.$

A geometric series converges if

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

Can you finish from here?
• Feb 8th 2011, 01:35 AM
Hellbent
$\displaystyle \displaystyle \bigg| \frac{2x-3}{x+4}\bigg| < 1$

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

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

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

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

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

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

$\displaystyle \displaystyle -\frac13 < x < 7$
• Feb 8th 2011, 02:22 AM
DrSteve
Looks good.

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

$\displaystyle x<-4$

$\displaystyle -4\leq x\leq 3/2$

$\displaystyle x>3/2$