# Quick help in Geometric series question

1. Dec 11, 2007

### rock.freak667

1. The problem statement, all variables and given/known data
The common ratio,ratio,r, of a geometric series is given by:
$$r=\frac{5x}{4+x^2}$$

Find all the values of x for which the series converges

2. Relevant equations

3. The attempt at a solution

For the series to converge |r|<1
so that
$$|\frac{5x}{4+x^2}|<1$$

this means that:

$$-1<\frac{5x}{4+x^2}<1$$

For

$$-1<\frac{5x}{4+x^2}$$
I get

{x:x<-4}U{x:x>-1}

and for

$$\frac{5x}{4+x^2}<1$$

I get
{x:x<1}U{x:x>-1}

The answer I want would be the region that satisfies both {x:x<1}U{x:x>-1} AND {x:x<-4}U{x:x>-1} right?

So I get

{x:x<1}U{x:x>-1} which is basically {x:-1<x<1}...(*answer I got)
is that correct? because I see that the answer is {x:-1<x<1}U{x:x<-4}U{x:x>4}
I am not sure if that is the same as what I have because I think that the answer I got includes the rest of {x:x<-4}U{x:x>4}

2. Dec 12, 2007

### wildman

no. You are incorrect.

The answer you got includes all reals: x > -1 in union with x < 1 includes all numbers less than one and greater than -1 or all reals. And all reals is not correct (take 3 as a counter example).

The answer in the book is correct: {x:-1<x<1}U{x:x<-4}U{x:x>4}

3. Dec 12, 2007

### dynamicsolo

I'm finding in this case (x+1)(x+4) > 0 , so either

x>-1 and x>-4 implies x>-1
or
x<-1 and x<-4 implies x<-4

So we are in agreement here.

In this case, I find (x-1)(x-4) > 0 , so either

x>1 and x>4 implies x>4
or
x<1 and x<4 implies x<1

Perhaps you "miswrote" yourself here?

Combining the two cases gives

x < -4 , -1 < x < 1 , x > 4 .

As a check, you'll find that values of x with absolute value between 1 and 4 don't work in the original inequality.

Last edited: Dec 12, 2007