limiting sum of a series??

**deej813** · December 7th 2009, 08:03 PM

the limiting sum of 3 + x + x^2 + ..... is 18. If |x|<1, find x
how do you do this??

**alexmahone** · December 7th 2009, 08:20 PM

Originally Posted by deej813

the limiting sum of 3 + x + x^2 + ..... is 18. If |x|<1, find x
how do you do this??

$3+x+x^2+...=18$

$3+\frac{x}{1-x}=18$

$\frac{x}{1-x}=15$

$x=15-15x$

$16x=15$

$x=\frac{15}{16}$

**deej813** · December 7th 2009, 08:27 PM

how does that work??
where does the x/1-x come from??

**Prove It** · December 7th 2009, 09:35 PM

Everything after the 3 is an infinite geometric series.

For any infinite geometric series of the form

$S_n = a + ar + ar^2 + ar^3 + \dots$

converges to $\frac{a}{1 - r}$ , provided $|r| < 1$ .

In your case, the infinite geometric series is

$S_n = x + x^2 + x^3 + x^4 + \dots$ .

Can you see what $a$ and $r$ have to be?

Can you see what it converges to if $|x| < 1$ ?

**deej813** · December 7th 2009, 09:46 PM

wouldn't it be 3/1-x cos isn't a=3??

**Prove It** · December 7th 2009, 09:56 PM

No.

I said the terms AFTER the 3.

In other words, your equation is

$3 + \left(x + x^2 + x^3 + \dots\right) = 18$ .

We are only dealing with the stuff inside the brackets.

Edit: It might help if you subtract the 3 to the other side.

So the equation becomes

$x + x^2 + x^3 + \dots = 15$ .

**deej813** · December 7th 2009, 10:01 PM

oh ok i get it
thanks

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