1. ## Convergent Geometric Series

Suppose we have two convergent geometric series ∑x^n and ∑y^n. Does it ever occur that ∑x^n∑y^n=∑[(x^n)(y^n)]?

THANKS.

2. Originally Posted by eg37se
Suppose we have two convergent geometric series ∑x^n and ∑y^n. Does it ever occur that ∑x^n∑y^n=∑[(x^n)(y^n)]?
Try $\displaystyle x=\frac{-1}{2}~\&~y=\frac{1}{4}$.

There is a general solution. Can you find it?

3. Originally Posted by eg37se
Suppose we have two convergent geometric series ∑x^n and ∑y^n. Does it ever occur that ∑x^n∑y^n=∑[(x^n)(y^n)]?

THANKS.
I would say yes. Assume that the lead term is $\displaystyle 1$ and that $\displaystyle n$ starts at $\displaystyle n = 0$ then

$\displaystyle \sum_{n=0}^\infty x^n = \frac{1}{1-x},$ $\displaystyle \sum_{n=0}^\infty y^n = \frac{1}{1-y}$ and $\displaystyle \sum_{n=0}^\infty x^n y^n = \frac{1}{1-xy}$

so what you're asking is are there soltions to

$\displaystyle \frac{1}{1-x} \cdot \frac{1}{1-y} = \frac{1}{1-xy} \;\; \text{on}\;\; (-1,1) \times (-1,1)$. I believe there are an infinite number of solutions to this equation on this interval.

4. Originally Posted by Plato
Try $\displaystyle x=\frac{-1}{2}~\&~y=\frac{1}{4}$.

There is a general solution. Can you find it?
Wouldn't it just be (1-(1/8)+(1/64)-(1/256)+...)
which is just ∑(-1/8)^n?

5. Originally Posted by eg37se
Wouldn't it just be (1-(1/8)+(1/64)-(1/256)+...)
which is just ∑(-1/8)^n?
If you substitute the x and y values that Plato suggests into the forumula I give

Originally Posted by Danny
$\displaystyle \frac{1}{1-x} \cdot \frac{1}{1-y} = \frac{1}{1-xy} \;\; \text{on}\;\; (-1,1) \times (-1,1)$.
I believe that you'll show they're identical.

6. Got it. Thank you.