Rearrangement of infinite series

1. Nov 5, 2005

happyg1

Hi,
I'm working on this problem:
Prove that if |x|<1, then
1 + x^2 + x + x^4 + x^6 + x^3 + x^10 + x^5 + ...= 1/(1-x).
I know that this is a rearrangement of the (absolutely convergent) geometric series, so it converges to the same limit. My trouble is proving that the rearrangement represents a 1-1 and onto mapping of the natural number onto themselves. I wrote out the terms and found the pattern, but I can't seem to get a formula for it. I'm stuck. Here's what I got:The first column represents the geometric series. The number after the colon is the rearrangement, and the last column is how I related the rearrangement to the origional goemetric series.

n=0:0==>n
n=1:2==>n+1
n=2:1==>n-1
n=3:4==>n+1
n=4:6==>n+2
n=5:3==>n-2
n=6:8==>n+2
n=7:10==>n+3
n=8:5==>n-3
n=9:12==>n+3
.
.
.
It has a definite pattern, and I can see it, but I can't write a formula down that works so that I can show it's bijective. I'm not sure that I am going about this correctly.
Any input will be appreciated.
CC

2. Nov 5, 2005

benorin

Hint, perhaps.

How do you construct a bijection from $$\mathbb{N}$$ to $$\mathbb{Z}$$?

3. Nov 5, 2005

happyg1

I let 0=1, -n=n+1 and n=2n+1...or something similar.
How does that help me map $$\mathbb{N}$$ to $$\mathbb{N}$$ in my situation?

4. Nov 5, 2005

happyg1

any help? Any thoughts?

5. Nov 5, 2005

Zurtex

Seems pretty simple to me, for n, some natrual number.

3n ==> 4n
3n + 1 ==> 4n + 2
3n + 2 ==> 2n + 1

Doesn't take much beyond that to work it out.

6. Nov 6, 2005

happyg1

thanks! I just had a block on getting the formula for the pattern. I now have it solved! AWESOME!!