We're using the definition and we're doing the proof from pages 113-114, Elementary Real Analysis by Bruckner et. al.
The definition is:
The proof requires that we show that $\left \lvert \sum_{-N \leq i \leq N} 2^{-|i|} - 3 \right \lvert< \epsilon/2$, which acts as the sum from the $I_0$ set, and that we also show the sum which extends it is also such that $\sum_{i \in J/ I_0} 2^{-|i|} < \epsilon/2$. Then we get:
$$\left \lvert \sum_{i \in J} 2^{-|i|} -3 \right \lvert = \left \vert \sum_{-N \leq i \leq N} 2^{-|i|} - 3 \right \lvert + \sum_{i \in J / I_0} 2^{-|i|} < \epsilon$$
which satisfies the definition.
First, we show, using properties of geometric sums, $$\left \lvert \sum_{-N \leq i \leq N} 2^{-|i|} - 3 \right \lvert = 2(2^{-N})< \epsilon/2$$
And now I've gotten stuck at the next part, show that $\sum_{i \in J/ I_0} 2^{-|i|} < \epsilon/2$. Now, I'm able to prove the statement only by assuming a certain structure, which prevents me from concluding about an unordered sum. My question is, how do I prove it without assuming such structure, or how do I prove for whatever structure the sum might have so that it holds for the unordered sum?
Here's my attempt, making some assumptions about the structure of the sum:
Suppose a finite $K \subset \mathbb{Z}$ with it's elements satisfying $|k| > N$. (What follows is the assumptions I need to get rid of:) We suppose $K$ has size $\|K\|$, and suppose there is an identical number of positive and negative elements, and so by letting $\beta := \frac{\|K\|}{2}$, we can write: $$2^{-|-N-\beta|} + \cdots + 2^{-|-N-1|}+2^{-(N+1)}+\cdots+2^{-(N+\beta)}$$ which is identical to: $$2^{-(N+\beta)} + \cdots + 2^{-(N+1)}+2^{-(N+1)}+\cdots+2^{-(N+\beta)}$$ so we can do: $$=2(2^{-(N+1)}+\cdots+2^{-(N+\beta)})$$ $$=2\cdot2^{-N}(2^{-1}+\cdots+2^{-\beta})$$
using the properties of geometric sums:
$$=2^{-N+1}\left(\frac{-1+2^{-\beta}}{1-2}\right)$$ $$=2^{-N+1} - 2^{-\beta}$$
From which we get that it is smaller than $2(2^{-N})$, and so smaller than $\epsilon/2$, which allows us to complete the proof.
But of course, we can't do that since there are other possibilities for the structure of the $\sum_{i \in J/ I_0} 2^{-|i|}$, e.g., an uneven number of positive elements and negative elements. I would need to somehow show that all of these are smaller than the one I've chosen, or something along those lines. Does anyone have suggestions? Thank you.
Let $M=max_{i\in J}\,|i|$. Then $\sum_{i\in J\setminus I_0}2^{-|i|}\le\sum_{N<|i|\le M}2^{-|i|}$, which you know how to compute.