Finite sub-sums of finite, countably infinite sums. Suppose that $I$ is a countable set and 
$$ \sum_{i \in I} X_i = X,$$
where $X \in \mathbb{R}$ (in particular $X$ is finite).  Does this mean that for all $\epsilon >0$ there exists a finite subset $J_{\epsilon} \subset I$ such that
$$X - \sum_{i \in J}X_i < \epsilon  ?
$$
So where did this question come from?  I'm currently studying analysis and I know that
$$\sum_{n=0}^\infty \frac{1}{2^n} = 2.$$
By the definition of a limit we have that for all $\epsilon >0$ there exists an $N \in \mathbb{N}$ such that for all $n>N$,
$$
2 - \sum_{n=0}^k \frac{1}{2^n} < \epsilon.
$$ Usually the proof of this relies (in my mind at least) on the fact that we've ordered the terms from largest to smallest.  
In terms of the first equation we have $I = \mathbb{N}_{\geq 0}$, $X_i = \frac{1}{2^i}$ and $X = 2$.  This is not the only possibility. We could have any other bijection $\alpha$ such that $$\alpha(X_i) = \frac{1}{2^{n_i}}.$$ 
Loosely speaking my question is: What do we do if we "didn't know the order" of the $X_i$? In my example this isn't really a problem because we know that there is some bijection but what happens if we have no additional information about the $X_i$ other than the infinite sum is finite?
 A: If $I$ is a countable set and $\{x_i:i\in I\}$ is a set of non-negative real numbers, then the sum can be defined unambiguously as
$$ \sum_{i\in I}x_i=\sup_{\substack{F\subset I\\|F|<\infty}}\sum_{i\in F}x_i $$
from which it it follows that if the sum is finite then for every $\varepsilon>0$ there is a finite set $F\subset I$ such that
$$ \sum_{i\in I}x_i-\sum_{i\in F}x_i<\varepsilon$$
Things are more complicated if some of the $x_i$ are negative. In that case, if
$$ \sum_{i\in I}|x_i|<\infty $$
then once again the order of summation doesn't matter. But otherwise, the order will be important. For instance, there's a theorem of Riemann that states that if a series $\sum_{n=1}^{\infty}a_n$ is conditionally convergent, then for any $\alpha\in[-\infty,\infty]$ there is a bijection $\sigma:\mathbb{N}\to\mathbb{N}$ such that
$$ \sum_{n=1}^{\infty}a_{\sigma(n)}=\alpha$$
A: Yes. Since the set is countable, you are really just talking about a sum
$$\sum_{i=1}^\infty x_i=x\in\Bbb R.$$
Since the sum converges, the sequence of partial sums converges, so
$$\lim_{n\to\infty}\left|\sum_{i=1}^n x_i-x\right|=0.$$
A: If at most finitely many of the $X_i$ are negative or at most finitely many are positive, it turns out that the order doesn’t matter: if there is an $X\in\Bbb R$ such that for each $\epsilon>0$ there is a finite $J_\epsilon\subseteq I$ such that 
$$\left|X-\sum_{i\in J_\epsilon}X_i\right|<\epsilon\;,$$
then every arrangement of $\sum_{i\in I}X_i$ as a standard series converges to $X$, and if there is no such $X$, then no arrangement as a standard series converges.
If there are infinitely many negative and infinitely many positive $X_i$, however, the order can matter. For instance, if $I=\Bbb Z^+$, and $x_i=\frac{(-1)^{i+1}}i$ for each $i\in I$, then in the natural order we get the alternating harmonic series
$$\sum_{i\ge 1}\frac{(-1)^{i+1}}i=1-\frac12+\frac13-\frac14+\ldots\;,$$
which converges, but the Riemann series theorem says that it can be rearranged to converge to any real number, to diverge to $-\infty$ or $+\infty$, or to do neither of these. This holds for any conditionally convergent series.
