Valid rearrangements of infinite series -- where is the bijection when you split up a series? Suppose that you have a sequence of non-negative real numbers $\{a_n\}_{n\in\mathbb N}$.  I know that the value of the series $\sum_{n=1}^\infty a_n$ is unchanged if you permute the indices.
However, it's unclear to me whether this implies that, for instance, $\sum_{n=1}^\infty a_n = \sum_{k=1}^\infty a_{2k}+\sum_{k=1}^\infty a_{2k-1}$.  What permutation is being used here?  It seems like the idea behind this is not so much a permutation as it is "summing the evens first" and then the odds.  But it's not a permutation of $\Bbb N$ to map each number to the evens ... and then later map them to the odds.
Clearly what's going on here is
$$ \sum_{n=1}^\infty a_n = \sum_{k=1}^\infty (a_{2k-1}+a_{2k}) $$
and I'm pretty sure an $\varepsilon,\delta$ argument would prove this equation true.

The question: But it's not really an instance of permutation is it? Or am I missing something?
 A: One way to view this is that if all terms of a series are nonnegative then the sum can be defined as follows:
$$
\sum_{n\,\in\,I} a_n = \sup \left\{ \sum_{n\,\in\,I_0} : I_0 \subseteq I,\, I_0 \text{ is finite} \right\}.
$$
Then the task remains to show that this is equal to the sums defined by limits of partial sums in the various rearrangements-that-are-not-permutations, etc.
A: 
I know that the value of the series $∑^∞_{n=1}a_n$ is unchanged if you permute the indices.

Wrong. Google "Riemann rearrangement theorem". What you say is only true if the series converges absolutely.

$\sum_{n=1}^\infty a_n = \sum_{k=1}^\infty a_{2k}+\sum_{k=1}^\infty a_{2k-1}$

Consider the partial sums up to $2N$ on the RHS (if the general term at least goes to zero, it isn't too hard to argue that the odd case is the same). This is true for all series(absolute and conditional convergence) for which both the summations on the RHS exist. Otherwise, we can say nothing.
EDIT: The theorem you state is true if $a_n$ are non negative(because the convergence must be absolute), sorry about that. Also of course in this case, both the series on the RHS exist by the comparison test. To show the summations are equal, we don't need to concern ourselves with series anymore, we just need to know $\lim(a_n+b_n)=\lim a_n+\lim b_n$ when both exist individually.
A: A note about this question which, if I had known this when I asked the question, would have allowed me to research the question better: Tonelli's theorem (or the Tonelli-Fubini theorem) would resolve this question easily.  So anyone interested in this can search for proofs of Tonelli's theorem.
