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If I have a sum $\sum_{n = -\infty}^{\infty} a_{n}$, I say that this equals $\lim_{N \rightarrow \infty} \sum_{n = -N}^{N} a_{n}$ when "the summation is absolutely convergent". I know that if we have $\sum_{n = 0}^{\infty} a_{n}$, then the series is said to be absolutely convergent if $\sum_{n = 0}^{\infty} |a_{n}| < \infty$. But how is absolute convergence defined when we have $\sum_{n = -\infty}^{\infty} a_{n}$?

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    It should be pointed out, however, that the limit $\lim\limits_{N \to \infty} \sum_{n=-N}^{N} a_n$ can exist without absolute summability: For a very simple example take $a_k = 1/k$ if $k \neq 0$ and $a_0 = 0$.2011-10-08

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If a sum is absolutely convergent, then the limit is independent of the order of summation. In particular, when a sum has only positive summands, then either it converges in any order or it does not converge in any order.

$\sum_{-\infty}^\infty a_n$ is absolutely convergent iff for every bijection $f:\mathbb N\to\mathbb Z$, $\sum_{i=0}^\infty a_{f(n)}$ is. But this is the same as saying that $\lim\sum_{-N}^N|a_n|<\infty$ which is the same as saying that there is $M\in\mathbb R$ such that for every finite set $S\subseteq\mathbb Z$, $\sum_{n\in S}|a_n|.

In general, if $I$ is an arbitrary index set, a sum $\sum_{i\in I}a_i$, with all $a_i$ real numbers, is absolutely convergent if there is $M\in\mathbb R$ such that for every finite set $S\subseteq I$, $\sum_{i\in S}|a_i|. In this case $\sum_{i\in I}a_i$ is defined to be $\sup\{\sum_{i\in S}a_i:S\subseteq I\mbox{ is finite }\}$.