# Rearrangements of absolutely convergent series

Accounts of absolute and conditional convergence always say that $$\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty a_{\sigma(n)}$$ if the series converges absolutely, if $\sigma$ is any bijection from the set of indices to itself.

Fubini's theorem tells us that $$\sum_{n=0}^\infty\sum_{m=0}^\infty a_{n,m} = \sum_{m=0}^\infty\sum_{n=0}^\infty a_{n,m}$$ if we have absolute convergence. (Tonelli's theorem tells us that these are equal if all terms are nonnegative, regardless of whether the series converges or not.)

But in this answer, I explained how to rearrange $$\sum_{n=0}^\infty\sum_{m=0}^\infty a_{n,m}$$ into $$\sum_{n=0}^\infty \sum_{k=0}^n a_{k,n-k}.$$ What known theorems apply to this case, and where is this case mentioned in the literature?

How exotic does the class of rearrangements of series get? What other kinds of rearrangements don't fit instantly into one of the cases covered by the two categories above that are covered by standard results?

(I might post my own answer to this.)

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I've posted my own answer, but other arguments for this result may also be of interest, so don't hesitate to add those. – Michael Hardy Jun 7 '13 at 22:09
@MinimusHeximus : I really don't think you should have more than two "$\sum$"s here. – Michael Hardy Jun 8 '13 at 15:46

It seems to me that you are alluding to the theory of unordered summation, which is discussed (for instance) in $\S$ 14.2.3 of these notes.

In particular it is discussed that unordered summation is equivalent to absolute convergence, and the application to double series is given as an exercise.

(In my opinion calling this phenomenon "Fubini's Theorem" is a bit fancier than necessary.)

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Fubini's theorem gives us more than equality of the two iterated sums, one of which is $$\sum_{n=0}^\infty {\Huge(} \underbrace{{}\qquad\sum_{m=0}^\infty a_{n,m}\qquad{}}_{\begin{smallmatrix}\text{n remains constant} \\ \text{as m goes through} \\ \text{the whole range.}\end{smallmatrix}} {\Huge)}.$$ Fubini's theorem tells us in addition that the two iterated sums are both equal to $$\sum \{ a_{n,m} : (n,m) \in (\mathbb Z^{\ge0})^2 \},$$ where that sum is defined as follows. Let $P=\{(n,m) : a_{n,m}\ge0\}$ and $N=\{(n,m) : a_{n,m}<0\}$. Then the sum is $$\sup\left\{ \sum_{(n,m)\in A} a_{n,m} : A\subseteq P\text{ and } A\text{ is finite} \right\} - \sup\left\{ \sum_{(n,m)\in A} -a_{n,m} : A\subseteq N\text{ and } A\text{ is finite} \right\}.$$ Then the first proposition on rearrangements stated above tells us that this is equal to $$\lim_{N\to\infty} \sum_{k=1}^N a_{n_k,m_k}$$ for every enumeration $(n_k,m_k)_{k=1}^\infty$ of $(\mathbb Z^{\ge0})^2$.

After that, observe that $$\sum_{n=0}^\infty\left(\sum_{k=0}^n a_{k,n-k} \right)$$ is simply one such enumeration.

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