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Whilst reading the answers to this question, one of the answers states:

"The problem of series that are not absolutely convergent is that you can't make arbitrary rearrangement of the terms."

However, my understanding was that due to the commutative law, we could write the following (c.f. Concrete Mathematics):

$$\sum_{k\in\mathbb{K}}{a_{k}}=\sum_{p(k)\in\mathbb{K}}{a_{p(k)}},$$

For any permutation function $p(k):\mathbb{Z}\to\mathbb{Z}, \space\forall k \in\mathbb{Z}$, in which all integers are mapped to an integer without duplicity, and for all sequences $a_{k}$.

Is this true even for conditionally converging series, if so, is my confusion based on the author of that answer's definition of "arbitrary rearrangement" (I'm assuming that a permutation is a form of "arbitrary rearrangement")?

Thanks in advance!

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Absolutely convergent and unconditionally convergent are identical. They will always converge to the same sum under arbitrary rearrangements. However, series that fail to be absolutely/unconditionally convergent can be rearranged to converge to any given sum, or to diverge.

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The distributive, associative, and commutative laws on p. 30 of Concrete Mathematics, formulas (2.15)-(2.17), are stated only for finite $K$. You cannot freely extend them to infinite index sets.

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  • $\begingroup$ So they are, my mistake. Is this because there is no function which maps the entirety of $\mathbb{N}$ to a rearrangement of $\mathbb{N}$ including every element once and only once? $\endgroup$
    – Thomas Russell
    Jun 17, 2012 at 14:43
  • $\begingroup$ @Shaktal: There are many such functions. The problem, as Cocopuffs and Cameron have noted, is that the commutative law simply doesn’t hold for all infinite series. Specifically, it fails violently for conditionally convergent series that aren’t absolutely convergent. $\endgroup$
    – Brian M. Scott
    Jun 17, 2012 at 14:45
  • $\begingroup$ @Shaktal: Not at all. There are in fact infinitely-many such functions. Some of them won't even alter the sum. Check out the link Cocopuffs gave for more details. $\endgroup$
    – Cameron Buie
    Jun 17, 2012 at 14:45
  • $\begingroup$ Okay, I will have to look further into this then, I'm still slightly amiss, thanks though guys! :) $\endgroup$
    – Thomas Russell
    Jun 17, 2012 at 14:47

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