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On page 168 of Mathematical Fallacies and Paradoxes, it states that the fact that the series

$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots $

has a sum depends on the Axiom of Choice. Where does the AC come in to play? I know that if the terms are permuted, we can get any sum we want, and I can see how the AC might be involved there, but just the fact that $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots $ converges?

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  • $\begingroup$ Perhaps the author means that one has to pick an arrangement out of the uncountably many possible arrangements. $\endgroup$
    – Ryan Thorngren
    Commented Apr 23, 2010 at 2:00
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    $\begingroup$ The text says that "The theorem that was called upon to show that [this series] has a sum depends on the axiom of choice" and this is a little different than saying that the convergence of this particular series depends on AC. $\endgroup$
    – Joel David Hamkins
    Commented Apr 23, 2010 at 2:04
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    $\begingroup$ It doesn't quite say that; it says that "the theorem that was used to show" that the series has a sum depends on the axiom of choice. It isn't clear from that passage what theorem they refer to. It may be possible to prove the fact without using that theorem. Later on that page, there is an allusion to some proof that the reals have greater cardinality than the integers that also made use of the axiom of choice, presumably a proof that was discussed earlier in the book. But unless I'm greatly mistaken, this fact doesn't rely on the axiom of choice either. $\endgroup$
    – Nate Eldredge
    Commented Apr 23, 2010 at 2:07
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    $\begingroup$ The statement in the book seems clearly false to me. On the other hand, the title is Mathematical Fallacies and Paradoxes, so caveat lector, I suppose. $\endgroup$
    – Pete L. Clark
    Commented Apr 23, 2010 at 2:13
  • $\begingroup$ Nate, yes, you don't need AC to prove that R is uncountable. $\endgroup$
    – Joel David Hamkins
    Commented Apr 23, 2010 at 2:20

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It doesn't seem to me that you need any choice principle at all to prove that this series converges. The Alternating Series Test that appears in any elementary calculus books seems to do the job, and doesn't seem to require any amount of AC. If $\Sigma_n (-1)^n a_n$ is an alternating series, with $a_n$ descending to $0$, then the finite partial sums up to positive term are descending and the partial sums up to a negative term are increasing, and that these two sequences converge to the same limit.

So I'm not sure which theorem had been "called upon to show" that the series converges, but unless I am mistaken, it must have been something other than what our students would call the Alternating Series Test.

Edit. More generally, I claim that the convergence of a series can never depend on the Axiom of Choice. Suppose that $r=\langle a_0,a_1,\ldots\rangle$ is a sequence of real numbers, and we are consider the series $\Sigma a_n$. The assertion "$\Sigma a_n$ converges" is a statement about $r$ having complexity $\Sigma^1_1(r)$, that is, an analytic fact about $r$, and therefore has the same truth value in the set theoretic universe $V$ as it has in the relativized constructible universe $L[r]$, where AC holds. In particular, the series converges in $V$ if and only if it converges in $L[r]$.

Conclusion. If you have a series $\Sigma a_n$ defined in a sufficiently concrete manner and you can prove in ZFC that it converges, then you can prove in ZF that it converges.

(The technical requirement here about sufficiently concrete is that the description of $r=\langle a_0,a_1,\ldots\rangle$ should be absolute from $V$ to $L[r]$. This would be true of any arithmetically definable series as in the question, defined by an arithmetic formula, or even a Borel definition or a $\Sigma^1_2$ definition.)

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  • $\begingroup$ I guess the complexity is even much better than $\Sigma^1_1(r)$, since the assertion that the sequence of partial sums is Cauchy is an arithmetic statement about $r$. $\endgroup$
    – Joel David Hamkins
    Commented Apr 23, 2010 at 4:22

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