axiom-of-choicelo.logicset-theory
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?
I have enjoyed the other answers very much. But perhaps it would be desirable to balance the discussion somewhat with a counterpoint, by mentioning a few of the counter-intuitive situations that can occur when the axiom of choice fails. For although mathematicians often point to what are perceived as strange consequences of AC, many of the situations that can arise when one drops the axiom are also quite bizarre.
The best answer I've ever heard --- and I think I heard it here on MathOverflow from Mike Shulman, which suggests that this question is roughly duplicated somewhere else --- is that you should care about constructions "internal" to other categories:
In all of these cases, the "axiom of choice" fails. In fact, from the internal-category perspective, the axiom of choice is the following simple statement: every surjection ("epimorphism") splits, i.e. if $f: X\to Y$ is a surjection, then there exists $g: Y \to X$ so that $f\circ g = {\rm id}_Y$. But this is simply false in the topological, Lie, and algebraic categories.
This leads to all sorts of extra rich structure if you do algebra internal to these categories. You have to start thinking about bundles rather than products, there can be "anomalies", etc.
Update:
In the comments, there was a request for a totally explicit example, where Axiom of Choice is commonly used but not necessary. Here's one that I needed recently. Let $\mathcal C$ be an abelian tensor category, by which I mean that it is abelian, has a monoidal structure $\otimes$ that is biadditive on hom-sets, and that has a distinguished natural isomorphism $\text{flip}: X\otimes Y \overset\sim\to Y\otimes X$ which is a "symmetry" in the sense that $\text{flip}^2 = \text{id}$. Then in $\mathcal C$ is makes sense to talk about "Lie algebra objects" and "associative algebra objects", and given an associative algebra $A$ you can define a Lie algebra by "$[x,y] = xy - yx$", where this is short-hand for $[,] = (\cdot) - (\cdot \circ \text{flip})$ — $x,y$ should not be read as elements, but as some sort of generalization. So we can makes sense of the categories of $\text{LIE}_{\mathcal C} = $"Lie algebras in $\mathcal C$" and $\text{ASSOC}_{\mathcal C} = $"associative algebras in $\mathcal C$", and we have a forgetful functor $\text{Forget}: \text{ASSOC}_{\mathcal C} \to \text{LIE}_{\mathcal C}$.
Then one can ask whether $\text{Forget}$ has a left adjoint $U: \text{LIE}_{\mathcal C} \to \text{ASSOC}_{\mathcal C}$. If $\mathcal C$ admits arbitrary countable direct sums, then the answer is yes: the tensor algebra is thence well-defined, and so just form the quotient as you normally would do, being careful to write everything in terms of objects and morphisms rather than elements. In particular, if $\mathfrak g \in \text{LIE}_{\mathcal C}$, then $U\mathfrak g \in \text{ASSOC}_{\mathcal C}$ and it is universal with respect to the property that there is a Lie algebra homomorphism $\mathfrak g \to U\mathfrak g$.
Let's say that $\mathfrak g$ is representable if the map $\mathfrak g \to U\mathfrak g$ is a monomorphism in $\text{LIE}_{\mathcal C}$. By universality, if there is any associative algebra $A$ and a monomorphism $\mathfrak g \to A$, then $\mathfrak g \to U\mathfrak g$ is mono, so this really is the condition that $\mathfrak g$ has some faithful representation. The statement that "Every Lie algebra is representable" is normally known as the Poincare-Birkoff-Witt theorem.
The important point is that the usual proof — the one that Birkoff and Witt gave — requires the Axiom of Choice, because it requires picking a vector-space basis, and so it works only when $\mathcal C$ is the category of $\mathbb K$ vector spaces for $\mathbb K$ a field, or more generally when $\mathcal C$ is the category of $R$-modules for $R$ a commutative ring and $\mathfrak g$ is a free $R$-module, or actually the proof can be made to work for arbitrary Dedekind domains $R$. But in many abelian categories of interest this approach is untenable: not every abelian category is semisimple, and even those that are you often don't have access to bases. So you need other proofs. Provided that $\mathcal C$ is "over $\mathbb Q$" (hom sets are $\mathbb Q$-vector spaces, etc.), a proof that works constructively with no other restrictions on $\mathcal C$ is available in
They give a reference to
in which the proof is given when $\mathcal C$ is the category of modules of a (super)commutative ring $R$, with $\otimes = \otimes_R$, and, importantly, $2$ and $3$ are both invertible in $R$. [Edit: I left a comment July 28, 2011, below, but should have included explicitly, that Corwin--Ne'eman--Sternberg require more conditions on $\mathcal C$ than just that $2$ and $3$ are invertible. Certainly as stated "PBW holds when $6$ is invertible" is inconsistent with the examples of Cohn below.]
Finally, with $R$ an arbitrary commutative ring and $\mathcal C$ the category of $R$-modules, if $\mathfrak g$ is torsion-free as a $\mathbb Z$-module then it is representable. This is proved in:
So it seems that almost all Lie algebras are representable. But notably Cohn gives examples in characteristic $p$ for which PBW fails. His example is as follows. Let $\mathbb K$ be some field of characteristic $p\neq 0$; then in the free associative algebra $\mathbb K\langle x,y\rangle$ on two generators we have $(x+y)^p - x^p - y^p = \Lambda_p(x,y)$ is some non-zero Lie series. Let $R = \mathbb K[\alpha,\beta,\gamma] / (\alpha^p,\beta^p,\gamma^p)$ be a commutative ring, and define $\mathfrak g$ the Lie algebra over $R$ to be generated by $x,y,z$ with the only defining relation being that $\alpha x = \beta y + \gamma z$. Then $\mathfrak g$ is not representable in the category of $R$-modules: $\Lambda_p(\beta y,\gamma z)\neq 0$ in $\mathfrak g$, but $\Lambda_p(\beta y,\gamma z)= 0$ in $U\mathfrak g$.
Best Answer
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.)