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.)
I am sympathetic to this question, which often arises for those first learning of Cohen's theorem, and I don't think it is an idle question. I recall my sophomore undergradatue self being confused about it when I first studied the set-theoretic independence phenomenon. And I think that Carl is right, that this particular issue is not addressed on the other CH questions.
I view the question as arising from the following line of thought: Cohen proved that it is possible that CH fails. Thus, it is possible that there is a set of reals whose cardinality is strictly between the integers and the continuum. But if we can decribe how such a set can exist, then haven't we actually described a set of such intermediate cardinality? That is, doesn't this mean that CH is simply false?
This line of thinking may be alluring, but it is wrong. The reason it is wrong, as Gerhard explains in his comment, is that it doesn't appreciate the role of models, or what might be called the set-theoretic background. What Cohen did was to show that if ZFC is consistent, then so is ZFC + ¬CH. (In contrast, Goedel proved that if ZFC is consistent, then so is ZFC + CH.) Thus, Cohen's intermediate-cardinality set has the property that it is intermediate in cardinality in the model that Cohen describes, with respect to that set-theoretic background, but it will not be intermediate-in-cardinality with respect to other set-theoretic backgrounds. The property of being intermediate in cardinality is dependent on the set-theoretic background in which this property is considered. For example, a set $X$ is uncountable if there is no function from the natural numbers onto it. But perhaps there is no such function mapping onto $X$ in a model of set theory $M$, but there is a larger model of set theory $N$, still having $X$, but for which now there IS a function mapping the natural numbers onto $X$. In fact, this very situation follows from Cohen's forcing method: any set can be made countable in a forcing extension.
Thus, whether a set forms a counterexample to CH cannot be observed looking only at that set---one must consider the set-theoretic universe in which the set is considered, and the possible bijective functions that might witness its countability or not. The very same set of reals can be countable in some models of set theory and bijective with the continuum in others.
Best Answer
The statement in question, frequently denoted $\mathsf{RT}^2_2$ in the context of reverse mathematics, is the instance of the infinite Ramsey theorem for unordered pairs and two colors. Specifically, the theorem is that given any graph on the set of natural numbers contains an infinite clique or an infinite anticlique. This is an infinitary theorem since the graph is infinite and so is the clique or anticlique.
One of the reasons why this result is of deep interest is that it has many finitary consequences. For example, the finite Ramsey theorem follows from the infinite case by a compactness argument. The theorem has many other seemingly unrelated consequences, for example Ramsey's theorem is a key step in the proof of termination of some algorithms.
The recent results of Patey and Yokoyama show (among other things) that $\mathsf{RT}^2_2$ is $\Pi^0_2$ conservative over $\mathsf{PRA}$, i.e. that every $\Pi^0_2$ statement provable using the infinitary theorem $\mathsf{RT}^2_2$ is already provable using only primitive recursive arithmetic. This is surprising because primitive recursive arithmetic is a completely finitary theory: primitive recursive functions have a finite description and they always terminate in finite time. Thus $\mathsf{PRA}$ is considered among the "safest" theories of arithmetic. It is much weaker than Peano Arithmetic, in fact Peano Arithmetic is far from conservative over $\mathsf{PRA}$.
$\Pi^0_2$ statements include statements of the form "this algorithm terminates on every input" so we now know that all algorithms that have been shown to terminate assuming $\mathsf{RT}^2_2$ have an alternate proof of termination that only uses $\mathsf{PRA}$ and therefore that these algorithms are primitive recursive themselves!