To each pair $(S,\mathcal{X})$ where $S=(s\_i)\_{i\in\mathbb{N}}$ is a decreasing sequence of positive real numbers and $\mathcal{X}\subseteq\mathbb{R}$, we can associate the **alternation game** $A\_S(\mathcal{X})$ as follows:


* Players $1$ and $2$ jointly build an increasing sequence of natural numbers $$a\_0<b\_0<a\_1<b\_1<...,$$ subject to the rule that $$\sum\_{i<a\_0}s\_i>\sum\_{a\_0\le i<b\_0}s\_i> ... >\sum\_{b\_k\le i< a\_{k+1}}s\_i>\sum\_{a\_{k+1}\le i<b\_{k+1}}s\_i>...$$
* Player $1$ wins iff the alternating series $$\left(\sum\_{i<a\_0}s\_i\right)-\left(\sum\_{a\_0\le i<b\_0}s\_i\right)+ ... +\left(\sum\_{b\_k\le i< a\_{k+1}}s\_i\right)-\left(\sum\_{a\_{k+1}\le i<b\_{k+1}}s\_i\right)+...$$ converges to an element of $\mathcal{X}$.


An [old question](https://mathoverflow.net/questions/426794/a-game-of-harmonic-seriess) of mine asked about the behavior of these games when we specifically take $S$ to be the harmonic sequence $s\_i={1\over i}$, but turned out to be difficult to attack. I'd like to get a better understanding of *why* this sort of question might be hard, and this seems like a good starting point:



> 
> Is it consistent with $\mathsf{ZFC}$ that *some* $A\_S(\mathcal{X})$ is undetermined?
> 
> 
> 


A positive answer to this question would help explain potential wildness, but would also seem to require us to be able to "code into" the alternation games (a la the [Banach game](https://mathoverflow.net/questions/281240/is-the-banach-game-quantifier-intractable-beckers-guess)), and at present I don't see how to do that.



Can these alternating series games be undetermined?
If we take $s\_i = 2^{-i}$, we should even get that $\mathrm{ZFC}$ proves the existence of some $\mathcal{X}$ with undetermined $A\_{(2^{-i})\_{i \in \mathbb{N}}}(\mathcal{X})$. The key parts are that different plays yield different reals (as nothing in the tail can overcome a difference in a prefix), and that each strategy is compatible with continuumsly many plays (as a move here doesn't constrain the future at all).


We should thus be able to do the standard trick of well-ordering both players strategies with the least possible order type, and then building $\mathcal{X}$ in stages to foil any particular winning strategy.


