Uncountable series without axiom of choice Consider a sequence of positive real numbers $(\alpha_i)_{i\in I}$ for some (suppose maybe wellordered for now) set $I$.
Using axiom of choice, it is easy to see that $\sum_i \alpha_i$ is infinite if uncountably many $\alpha_i$ are nonzero, and that its value, finite or infinite, does not depend on the ordering of $I$ (so in fact it need not be ordered for the sum to make sense). I'm pretty sure countable choice is enough for that.
From that it follows, for example, that in a Hilbert space with a given orthonormal basis, every vector can be expressed as a countable linear combination of basis vectors.
But what happens if we drop choice? Can there be an essentially uncountable but summable series? Or does the wellorderability of the index set force the same fact to hold in ZF?
If it does, then is the paragraph about Hilbert spaces also true in ZF (I believe it's enough for Parseval's identity to hold)?
 A: Let's analyze this situation.
The argument that if $\langle r_i\mid i\in I\rangle$ is a sequence of positive real numbers, and $\sum r_i<\infty$, then there are only finitely many $i$'s such that $r_i\geq\frac1n$, for any $n\in\Bbb N$, holds without the axiom of choice, of course. And so we conclude that $|\{r_i\mid i\in I\}|\leq\aleph_0$, since the countable union of finite sets of real numbers is countable, as the order on the real numbers well-orders them uniformly (the finite sets).
But of course, $I$ can still be uncountable. More specifically, if $I$ is a set such that for some $f\colon I\to\omega$, the preimage of each $n$ is finite, then $I$ can index a convergent sequence. For example, by having $x_i=\left(\frac12\right)^{n\cdot m+1}$ whenever $f(i)=n$ and $|f^{-1}(n)|=m$. Or some other suitably small number.
And of course, it is consistent that there are such sets. We can engineer them as freely as we want. Namely, if $g\colon\omega\to\omega$ is any function whose values are all positive, then we can arrange for a set $A$ which is the countable union of finite sets $A_n$, where $|A_n|=g(n)$, and $A$ is uncountable, or more specifically, Dedekind-finite (it's not nearly as impressive when $A$ is Dedekind-infinite, although we can do that to some degree).
On the other hand, if $I$ is such that there is no function from $I$ to $\omega$ with finite fibers, then $I$ cannot possibly index a convergent series, since for some $n$, numbers larger than $\frac1n$ will appear infinitely often. And therefore the sum must be infinite.
One last remark is that countable choice is not needed to show that the sum is independent of the order. In the cases above where the set is the countable union of finite sets and it is Dedekind-finite, then the index set cannot even be linearly ordered. And so there is no reasonable way to order such a set anyway. Not to mention that by counting how many times each number appears, you can always rearrange the series to be indexed by $\omega$ as a plain ol' series of positive numbers.
And if you really insist on overkilling the issue, then you can notice that this means that you can encode the sequence of reals into a single real $x$, and in $L[x]$ which is a model of choice the sum converges.
The following papers might interest you:

Brunner, Norbert. "Sequential compactness and the axiom of choice." Notre Dame Journal of Formal Logic 24, no. 1 (1983) 89-92.
Brunner, Norbert. "Garnir's dream spaces with Hamel bases." Arch. Math. Logik Grundlag. 26 (1987), no. 3-4, 123–126.
Brunner, Norbert. "Linear operators and Dedekind sets." Math. Japon. 31 (1986), no. 1, 1–16.

