sequences and series - What, if anything, is the sum of all complex numbers?

If this question is ill-defined or is otherwise of poor quality, then I'm sorry.

What, if anything, is the sum of all complex numbers?

If anything at all, it's an uncountable sum, that's for sure.

I'm guessing some version of the Riemann series theorem would mean that there is no such thing as the sum of complex numbers, although - and I hesitate to add this - I would imagine that

$$\sum_{z\in\Bbb C}z=0\tag{$\Sigma$}$$

is, in some sense, what one might call the "principal value" of the sum. For all $w\in\Bbb C$, we have $-w\in\Bbb C$ with $w+(-w)=0$, so, if we proceed naïvely, we could say that we are summing $0$ infinitely many times${}^\dagger$, hence $(\Sigma)$.

We have to make clear what we mean by "sum", though, since, of course, with real numbers, one can define all sorts of different types of infinite sums. I'm at a loss here.

Has this sort of thing been studied before?

I'd be surprised if not.

Please help :)

---

$\dagger$ I'm well aware that this is a bit too naïve. It's not something I take seriously.

Answer

Traditionally, the sum of a sequence is defined as the limit of the partial sums; that is, for a sequence $\{a_n\}$, $\sum{a_n}$ is that number $S$ so that for every $\epsilon > 0$, there is an $N$ such that whenever $m > N$, $|S - \sum_{n = 0}^ma_n| < \epsilon$. There's no reason we can't define it like that for uncountable sequences as well: let $\mathfrak{c}$ be the cardinality of $\mathbb{C}$, and let $\{a_{\alpha}\}$ be a sequence of complex numbers where the indices are ordinals less than $\mathfrak{c}$. We define $\sum{a_{\alpha}}$ as that value $S$ so that for every $\epsilon > 0$, there is a $\beta < \mathfrak{c}$ so that whenever $\gamma > \beta$, $|S - \sum_{\alpha = 0}^{\gamma}a_{\alpha}| < \epsilon$. Note that this requires us to recursively define transfinite summation, to make sense of that sum up to $\gamma$.

But here's the thing: taking $\epsilon$ to be $1$, then $1/2$, then $1/4$, and so on, we get a sequence of "threshold" $\beta$ corresponding to each one; call $\beta_n$ the $\beta$ corresponding to $\epsilon = 1/2^n$. This is a countable sequence (length strictly less than $\mathfrak{c}$). Inconveniently, $\mathfrak{c}$ is regular: any increasing sequence of ordinals less than $\mathfrak{c}$ with length less than $\mathfrak{c}$ must be bounded strictly below $\mathfrak{c}$. So that means there's some $\beta_{\infty}$ that's below $\mathfrak{c}$ but greater than every $\beta_n$. But by definition, that means that all partial sums past $\beta_{\infty}$ are less than $1/2^n$ away from $S$ for every $n$. So they must be exactly equal to $S$. And that means that we must be only adding $0$ from that point forward.

This is a well-known result that I can't recall a reference for: the only uncountable sequences that have convergent sums are those which consist of countably many nonzero terms followed by nothing but zeroes. In other words, there's no sensible way to sum over all of the complex numbers and get a convergence.