Qfyilyi

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

$$sum_{zinBbb C}z=0tag{$Sigma$}$$

is, in some sense, what one might call the "principal value" of the sum. For all $winBbb C$, we have $-winBbb C$ 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.

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.

If ${z_i:iin I}$ is any indexed set of complex numbers, then the series $sum_{iin I}z_i$ is said to converge to the complex number $z$ if for every $epsilon>0$ there is a finite subset $J_epsilon$ of $I$ such that, for every finite subset $J$ of $I$ with $J_epsilonsubseteq J$, $vert sum_{iin J}z_i-zvert<epsilon$. In other words, to say that the series converges to $z$ is to say that the net $Jmapstosum_{iin J}z_i$, from finite subsets of $I$ directed by inclusion, converges to $z$. This is a special case of the standard definition of unordered summation which works in an arbitrary commutative (Hausdorff) topological group. A comprehensive reference for this kind of summation is section 5 of Chapter III of Bourbaki's General Topology. In particular, because $mathbf{C}$ is first-countable, the corollary on page 263 of loc. cit. implies that, if such a $z$ exists, then the set ${iin I:z_ineq 0}$ is countable (such a result was mentioned in the answer by Reese, but it is not necessary to use any kind of transfinite recursion to make sense of this kind of summation).

So, unless you want to introduce a nonstandard notion of summation (which necessarily must lack some of the features one would hope for based on the case of finite sums), the series in question cannot be meaningfully said to converge to anything.

The obvious answer is that no such sum can be defined uniquely. If one limits to summation methods that are symmetric around the origin, the conditional limit could be zero. (The process would be to sum over squares (-x,-ix; x, -ix; x, ix, -x, ix) such that each set of 4 points sums to zero (as does the origin.) Of course,summing over (-x,-ix; 2x, -ix; 2x, ix, -x, ix) would give a result that becomes arbitrarily large. By playing around with summation methods one could generate any result.