18
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Since each $x \in \mathbb{C}$ has its corresponding $-x$, the sum must vanish. $\endgroup$ – David G. Stork Dec 25 '18 at 16:59
  • 2
    $\begingroup$ There's no order to complex numbers, so is there any sense of it converging in the first place? I'm not sure, @DietrichBurde $\endgroup$ – Shaun Dec 25 '18 at 17:00
  • 13
    $\begingroup$ I am not aware of any definition of a sum of an uncountable set. $\endgroup$ – Lee Mosher Dec 25 '18 at 17:01
  • 11
    $\begingroup$ @DavidG.Stork This is no argument, see Grandi's series. $\endgroup$ – Dietrich Burde Dec 25 '18 at 17:04
  • 4
    $\begingroup$ @LeeMosher My point was that you'd said you were not aware of any definition. There is a fairly standard and natural definition of $\sum_{z\in\Bbb C} c_z$, according to which $\sum_{z\in\Bbb C} z$ does not exist - that seems more satisfactory then jjust saying we don't have a definition. (It really is the "right" definition imo, being exactly the limit of the net of partial sums, $\lim_{F\to\Bbb C}\sum_F$. Turns out to be equivalent to integrability with respect to counting measure...) $\endgroup$ – David C. Ullrich Dec 25 '18 at 20:48
25
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ How does one mentally reconcile this information with the idea of an integral? You could view that as a continuous sum. $\endgroup$ – orlp Dec 25 '18 at 20:21
  • 4
    $\begingroup$ The regularity of $\mathfrak{c}$ is really a red herring here (and in fact, $\mathfrak{c}$ may not be regular!). You could just as well pick an enumeration of $\mathbb{C}$ whose length has countable cofinality, and then there might not be any point beyond which all the terms are $0$. What you can prove though (independent of the ordering used) is that only countably many terms can be nonzero. (Proof: if there are uncountably many nonzero terms, apply your argument to the first partial sum which contains uncountably many nonzero terms, which must have cofinality $\omega_1$.) $\endgroup$ – Eric Wofsey Dec 25 '18 at 23:40
  • 3
    $\begingroup$ As for the regularity of $\mathfrak{c}$, it is consistent with ZFC that $\mathfrak{c}$ is singular (for instance, it could be $\aleph_{\omega_1}$). However, it follows from König's theorem that it has uncountable cofiinality so your argument still does work for it. $\endgroup$ – Eric Wofsey Dec 25 '18 at 23:44
  • 5
    $\begingroup$ @orlp An integral isn't an infinite sum; it's an infinite average. You're effectively dividing each term by a measure of the "number" of terms, rendering it infinitesimal; in this case, the uncountable analogue of infinitesimal. That means it doesn't really resemble a "sum" in the sense we usually mean (for example, the integral of the function $f(x)$ which is $1$ at $0$ and $1$ and zero elsewhere is not $2$). $\endgroup$ – Reese Dec 26 '18 at 5:30
  • 2
    $\begingroup$ I think that the transfinite argument is overkill: the fact that non-zeros are countable follows easily from the fact that given any $\epsilon >0$, the number of numbers with magnitude greater than $\epsilon$ must be finite, and we have a countable sequence of $\epsilon$s going to zero. $\endgroup$ – Acccumulation Dec 29 '18 at 18:03
15
$\begingroup$

If $\{z_i:i\in I\}$ is any indexed set of complex numbers, then the series $\sum_{i\in I}z_i$ 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 $J_\epsilon\subseteq J$, $\vert \sum_{i\in J}z_i-z\vert<\epsilon$. In other words, to say that the series converges to $z$ is to say that the net $J\mapsto\sum_{i\in J}z_i$, 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 $\{i\in I:z_i\neq 0\}$ 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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.