A fairly well-known (and perplexing) fact is that the sum of all natural numbers is given the value -1/12, at least in certain contexts. Quite often, a "proof" is given which involves abusing divergent and oscillating series, and the numbers seem to fit at the end. (There are also other proofs involving the zeta function or other things, but that's not my concern right now).

However, I tried to calculate it my own way using similar methods, and this is what I got:

$$\begin{align} S = 1 &+ 2 + 3 + 4 + …\\\\ 2S = 1 &+ 2 + 3 + 4 + 5 + \ldots\\ &+ 1 + 2 + 3 + 4 + \ldots\\\\ = 1 &+ 3 + 5 + 7 + 9 + \ldots\\ \end{align}$$ Also, $2S = 2 + 4 + 6 + 8 + 10 + \ldots$

Adding the above together,

$$4S = 1 + 2 + 3 + 4 + 5 + … = S$$

Which means $3S = 0$, therefore $S = 0$

Obviously, there must be some reason why this is not ok, otherwise we'd have $-1/12 = 0$.
But why is my method wrong while the one involving oscillating series is considered acceptable?

Additional clarification: I was wondering if there are specific ways to manipulate this kind of series such that by assuming that the result is finite and performing only certain types of "permitted" operations, one could be confident to get to the same result as the rigorous way of assigning a finite value to the sum of the series. So far, the consensus seems to be negative.

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    A fairly well-known (and perplexing) fact is that the sum of all natural numbers is -1/12. No, this is not true. It can, if you really want, be seen as an assigned value to a divergent series. – Eff Jul 25 '16 at 20:45
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    The simple answer is that the usual arithmetic of the integers does not apply to the "sum" 1+2+3+... . – Eigenfield Jul 25 '16 at 20:46
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    You and an infinite number of friends need to go take a vacation in Hilbert's Hotel. It's full, but they'll find room. – Joffan Jul 25 '16 at 20:50
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    "But why is my method wrong..": maybe, with all due respect, because you do not know what you are talking about. There is a lot of work that one needs to do in order to really understand how $1 + 2 + ... = -1/12$ would make sense in some contexts and why it is rigorously justifiable within these contexts. – user258700 Jul 25 '16 at 20:54
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    For once, here is a -1/12 question which is actually precise enough to have an answer, and an interesting one (which, unfortunately, no comment or post below is mentioning). To sum up, everything you did is fine, until and included the step asserting that 4S=S. The faulty step is which means 3S = 0. To get this, one needs to substract S, and this is where you assume that S is finite! The correct conclusion to your proof is that either S is infinite, or 3S=0, that is, S=0. And this conclusion is perfectly valid... :-) – Did Jul 25 '16 at 21:17

The simple answer to what seems to be your question: "why is my method wrong?"

You are assuming that a clearly divergent series converges to some number $S$; this provides a contradiction, if you assume a contradiction to be true then you can essentially use it to prove whatever nonsense you want.

Showing that this series sums to $-1/12$ and using this result does not have much to do with what you are thinking of as the sum of an infinite series. The Numberphile video is entertaining, but it is trash as far as actual mathematics. They don't explain this result, the meaning behind it, the context in which it is used, or the justification for the steps they use in their "proof". There is nothing rigorous about the proof they present, it's all smoke and mirrors. So it is natural that, by imitating their proof, you are ending up with a wrong result.

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    @par Yeah, actually after watching the video I'm considering modifying my last paragraph to say "the proof you are trying to imitate is trash, so you are also ending up with trash". – Morgan Rodgers Jul 25 '16 at 21:25
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    It's true that the video doesn't give justification for what it does, but that does not mean a justification does not exist. Most of their manipulations can be explained by considering formal power series. Ramanujan himself used similar techniques to prove the sum. – Meni Rosenfeld Jul 25 '16 at 23:44
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    @DavidK: All reliable sources on string theory that I've been able to find seem to corroborate the fact that there are too many parameters for it to be a testable theory. A mathematician once said: With four parameters I can fit an elephant, and with five I can make him wiggle his trunk. – user21820 Jul 26 '16 at 4:17
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    @MeniRosenfeld I'm not saying there aren't justifications, but those justifications are important and require a specific interpretation of an infinite series. When you just say "$1-1+1-1\ldots = 1/2$" (!) without explaining what you mean by this, then add this series to itself shifted over (!!) without explaining why or when you can do this, you give viewers the impression that "hey, we can do whatever we want!". And with this attitude, you can easily prove that $0=1$. – Morgan Rodgers Jul 26 '16 at 7:33
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    I'd have to agree that Morgan Rodgers is right in his own defense. Evaluating divergent series must be given context, but within its context, they can be useful. Mainly string theory, yes, which is indeed difficult to say about its usefulness, but still, divergent series will most likely find practically uses in the end. Looking at it from one end, this is almost like set theory, something hard to accept at first, but nowadays, most people accept it for what it is, given time and applications. – Simply Beautiful Art Jul 26 '16 at 12:10

The most common way to define $\sum_{n\geq1}n$ is by its partial sums: $$ \sum_{n\geq1}n\equiv\lim_{N\rightarrow\infty}s_{N}\text{ where }s_{N}\equiv\sum_{n=1}^{N}n=\frac{1}{2}N\left(N+1\right). $$ Clearly, this is divergent.

As you seem to have noticed, there are other ways to define this sum. This causes all sorts of confusion, since people like to also use the notation $\sum_{n\geq1}n$ in these alternative definitions.

Let's now look at "zeta function regularization", which is the interpretation which yields $-1/12$. In particular, note that $$ \zeta(s)=\sum_{n\geq1}n^{-s}\text{ for }\operatorname{Re}(s)>1 $$ by definition, where we are in the above using the usual notion of summation by a limit of partial sums. The $\zeta$ function, however, is also defined for other values of $s$ by analytic continuation. We can thus reinterpret $$ \check{\sum_{n\geq1}}n\equiv\zeta(-1)=-1/12. $$ However, note that the two interpretations $$\sum\text{ and }\check{\sum}$$ are completely different!

As for the method which you refer to as "abusing divergent and oscillating series", perhaps you are thinking of this. This is not rigorous.

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    @par I see. But then that is incorrect: the zeta function at $\;z=-1\;$ is not defined by the well known series $\;\sum n^{-s}\;$ but by its analytic continuation, which definitely is not that Dirichlet series. – DonAntonio Jul 25 '16 at 21:14
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    @DonAntonio: please reread carefully, I mention that the Dirichlet series and the $\zeta$ function agree on $\operatorname{Re}(s)>1$, and that we are simply interpreting the summation as the value $\zeta(-1)$ even though the corresponding Dirichlet series diverges. Perhaps I did not do a great job of explaining. – parsiad Jul 25 '16 at 21:17
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    @aditsu: yes, you should ignore that video as it was made solely for the purposes of "blowing people's minds, dude" ...and for presumably getting more YouTube subscribers and the cash benefits that come with that. That video has, unfortunately, confused a lot of people. Established scientists should know better. (gets off soap box) – parsiad Jul 25 '16 at 21:18
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    Well, it's arguably not a fact. It's just a fun thing. You can derive anything your heart desires from falsehoods. – parsiad Jul 25 '16 at 21:28
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    @aditsu: As I explain in my own answer, it's more than just a fun coincidence. But on the other hand, it's very tricky to know what can be done and what cannot, and the video doesn't even try to explain why its manipulations are valid. – Meni Rosenfeld Jul 26 '16 at 9:15

The error in your derivation - I kid you not - is the assertion that 1+2=3 (and similarly 2+3=5, 3+4=7 etc.)

Sums like this get their meaning from Zeta function regularization, and the intuitive algebraic manipulations are only shorthand for correct manipulations on corresponding Dirichlet series. It so happens that the manipulations in the linked proof can be translated to proper ones, while your manipulations cannot.

In particular, it's easier to work with alternating sums because they can be thought of as formal power series in which you then substitute 1 (and since shifting is equivalent to multiplying by a power of $x$, you can do that freely). But you can't do that for sums like 1+2+3+... because you would get a 0 denominator, so other techniques are needed.

If your manipulations are translated to Zeta Function regularization, you'll get that the 1 and 2 you try to add are not really 1 and 2, they are $1^{-s}$ and $2^{-s}$, and these do not add up to $3^{-s}$, rendering the rest of the derivation invalid.

See also https://en.wikipedia.org/wiki/1_+2+3+4+_%E2%8B%AF#Heuristics.

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    It's customary to explain downvotes. – Meni Rosenfeld Jul 26 '16 at 9:11
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    Not my vote, but it is fairly elementary that 1+2=3. That isn't the problem at all. Binary addition is perfectly well defined. And a sum over a finite number of terms is perfectly well-defined as a generalization of binary addition, even for less than 2 terms. Nice properties like a+b=b+a and (a+b)+c = aa+(b+c) also generalize for sums of finite length. No, the problem here is that these properties no longer hold for sums with an infinite number of terms. – MSalters Jul 26 '16 at 12:36
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    @MSalters: That's one perspective on the problem, which is of course valid but does not address the OP's question at all - "You can't operate on infinite series" doesn't explain why some operations do give correct results. I've offered a different perspective which offers more context. It's true that $1+2=3$ but, as I explained, when the OP writes 2 it doesn't really mean 2 and when he writes 3 it's not really 3. It means $2^{-s}$ and $3^{-s}$ and it is not true that $1^{-s}+2^{-s}=3^{-s}$. So the OP's derivation fails when the naive symbols are translated to their proper interpretations. – Meni Rosenfeld Jul 26 '16 at 16:38
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    @MSalters: In general, I take issue with any answer of the form "According to the series summation definition we learned in freshman calculus, this series diverges and all your operations are illegal! Nothing to see here.". A more healthy mathematical approach is to consider how we can expand and generalize the definitions to obtain new, interesting and consistent results. Ramanujan summation is a thing, and it has its own set of rules - different from the rules of classical summation. I've tried to give a glimpse into what those rules are. – Meni Rosenfeld Jul 26 '16 at 16:46
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    @MeniRosenfeld: I also absolutely hate that kind of answer as well, and how it is so prevalent on SE sites; it drives me nuts. Thanks for being someone who actually cares to give a satisfying answer instead of shutting the OP down. Wish I could +100. – Mehrdad Jul 26 '16 at 20:02

In some sense, the error is in the subtraction. You're trying to do some arithmetic on things which are no longer proper numbers, and not everything behaves the same.

For a mild simplification, think about the sum $$ S = 1 + 1 + 1 + \cdots $$ Then by the same reasoning as yours, we can consider $$ 2S := S + S = 1 + 1 + 1 + \cdots $$ Here I've interleaved sums, and this is okay because all terms are positive, though not valid in general. This addition of $S$ with itself in some sense is fine and we see $2S = S$. Then you'd like to say that $$ 2S = S \implies S = 0, $$ but the problem it doesn't for arbitrary "quantities" S, including the case that $S = +\infty$, which is the value of $S$ here. (The above statement doesn't even hold in many finite algebraic systems like $\mathbb Z/6 \mathbb Z$.)

Weird things happen when you try to work with infinities, and not everything commutes. In particular, you need to be careful about subtracting infinite sums because conditionally convergent sums can be rearranged to take arbitrary values.

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    Thanks for mentioning that conditionally convergent sums can be rearranged to take arbitrary values; this isn't just an issue of divergent "sums". – PM 2Ring Jul 26 '16 at 8:21
  • Yes, I think that the last line here is the real answer at heart. You can't rearrange conditionally convergent series; even more so, you can't rearrange divergent series! (At least, not always.) Yet that is what the OP does when going from $(1 + 2 + 3 + 4 + 5 + \cdots) + (1 + 2 + 3 + 4 + \dots)$ to $1 + 3 + 5 + 7 + 9 + \cdots$. – Toby Bartels Jul 26 '16 at 19:42
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    @TobyBartels Well, you can formally rearrange series with positive terms, and if one is divergent ($+\infty$) they all are. So I was saying in some sense this part is alright, with the convention that $\infty + \infty = \infty$. However $\infty - \infty$ in an indeterminate form. – Kimball Jul 27 '16 at 1:06

If you assume the sum of all naturals is finite and the "usual arithmetic" rules here, you can get pretty awesome results, like:

$$S=1+2+3+\ldots\implies 2+4+6+\ldots=2(1+2+3+\ldots)=2S\implies$$



And a lot more of whatever else you want, say:


$$=2+4+6+\ldots=2S\implies 3S=1+1+1+\ldots\;,\;\;\text{etc.} $$

  • That doesn't exactly answer the question. I'm sure each of your infinite sums has a certain corresponding value given by the same theories that give -1/12 (some values may be infinite). Some of them might even match the intuitive arithmetic, others may not. It would be good to analyze that. – aditsu Jul 25 '16 at 21:06
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    @aditsu I don't think there is "an answer" to the question. I'm remarking the fact that assuming a divergent series converges cna lead us to many, many contradictions. – DonAntonio Jul 25 '16 at 21:07
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    @aditsu, to add to what DonAntonio said in the comment, you are using laws of finite arithmetic to a divergent series. That can lead to trouble. You mention that $S=-1/12$. That is true in a very particular context, in which $S$ is not interpreted as a sum, but more as the value of the extension (in some sense) of the function $z(s)=1+2^{-s}+3^{-s}+4^{-s}+\ldots$ to the complex numbers $s$ with real part less than 1. This is very different from saying that $1+2+3+4+\cdots=-1/12$... – bartgol Jul 25 '16 at 21:13

A whole book could be written about it. In fact, it has:

G.H Hardy, Divergent Series, (Oxford, 1949).

Euler (to name but one) was far less uptight about assigning values to divergent series (such as $1+2+4+8+16…=-1$) than today's mimsy pedants. Cauchy, Abel and others took the subject further and with more rigour.

But I am not going to summarize the entire book here.

I think such issues, which superficially seem to be about symbol-manipulation, are more genuinely about operations within a context in which "infinite sums" (for example) are characterized in various ways. That is, to say $\sum_n a_n=A$ should mean what? It is certainly true that the "populist" answer is convergence-of-sequence-of-finite-subsums, where convergence of a sequence is fairly intuitive (whatever its formal "definition/characterization" may be).

Thus, in one context (or more), the infinite sum has no sense as a real or complex number. Ok.

In other senses, going back at least to Euler, and resurrected by Ramanujan post-Weierstrass, when one a-priori acknowledges the non-convergence (=lack of useful, manipulable sense, in terms of many characterizations...), then one might aggressively ask about other interpretations (... contexts...)

So Euler already knew that, in some very particular sense, the values of zeta(s) at negative integers are what we know understand them to be... by analytic continuation, although the latter notion did not exist in Euler's time (so far as I know).

The genuine operational point is to maybe not just play games with symbols so much, but, rather, to try to say what real thing one's symbols encapsulate, so that the symbol-motion is just book-keeping about a real thing, rather than a fiction with nothing underlying.

To say that manipulation of "non-convergent sums" is not legitimate is too glib, since it assumes a certain interpretation which is in many regards the strictest possible. Much like declaring as much stuff as possible "nonsense" just to avoid having to grapple with it.

Still, again, to just shuffle symbols is not so interesting... not at all because of possible "lack of rules [sic]", but because it might be without context, and/or without characterization of the phenomena purportedly described by the symbols.

(My colleagues who describe various ideas as "just marks on the page" dismay me...)

Note that 4S = S may also mean both are Infinite as

N*infinite = Infinite ; N being a +Ve Real Value  
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    $4S=S$ would in my world mean $S=0$.. – Mårten W Jul 26 '16 at 11:53
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    @MårtenW: That obviously depends on the domain of S. A classical example is X*X*X*X=1. In the domain of non-negative integers, X=1 is the only solution; and with complex numbers we have 4 solutions. Modulo 10, X=3 is a solution. The original question fails because it assumes rules that only hold for finite rationals, yet it contains a sum S whose value is not a finite rational. – MSalters Jul 26 '16 at 12:45
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    @MSalters I agree with Eduru, if $S=\infty$ then $4S=S$. That's because the extended real numbers is not a proper domain where cancellation holds - in fact it's not even a proper ring because the binary operations are not defined on all pairs. But nonetheless you work in that space in real analysis and the value of a divergent series of positive numbers is equal to $\infty$. But we do not cancel $\infty$ in equations because that's not justified by the algebraic properties of the space. This is basic abstract algebra. – Gregory Grant Jul 26 '16 at 13:28
  • @MSalters: $4S=S$ has the solution $S=0$ over the complex numbers as well. I'm not sure what relevance your discussion about $X^4=1$ has to do with it all. – Mårten W Jul 26 '16 at 13:53
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    @MårtenW I'm pretty sure it was an example of a situation where the existence of additional solutions depends on the domain. – djechlin Jul 26 '16 at 19:24

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