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I recently saw the following "proof" online, and couldn't pinpoint where the mistake was made:

From a well known property, $$1+2+3+\cdots = -\frac{1}{12}.$$

Multiplying both sides by $-1,$ we get $$-1-2-3-\cdots = \frac{1}{12}.$$

We can thus rearrange these equations as follows:

\begin{align*} 1+2+3+4+\cdots= \, -\frac{1}{12} \\ -1-2-3-\cdots= \; \: \, \, \frac{1}{12} \\ -1-2-3-\cdots= \; \: \, \, \frac{1}{12} \\ 1+2+\cdots = -\frac{1}{12} \end{align*} Adding, the RHS clearly sums to $0$, while the LHS yields $1$, seemingly yielding that $0=1$. Where did this proof go wrong?

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The “well known” property is derived from Ramanujan summations, in which divergent sums are treated differently than normal, but then you treat the divergent sum in a typical manner. Plus being able to rearrange the terms and get the same sum is something reserved for absolutely convergent series, which the LHS is not.

Laars Helenius
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    This is the most insightful answer! To add some precision: @AlbonWu implicitly assumes that there is a summability method that meets three criteria: (1) It assigns $-1/12$ to the series. (2) It is linear, respecting adding series term by term. (3) It is stable, respecting shifting the series to the right. But there is no such method. In particular, Ramanujan summation satisfies (1) and (2) but not (3). Meanwhile, the "typical manner" summation of convergent series satisfies (2) and (3) but not (1). – Chris Culter Nov 27 '18 at 23:12
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I think a lot of people have been confused by the "well known property" that you are mentioning, so I am going to elaborate a little bit.

First of all, it is not true that $$ 1+2+3+\dots = -\frac{1}{12}. $$ And the above shouldn't be true. After all, it doesn't make any sense!

I will briefly explain why you see this identity in a lot of places. Basically, there is a function known as the zeta function which is of particular importance in number theory. For all numbers $s>1$, the zeta function is given by the formula $$ \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$ If you were to pick $s\leq 1$, the the above sum would be divergent and $\zeta$ would not be well defined.

On the other hand, the function $\zeta$ can be extended to the real line in a meaningful way (this is known as analytic continuation and if taught in undergraduate complex analysis courses). So, we have function $\zeta$ which is defined for all numbers $s$ and such that $$ \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$ whenever $s>1$.

Now, it turns out that $\zeta(-1) = -1/12$. Plugging $s=-1$ into $$ \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$ one would obtain $$ 1+2+3+\dots $$ This is where your "well known identity" comes from.

Quoka
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  • +1: nice answer: a summary is that we have two limiting processes here (limit of a series and limit of a continuous function as its argument tends to a point) and the order of carrying out the processes is important: sum-then-limit-of-function converges but limit-of-functions-then-sum diverges. – Rob Arthan Nov 27 '18 at 23:15
  • @Quoka "And the above shouldn't be true. After all, it doesn't make any sense!" It does make sense, unless you dogmatically assume Cauchy summation is the only kind of summation that "makes sense". – user76284 Jul 17 '19 at 21:36
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You start from a Ramanujan summation. In the next steps you use linearity (multiplication with $-1$ as well as termwise addition), stability (extracting finitely many summands is allowed), and regularity (for the convergent series of all $0$ terms, Ramanujan summation yields $0$). Can you justify that Ramanujan summation is linear, stable, and regular?

There are summation methods that are stronger than standard summation (=limit of partial sum), for example Cesàro summation. However, none of these assigns a finite value to $\sum n$, and your argument shows exactly the reason why a finite value is not possible for such a summation method.

Hagen von Eitzen
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I couldn't pinpoint where the mistake was made...

In order to understand the original fake proof, I suggest you start with this one, which is simpler:

From a well known property, $$1+2+3+\cdots = \infty.$$

Multiplying both sides by $-1,$ we get $$-1-2-3-\cdots = -\infty.$$

We can thus rearrange these equations as follows:

\begin{align*} 1+2+3+4+\cdots&= \infty\\ \\ -1-2-3-\cdots&= -\infty\\ \\ -1-2-3-\cdots&= -\infty\\ \\ 1+2+\cdots &= \infty \end{align*} Adding, the RHS clearly sums to $0$, while the LHS yields $1$, seemingly yielding that $0=1$.

Can you find the mistake for this one?

Community
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Pedro
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The "well-known property" that you mention in the beginning does not hold. Therefore, neither does anything that you state after that.

Daniel Robert-Nicoud
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Well, at least we have that $$\sum_{n \in \mathbb{N}} n=-\frac{1}{12} \implies 1=0$$ Is true.

Botond
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  • True I suppose since $p \implies ((\mbox{ not } p) \implies q)$. – Michael Nov 27 '18 at 22:59
  • Even this is hard to argue because it uses a false statement in the arrangement of the elements of the series. I just didn't call it out explicitly because it's possible the very same fallacy is used to prove the erroneous starting statement, which leaves me conflicted. – Git Gud Nov 27 '18 at 23:00
  • What do you mean, @Michael? – Botond Nov 27 '18 at 23:02
  • @Botond : Oh, well I suppose I should then ask what do you mean in your above answer? I assumed you were using the logical fact $p \implies ((not p) \implies q)$ (if we know something is both true and false, then anything is true). In this case I thought you were taking $p,q$ propositions defined as $p = {\sum_{n\in\mathbb{N}} n \neq -1/12}$ and $q = {1=0}$. – Michael Nov 27 '18 at 23:06
  • @GitGud I think the correctness of the manipulations do not matter. The base statement is false, so he can "derive everything from it". – Botond Nov 27 '18 at 23:06
  • Of course, but that lies on the knowledge that the antecedent is false. I'm speaking from the POV of someone who doesn't know that, like the OP (doubt it, TBH). Think of all the statements of the form "If the RH holds, then yada, yada, yada". – Git Gud Nov 27 '18 at 23:09
  • @Michael I just wanted to point out that the truth value of $p \implies q$ is always true when the truth value of $p$ is false. – Botond Nov 27 '18 at 23:10
  • @Botond : Yes that is in the same spirit as my first comment, which means I correctly got your point after all. – Michael Nov 27 '18 at 23:12
  • @GitGud I'm not certain if I understand you for 100% and I don't want to waste your time, but I think the others pointed out the problems I did not. My goal is/was just to point out the potential danger on $\implies$. – Botond Nov 27 '18 at 23:32
  • @Michael Yes, sorry. It confused me a bit. Sadly, $\text{I'm good at mathematical logic} \implies 1=0$ as well. – Botond Nov 27 '18 at 23:33
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    @Botond : That is funny. I liked this answer best, I'm sorry somebody downvoted it (likely they didn't understand the logic). – Michael Nov 27 '18 at 23:51
  • @Michael I downvoted it because the the implication is false. The antecedent is true but the consequent is false. – user76284 Jul 17 '19 at 21:38
  • @user76284 $\sum_{n \in \mathbb{N}} n=-\frac{1}{12}$ is clearly false. – Botond Jul 18 '19 at 04:40
  • @Botond What do you mean? It’s false if you’re restricting yourself to Cauchy summation, but more powerful summation methods exist. – user76284 Jul 18 '19 at 05:49
  • @user76284 $\sum_{n \in \mathbb{N}} a_n=\lim_n \sum_{i=0}^n a_i$, by definition. But we could say the same about the rhs as well, couldn't we? $1=0$ is false if we restrict ourselves to the standard equality, but we surely can define a new one where it is true. – Botond Jul 18 '19 at 05:57
  • “by definition” Not really. This is like saying “$2^\pi$ doesn’t exist because $2^x$ is defined as $\prod_{n=1}^x 2$ which doesn’t make sense when $x=\pi$”. The point is that there’s a natural extension of this operation and it makes perfect sense to refer to it likewise. – user76284 Jul 18 '19 at 06:08
  • “But we could say the same about the rhs as well, couldn't we?” The difference is that stronger summation methods are consistent with Cauchy summation (they assign the same values to Cauchy-convergent series), whereas $0=1$ isn’t. – user76284 Jul 18 '19 at 06:09
  • Imagine if you insisted on using a different symbol for addition of natural numbers, integers, real numbers, complex numbers, vectors, functions, etc. – user76284 Jul 18 '19 at 06:13
  • @user76284 Why would Ramanujan summation be a "natural extension"? – Botond Jul 18 '19 at 06:25
  • @Botond You don't need Ramanujan summation to derive $\sum_{n=1}^\infty n = -1/12$. You only need to observe that -1/12 is the (unique!) value obtained by analytically continuing $\sum_{n=1}^\infty n^s$ to $s=1$. You can also use the limit approach I describe here. – user76284 Jul 18 '19 at 07:05
  • @user76284 Yes, if we define $f$ for $\Re(x)>1$ as $f(x)=\sum_{n \geqslant 1} \frac{1}{n^x}$, then we, in fact, can assign a value to $f(-1)$ with analytical continuation. But it will not be the value of $\sum_{n \geqslant 1} \frac{1}{n^{-1}}$, because the sum does not converge. – Botond Jul 18 '19 at 07:09
  • @Botond Like I said, the fact that a series doesn't converge in the Cauchy sense doesn't mean it doesn't have a value. A series can have a value it doesn't Cauchy-converge to. – user76284 Jul 18 '19 at 07:13
  • @user76284 Yes, you can extend it as you wish. You can do whatever you want. What if I'd like to assign the value $42$ to all of the divergent sums? I could do it, couldn't I? But it would no longer be the "casual sum", so it would not necessarily possess the properties of the "casual sum". And I'd say that about 99% of the people are using the $\sum$ symbol to denote the limit of the partial sums, so it's not a great idea to use it for a different thing without saying it. – Botond Jul 18 '19 at 07:23
  • @user76284 And if the $\sum$ is not the limit of the partial sums, then it's also bad to say that "$1+2+3+...=-\frac{1}{12}$", because as you said, the $\sum$ is not the limit of the partial sums, but the lhs is meant to be that. It's like saying that I define $\sum_n a_n$ as $\lim_n a_n$, and then complaining about $\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+...=0$ is not true. – Botond Jul 18 '19 at 07:27
  • @Botond "What if I'd like to assign the value 42 to all of the divergent sums?" That definition is not consistent with Cauchy-summation. "the lhs is meant to be that" Only if you restrict yourself to Cauchy-summation. "It's like saying that I define $\sum_n a_n$ as $\lim_n a_n$" That definition is also not consistent with Cauchy-summation. – user76284 Jul 18 '19 at 07:32
  • @user76284 Why would my "42-sum" be inconsistent with the Cauchy-summation? It'd assign the same value to the convergent sequences and $42$ to the divergent ones. Why is it different from your Ramanujan sum? – Botond Jul 18 '19 at 07:38
  • @Botond $0 = \sum_n 0 = \sum_n (n + (-n)) = \sum_n n + \sum_n (-n) = 42 + 42 = 84$ – user76284 Jul 18 '19 at 07:38
  • @Botond "Why is it different from your Ramanujan sum?" I never mentioned a Ramanujan sum. – user76284 Jul 18 '19 at 07:39
  • @user76284 I don't see the "problem" with your derivation of 0=84. $\sum_{n \in \mathbb{N}} n = -\frac{1}{12}$ would also mean that $0=1$, as you can see in the question because you are using the properties of a different function. Why do you expect that it will work? – Botond Jul 18 '19 at 07:42
  • @Botond $\sum_{n=1}^\infty n = -1/12$ does not imply $0=1$. In fact, your method is even worse than I described because it violates homogeneity: $42 = \sum_n 2n = 2 \sum_n n = 2 \cdot 42 = 84$. – user76284 Jul 18 '19 at 07:44
  • @user76284 I think I've had enough of your sigmas. Have a nice day! – Botond Jul 18 '19 at 07:47
  • @Botond Could you explain why you think my manipulation can be used to derive $0=1$ from $\sum_{n=1}^\infty n = -1/12$? To derive the contradictions, I only used additivity (in fact, subadditivity alone suffices) in my first example and positive homogeneity in my second example. The OP, on the other hand, is using the property of stability, which cannot be satisfied by any method capable of summing certain kinds of series (like the one in the OP) and thus can never belong to any "complete" or "sufficiently powerful" theory of summation. – user76284 Jul 18 '19 at 08:25