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This question already has an answer here:

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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marked as duplicate by Saad, Brahadeesh, Lord Shark the Unknown, Rebellos, I am Back Nov 28 '18 at 11:47

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    $\begingroup$ So you have no problems with your starting assumption? $\endgroup$ – Git Gud Nov 27 '18 at 22:47
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    $\begingroup$ "From a well known property": $1+2+3+\cdots = -\frac{1}{12}.$. What property is that? $\endgroup$ – Namaste Nov 27 '18 at 22:47
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    $\begingroup$ Wow... Amazing 'well-known property' $\endgroup$ – DavidPM Nov 27 '18 at 22:49
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    $\begingroup$ This is the first $0=1$ proof that I have seen with a starting assumption that is as unconvincing as the conclusion. $\endgroup$ – Michael Nov 27 '18 at 22:58
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    $\begingroup$ @Michael and yet it is well-known $\endgroup$ – Hagen von Eitzen Nov 27 '18 at 23:05
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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.

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    $\begingroup$ 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). $\endgroup$ – 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.

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  • $\begingroup$ +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. $\endgroup$ – Rob Arthan Nov 27 '18 at 23:15
  • $\begingroup$ @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". $\endgroup$ – user76284 Jul 17 at 21:36
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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?

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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.

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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.

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

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  • $\begingroup$ True I suppose since $p \implies ((\mbox{ not } p) \implies q)$. $\endgroup$ – Michael Nov 27 '18 at 22:59
  • $\begingroup$ 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. $\endgroup$ – Git Gud Nov 27 '18 at 23:00
  • $\begingroup$ What do you mean, @Michael? $\endgroup$ – Botond Nov 27 '18 at 23:02
  • $\begingroup$ @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 $q = \{1=0\}$. $\endgroup$ – Michael Nov 27 '18 at 23:06
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    $\begingroup$ @Botond : That is funny. I liked this answer best, I'm sorry somebody downvoted it (likely they didn't understand the logic). $\endgroup$ – Michael Nov 27 '18 at 23:51