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This is not a duplicate question because I am looking for an explanation directed to a general audience as to the mistakes (if any) in Numberphile's proof (reproduced below). (Numberphile is a YouTube channel devoted to pop-math and this particular video has garnered over 3m views.)

By general audience, I mean the same sort of audience as the millions who watch Numberphile. (Which would mean, ideally, making little or no mention of things that a general audience will never have heard of - e.g. Riemann zeta functions, analytic continuations, Casimir forces; and avoiding tactics like appealing to the fact that physicists and other clever people use it in string theory, so therefore it must be correct.)

Numberphile's Proof.

The proof proceeds by evaluating each of the following:

$S_1 = 1 - 1 + 1 - 1 + 1 - 1 + \ldots$

$S_2 = 1 - 2 + 3 - 4 + \ldots $

$S = 1 + 2 + 3 + 4 + \ldots $

"Now the first one is really easy to evaluate ... You stop this at any point. If you stop it at an odd point, you're going to get the answer $1$. If you stop it at an even point, you get the answer $0$. Clearly, that's obvious, right? ... So what number are we going to attach to this infinite sum? Do we stop at an odd or an even point? We don't know, so we take the average of the two. So the answer's a half."

Next:

$S_2 \ \ = 1 - 2 + 3 - 4 + \cdots$

$S_2 \ \ = \ \ \ \ \ \ \ 1 - 2 + 3 - 4 + \cdots$

Adding the above two lines, we get:

$2S_2 = 1 - 1 + 1 - 1 + \cdots$

Therefore, $2S_2=S_1=\frac{1}{2}$ and so $S_2=\frac{1}{4}$.

Finally, take

\begin{align} S - S_2 & = 1 + 2 + 3 + 4 + \cdots \\ & - (1 - 2 + 3 - 4 + \cdots) \\ & = 0 + 4 + 0 + 8 + \cdots \\ & = 4 + 8 + 12 + \cdots \\ & = 4S \end{align}

Hence $-S_2=3S$ or $-\frac{1}{4}=3S$.

And so $S=-\frac{1}{12}$. $\blacksquare$

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    $\begingroup$ If you force yourself to ignore any advanced math, then the only interpretation of the infinite series is as the limit of partial sums of the series, in which case the error is in assuming the sums are convergent from the outset. We can't do that. Otherwise e.g. $(1+1+1+\cdots)-(0+1+1+\cdots)$ gets you $1=0$. $\endgroup$
    – anon
    Sep 19, 2015 at 3:03
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    $\begingroup$ Grrr... So tempted to downvote any question containing "$-1/12$"... $\endgroup$
    – MPW
    Sep 19, 2015 at 4:08
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    $\begingroup$ Ok, the whole "We don't know, so we take the average of the two" is ludicrous. How can anyone possibly accept that as sound mathematics?! Poppycock. Why not "We don't know, so we take the result to be 42" instead? $\endgroup$
    – MPW
    Sep 19, 2015 at 4:44
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    $\begingroup$ Although the proof as stated is nonsense, there is a theory of divergent series under which the value $-1/12$ is indeed assigned to this series. You may find this of interest: en.wikipedia.org/wiki/Ramanujan_summation $\endgroup$
    – user169852
    Sep 21, 2015 at 5:15
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    $\begingroup$ 25 centuries ago a Greek mathematician would have said that he knows very well what a sum of finitely many numbers is,but that if you want to assert that a sum of an infinite number of terms means anything,you had better define it before saying anything more about it.And stopping partway through a finite sum tells you nothing about the total,so why should it tell you anything in an infinite sum? Considering that you don't even know if an infinite sum exists until you have defined it. You cannot prove that a widget is blue without defining "widget.". $\endgroup$ Sep 22, 2015 at 3:44

5 Answers 5

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This is a very good question. Most responses to the video instruct the viewer to use zeta function regularization or Ramanujan summation, without actually engaging the proof and showing why it's wrong. Or their demonstrations why it's wrong ignore the entire body of summability methods for divergent series.

Although they are not careful in the video to discuss notions of convergence, and which steps are compatible with which classes of divergent series, some of their steps can be justified. For example, the series $1-1+1-1+\dotsb$ is Cesàro summable, which is a stable and linear summation method, which means that the computation $1-1+1-1+\dotsb = 1 - (1-1+1-1+\dotsb)$ is justified, which shows that $1-1+1-1+\dotsb = 1/2$ (that computation isn't in the Numberphile video you linked, where they just wave their hands and say "average it", but there is also an embedded link to another Numberphile video where they do show this computation).

Similarly the computation showing $1-2+3-4+\dotsb=1/4$ can be justified by appealing to Cesàro summability.

However the series $1+2+3+\dotsb$ is not Cesàro summable, nor indeed can any stable and linear summation method sum this series without giving rise to inconsistencies. See for example this wikipedia heading or this example computation by robjohn. So when they do the subtraction $(1+2+3+\dotsb) - (1-2+3-4+\dotsb) = 0 + 4 + 0 + 8 + \dotsb,$ they are invoking linearity; they assuming a linear summation method. Then when they write $0 + 4 + 0 + 8 + 0 + \dotsb = 4 + 8 + 12 + \dotsb$ they are assuming stability. This is the error. It is inconsistent to assume both properties.

Edit: As pointed out by Mathologer, in fact stability allows you to insert only finitely many zeros into a series. So this would be an error even in the presence of Cesàro summable series.

I wonder whether their proof can be repaired by dropping the use of one of the two axioms, linearity or stability. I couldn't do it, so I asked here.


Edit: Today (13 Jan 2018) Mathologer posted a new video which I liked very much. I hope it doesn't sound like tooting my own horn if I say it seems to be a lengthy recapitulation and explication of exactly all the points in my answer here. He does improve on my answer in one way: he points out that to insert infinitely many zeros into a sum, as is done in the Numberphile video, is not valid even with a stable summation method. I called out that step as a mistake, but only because it was invoked on a sum not known to admit a stable summation method. In fact it would be a mistake anyway.

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    $\begingroup$ What is wrong with ignoring "the entire body of summability methods for divergent series"? Ultimately, the key error in the argument is that it doesn't start with a definition of what "infinite sum" means; it just assumes one exists and has certain properties. Whether or not such a definition does exist for which the argument is valid, the argument cannot be valid without indicating what definition you have in mind. And if you don't specify a definition, it can reasonably be assumed you mean the standard definition. $\endgroup$ Jul 31, 2016 at 20:20
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    $\begingroup$ To put it another way, the real reason the argument is wrong isn't that it is inconsistent to assume both linearity and stability. The most basic error is assuming those properties in the first place, without a proof (and you can't give a proof without giving a definition of "infinite sum"). It is interesting and relevant that it actually is impossible to give a definition with both properties, but even if it weren't the argument would still be flawed (even if it is possible to fix it by setting it up in a different context). $\endgroup$ Jul 31, 2016 at 20:25
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    $\begingroup$ @EricWofsey sure, since the guys didn't start by specifying a definition of summation. For that reason, their argument is not rigorous. I'm not saying that. I'm saying that (some steps of) their argument can be made rigorous. It's an error to assume a linear summation method if you haven't shown one exists. If it does exist and you forgot to show it, it's a minor error. If it doesn't exist and there's a proof it cannot exist, it's a major error. $\endgroup$
    – ziggurism
    Nov 5, 2017 at 21:22
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The operations with series are not legitimate. In modern mathematics, what it means for a series to have a sum is clear enough.

If the author would like to operate series ad hoc like that, then he or she is accountable for creating a new set of axioms to justify the operations. For, if not, then one series may admit two different sums, which makes the theory of series opinion-based.

The operation process interesting you is like saying that $\infty - \infty = 0$. People that convince themselves that this is valid forget that subtraction operation by definition does not eat $\infty$ at all. If one, for some reason, does want to write $\infty - \infty = 0$, then he or she should give a reasonable definition of subtraction.

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The first (and fatal) error is the argument that $S_1=1/2$. That's simply not what the definition of an infinite sum is; the definition is as a limit. To show that $S_1=1/2$ (as opposed to not existing at all), you have to prove that the limit of the partial sums is $1/2$; you're not just allowed to make up whatever plausible-sounding argument you want. (There are also similar errors in the computations of $S_2$ and $S$--those computations use manipulations that can be proven to be valid if you already know the sums converge to some number, but they cannot be used to prove that the sums do converge.)

However, I think there's a larger teachable moment here, which is that ultimately, we're allowed to make whatever definitions we want in math (as long as we clearly communicate to others what our definitions are). In particular, you might think that the inability to say that $S_1=1/2$ is a flaw with the standard definition of infinite sums, so you should make a better one. Maybe you don't know how to give a completely precise definition of what your better definition of infinite sums should be, but you can write down some properties it ought to have. Maybe one of them is that if the partial sums go back and forth between two numbers, the infinite sum should be the average of them. You can carefully go through the argument and write down a list of axioms that infinite sums need to satisfy in order for the argument to go through (among them are: a sum doesn't change when you shift its summands to the right and put a $0$ at the beginning, if you add the terms of two sums together then you get a new sum whose value is the sum of the values of the two original sums, etc.).

In the end, you get a valid proof that for any notion of "infinite sum" satisfying some reasonable-looking axioms, the sum of $1+2+3+\dots$ must be equal to $-1/12$. At this point, you might point out another reasonable axiom for sums, which is that an infinite sum, all of whose terms are positive, must itself be positive. If you add this axiom, you've reached a contradiction. Thus you've proved that your seemingly reasonable list of axioms (including this last one) is self-contradictory: that is, there is no possible definition of "infinite sum" that can satisfy all your axioms. This is good evidence that maybe some of your axioms aren't quite as reasonable as you thought, and maybe you should stick to the standard definition. (In fact, it turns out that you don't even need to assume the last positivity axiom to run into problems. See the discussion in ziggurism's answer, in particular this nice argument on Wikipedia that it links to.)

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I've argued this with others unsuccessfully, but I'm a glutton for punishment.

What you are saying is

assume that

$$1 - 1 + 1 - 1 + 1 - 1 + \ldots$$ is a convergent infinite series

and let $$S_1 = 1 - 1 + 1 - 1 + 1 - 1 + \ldots$$

The assumption that

$$ 1 - 1 + 1 - 1 + 1 - 1 + \ldots$$ is a real number is a false hypothesis. Convergent series are very rigorously defined and $1 - 1 + 1 - 1 + 1 - 1 + \ldots$ isn't one.

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  • $\begingroup$ Undefined and false are not the same. $\endgroup$
    – DanielV
    Sep 22, 2015 at 2:40
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    $\begingroup$ @DanielV. So there is nothing wrong with saying "Let $x = \frac 10$." $\endgroup$ Sep 22, 2015 at 2:52
  • $\begingroup$ @StevenGregory did those "others" include mathematicians? Just curious. I find your answer the best. Short and crisp, yet true. $\endgroup$
    – SAJW
    Nov 2, 2016 at 7:27
  • $\begingroup$ There are times where a value can be assigned to a divergent series, for example see $1+2+3+\cdots$ $\endgroup$ Nov 2, 2016 at 9:38
  • $\begingroup$ @stevengregory of course you can start saying "let $x=\frac10$". The problem with that it that then many common operations doesnt work anymore, but you absolutely can do it. $\endgroup$
    – Masacroso
    Jan 17, 2018 at 11:38
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The theory of assigning a value to a series that normally is considered to diverge is called the study of "Divergent Series".

G.H.Hardy wrote a well-known book with this title.

A simple way to assign a value is to have a series that converges for a limited range of its parameter, and see what happens when a value is assigned where the series does not converge.

For example: $\sum_{n=0}^{\infty} (-1)^nx^n =\frac1{1+x} $. If we set $x=1$, we get $1-1+1-1+1... = \frac12 $. If we set $x=2$, we get $1-2+4-8+16... = \frac13 $.

There are lots more advanced topics in this field. Ramanujan did a bunch.

It can be fun.

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    $\begingroup$ I'm puzzled by the downvotes on this answer. $\endgroup$ Sep 21, 2015 at 22:42
  • $\begingroup$ My guess is that they do not believe in divergent series. $\endgroup$ Sep 21, 2015 at 22:46
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    $\begingroup$ It does not answer the question, which was, "What mistakes, if any, are there in Numberphile's proof?" $\endgroup$
    – user46234
    Sep 22, 2015 at 2:01
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    $\begingroup$ OK. The answer is "none" if you allow all the operations. $\endgroup$ Sep 22, 2015 at 2:16
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    $\begingroup$ What do you mean by "allow" and "operations"? What are the "operations" you speak of? And how do I go about "allowing" or "disallowing" such operations? I don't know if your above comment was meant to be a serious one, but as it stands, it's as informative as saying that there's no mistake in $2+2=5$ if I allow all the operations. $\endgroup$
    – user46234
    Sep 22, 2015 at 7:40

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