What do we get if we sum all the natural numbers? This was the question we asked in our recent Numberphile video. The answer we gave was, to the surprise of many I'm sure, $-1/12$. It's by no means obvious, but this is the only sensible value one can attach to this divergent sum. Infinity is not a sensible value. In my opinion, as a physicist, infinity has no place in physical observables, and therefore no place in Nature. David Hilbert, one of the founding fathers of quantum mechanics, described infinity as "a mathematical abstraction that does not have a physical content". I think most physicists would firmly agree with this sentiment. The trouble is that divergent sums like the one we discuss in the video do appear in calculations of physical observables, such as the Casimir energy, or in the dimensionality of the Universe in bosonic string theory. Therefore, only a very brave individual would dream of attaching the value infinity to sums like this. Minus a twelfth is far less crazy a value when you start talking about Physics.
Nevertheless, having read the comments, tweets and blogs, if I were to ask again "what do we get if we sum the natural numbers?" I think my answer would now be "we upset people". From what I can tell, there have been two levels of objection: those that simply do not accept the result because of its apparent absurdity, and those that have taken issue with the "proof" presented in the video. To the former I assure them that while this result is utterly counterintuitive, there is a very clear mathematical sense in which this sum should be assigned this rather counterintuitive value. And most importantly to me, it makes sense when we start talking Physics. The corresponding Mathematics is deep and complicated and I won't go into it here. Instead, let me point you to a few resources online and you can take it from there:
To the second class of viewers who objected to my "proof" in the video, let me say this. I do not dispute that generically one cannot and should not manipulate divergent series in a cavalier manner. Indeed, as Abel said, "Divergent series are the invention of the devil, and it is shameful to base on them any demonstration whatsoever". However, the fact is that the manipulations I did in the video can each be mapped by analytic continuation to perfectly legitimate manipulations of convergent series, as I will show shortly. Analytic continuation is an extremely powerful mathematical tool that allows us to extend the domain of analytic functions. You think they are only defined in a given region, but analytic continuation allows you to extend them into new regions. The spirit behind analytic continuation is actually quite simple. Let me illustrate. Imagine I know the exact position of a particle for the whole of the next five seconds (quantum mechanically this is impossible, of course, but lets keep things classical). Let me also assume that the particle's motion is not subject to sudden changes - its motion is nice and smooth (thats the analytic part). Then if I know its position so precisely over this time interval, at any given point in time inside that interval, I know its position, velocity, acceleration, rate of change of acceleration, and so on. In fact, I know all of its derivatives at that point. And if I know that, and the motion is indeed sufficiently smooth, then I can infer the position of the particle at any time, beyond the five seconds that I originally knew about. I have, at least in some intuitive sense, analytically continued the particle's position beyond those original five seconds. Of course, I'm only giving an intuitive feel for analytic continuation here. In truth it really involves analytic functions defined on open subsets of the complex plane, as opposed to the real line, but you get the gist.
Analytic continuation is used all over Physics. 't Hooft and Veltmann used analytic continuation to regulate seemingly divergent integrals that appear when we study Yang Mills theory, and this won them the Nobel Prize in 1999. The notion of a white hole, as the intuitive opposite of a black hole, relies on analytically continuing the black hole beyond the initial domain. I have a slightly less ambitious goal. I want to use analytic continuation as some justification for what I did in the video. Like I have already said, the manipulations I did can each be mapped by analytic continuation to perfectly legitimate manipulations of convergent series. Because of this, the formulae I extract do in fact hold, and the manipulations are legitimate because all the while I am concealing the weapon of analytic continuation. Let me prove my claim.
Define three series in the complex plane, $z \in \mathbb{C}$ $$S_1(z)=1+\sum_{n=1}^\infty (-1)^n \left[(n+1)^{-z}-n^{-z} \right]$$ $$S_2(z)=\sum_{n=1}^\infty (-1)^{n+1} n^{-z}$$ $$S(z)=\sum_{n=1}^\infty n^{-z}$$ Now each of these series is absolutely convergent for $Re(z)>1$, so I can manipulate them pretty wantonly in this domain. The last of these corresponds to the Riemann-Zeta function, $\zeta(z)$, and it is well known that it may be analytically continued into the domain $Re(z)<0$ using a relation $\zeta(1-z) = 2(2\pi)^{-z} cos(\pi z/2) \Gamma(z)\zeta(z)$. Of course this relation itself is the key to proving that the sum of the natural numbers is $-1/12$. However, here we will use it to analytically continue general expressions into the manipulations performed in the video. To this end we note that at $z=-1$, the seriees reduce to the ones discussed in the video $$S_1(-1)=1-1+1-1+ \ldots$$ $$S_2(-1)=1-2+3-4+ \ldots$$ $$S(-1)=1+2+3+4+ \ldots$$ In the first step of my video, I write $$\begin{align} 2S_2 =&1-2+3-4+\ldots \\ +&0+1-2+3-\ldots \\=&1-1+1-1+\ldots \\=&S_1\end{align}$$ What we are really doing here, however, is the following, $$\begin{align} 2S_2(z)=&\sum_{n=1}^\infty (-1)^{n+1} n^{-z}+\sum_{n=2}^\infty (-1)^{n} (n-1)^{-z} \\ =&\sum_{n=0}^\infty (-1)^{n+2} (n+1)^{-z}+\sum_{n=1}^\infty (-1)^{n+1} n^{-z} \\ =&1+\sum_{n=1}^\infty (-1)^n \left[(n+1)^{-z}-n^{-z} \right]\ \\ =&S_1(z) \end{align}$$ The next step in the video was to write $$\begin{align} S-S_2=&1+2+3+4+\ldots\\ -&(1-2+3-4+\ldots )\\ =&4+8+\ldots \\ =&4S\end{align}$$ Of course, what we are really doing here is what Ed showed in the extra footage. Specifically, $$\begin{align} S(z)-S_2(z)=& \sum_{n=1}^\infty n^{-z}- (-1)^{n+1} n^{-z} \\ =& \sum_{m=1}^\infty (1+(-1)^{2m}) (2m)^{-z}+(1+(-1)^{2m-1}) (2m-1)^{-z} \\ =& \sum_{m=1}^\infty (1+(-1)^{2m}) (2m)^{-z} \\ =&2^{1-z} \sum_{m=1}^\infty m^{-z} \\ =&2^{1-z}S(z) \end{align}$$ Remember, these manipulations are perfectly OK for $Re(z)>1$. And now I appeal to analytic continuation because $S(z)$ is just the Riemann-Zeta function, which I know how to extend into $Re(z)<0$. This means that all the expressions above can be considered valid also for $Re(z)<0$, and in particular at $z=-1$, which are the manipulations presented in the video. I accept that manipulating divergent series in a cavalier manner is fraught with danger, but in this particular instance, backed by analytic continuation, I could get away with what I did. I wanted a simple presentation that people with limited mathematical training could follow, so I took the plunge and I went with it. I also took Grandi's series, $S(-1)=1/2$, as my starting point. Of course, I obtained this value by reverse engineering. I knew that zeta function regularisation allowed me to directly assign the value $S(-1)=-1/12$, from which it follows $S_1(-1)=1/2, ~S_2(-1)=1/4$. When I was thinking about how to make this video I looked at these three values, and I asked, which of these is intuitively the most plausible? Which will a layman get least upset about? Thus I settled on Grandi's series, $S_1(-1)=1/2$. For Grandi's series, the partial sums oscillate between 0 and 1, so the average of these two does not come across as a completely implausible assignment to a non-expert. It certainly seems much more plausible than assigning a negative number to the sum of positive integers, thats for sure!
I should also add that I am not saying that this "averaging" is the precise mathematical reason we assign the value $1/2$ to Grandi's series. As James Grime discussed in his video, one can use Cesaro summation to justify this assignment. Actually, for me, because of the manipulations I wanted to perform, I was really using zeta function regularisation to justify the assignment. But there is no way I was going to go into all of this in the video. I just wanted to give non-expert viewers a starting point they might have some confidence with, and I didn't want to get bogged down with details. I knew I could get from this to something crazy like $\sum_{n=1}^\infty n =-1/12$ using the manipulations described above, but, assigning the value $z=-1$ at each step. The validity of the steps in $Re(z)>1$ and analytic continuation meant I could get away with this, and for me it kept the discussion cleaner and more accessible for the non-experts, even if it upset (some of) the so-called experts.
There is an enduring debate about how far we should deviate from the rigorous academic approach in order to engage the wider public. From what I can tell, our video has engaged huge numbers of people, with and without mathematical backgrounds, and got them debating divergent sums in internet forums and in the office. That cannot be a bad thing and I'm sure the simplicity of the presentation contributed enormously to that. In fact, if I may return to the original question, "what do we get if we sum the natural numbers?", I think another answer might be the following: we get people talking about Mathematics.
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