8

The formal series

$$ \sum_{n=1}^\infty 1 = 1+1+1+\dots=-\frac{1}{2} $$

comes from the analytical continuation of the Riemann zeta function $\zeta (s)$ at $s=0$ and it is used in String Theory. I am aware of formal proofs by Prof. Terry Tao and Wikipedia, but I did not fully understand them. Could someone provide an intuitive proof or comment on why this should be true?

Willie Wong
  • 73,139
Tunk-Fey
  • 24,849
  • Related: http://math.stackexchange.com/q/39802/ (and links there) – Grigory M Jan 23 '14 at 12:10
  • Did you read the links? You cannot add a divergent series. – John Alexiou Jan 23 '14 at 16:33
  • 10
    @ja72: Whether an infinite series is divergent or not depends on what notion of summation one uses. – Siva Jan 23 '14 at 16:39
  • 3
    @Tunk-Fey: Lumo has a few posts about this, on his blog this month: http://motls.blogspot.com/2014/01/?m=1 – Siva Jan 23 '14 at 16:41
  • @ja72 Yup! I read it. I just wonder why does $1+1+1+\dots=-\frac{1}{2}$? Can anyone prove it by using a simple method as is shown on this video (although it's different type of number series). – Tunk-Fey Jan 23 '14 at 16:42
  • 1
    Although methods like that video seem attractive (because they're not "fancy") it's better to learn the real methods (for example zeta functions). When you need to work with naively divergent series, you should understand how regulators work and in what sense your result is meaningful. In the video, the guy writes down $1 - 1 + 1 -1 + \ldots =$ something, but give an solid argument why that must be true. – Vibert Jan 23 '14 at 16:51
  • 8
    We have been bombarded by questions on this series and $1+2+3+4\dots=\frac {-1}{12}$ lately at math. You could search for "1+2+3+4+5" and "1+1+1+1+" and "1-1+1-1+1" to find them. I don't know why, but people are seeing the video like crazy. – Ross Millikan Jan 23 '14 at 18:29
  • 3
    Five more articles with 8 different ways to compute the sum of integers (most of the methods extend to the "sum of ones", too) etc.: http://motls.blogspot.com/2007/09/zeta-function-regularization.html?m=1 http://motls.blogspot.com/2011/07/why-is-sum-of-integers-equal-to-112.html?m=1 http://motls.blogspot.com/2014/01/sum-of-integers-and-oversold-common.html?m=1 http://motls.blogspot.com/2014/01/eta-function-and-sum-of-positive.html?m=1 http://motls.blogspot.com/2014/01/a-recursive-evaluation-of-zeta-of.html?m=1 – Luboš Motl Jan 23 '14 at 18:31
  • 1
    I do not agree that such identities are more important in physics than in math. There is nothing more central to mathematics than the zeta function. – WillO Jan 24 '14 at 13:42
  • I did a little bit of clean-up in order to merge the version migrated here from Physics.SE with the version that was already present in Math.SE. – Willie Wong Jan 24 '14 at 14:37
  • 1
    @WillieWong why is the question still on hold after the merger? And why did the OP of the migrated question not keep the highly deserved upvotes? At least in the migrated question it was clear what he was asking. So can the merger be done in a more just way and the final merged question kept open? Imho it really is a legitimate question, even though it should have stayed on physics. Open ... – Dilaton Jan 24 '14 at 21:30
  • 1
    Ok as soon as my WLAN stops sucking, I will write an corresponding answer to the "Request for Reopen Votes" meta list ... – Dilaton Jan 24 '14 at 21:39
  • 1
    @WillieWong could it be that the older duplicate question was closed as "unclear what you are asking" whereas the migrated one was legitimate? The merged onr now is the migrated one just with an additional answer, so people probably forgot to reopen the now legitimate merged question...? – Dilaton Jan 24 '14 at 21:49
  • Have you watched Numberphile video? lol... – Metin Y. Jan 25 '14 at 06:47
  • 2
    @Dilaton: Site policy is always to merge the migrated version into the version we already have on file in the case of cross postings. The community voted to close the original, and the community voted to mark the migrated question as a duplicate. It should thus also be up to the community to decide whether to re-open this question. – Willie Wong Jan 27 '14 at 08:25
  • 1
    @WillieWong I see. But is it not a bit unfair to mark a legitimate question as a duplicate of a closed question, if the first one is much better and has quite some upvotes originally? However I unfortunately I have to admit that the mistake which triggered the whole issue was clearly made on the physics site: The question should never have been migrated here, since as Lubos Motl said such series are very important in QFT, ST, etc... Unfortunately moderation on physics is dominated by people who are not knowledgeable about such advanced topics at a deeper technical level, so mistakes happen ... – Dilaton Jan 27 '14 at 09:17
  • The last time I looked, analytic continuation never says that a divergent series converges. – Balarka Sen Sep 01 '14 at 10:40
  • Does this answer your question? Why does $1+2+3+\cdots = -\frac{1}{12}$? – Hans Lundmark Apr 23 '22 at 07:46

3 Answers3

22

Let me walk you through the Riemann zeta computation. Call $S$ your original sum. Let's regulate the sum as follows: $$S_s \equiv \sum_{n \geq 1} \frac{1}{n^s}.$$ Fix $n \geq 1.$ Then $n^{-s} \rightarrow 1$ as $s \rightarrow 0,$ so if we can assign a meaning to $S_s$ as $s \rightarrow 0$, we can interpret $S$ as this limit.

Now, for $s > 1$ the above sum exists and it equals the Riemann zeta function, $\zeta(s).$ $\zeta$ has a pole at $s=1$, which is just the statement that the (non-regulated) sum $\sum 1/n$ diverges. But we can analytically continue $\zeta$ if we take care to avoid this pole. Then we can Taylor expand around $s=0$

$$\zeta(s) = -\frac{1}{2} - \frac{1}{2} \ln(2\pi) s + \ldots$$ which implies that

$$S = \lim_{s \rightarrow 0} S_s = -\frac{1}{2}.$$ (The equality sign is to be understood in the regulated sense.)

There are many other ways to regulate the sum. You can e.g. suppress the tail as $\sim \exp(-\epsilon n)$, but then you need to add a counterterm to absorb a pole as $\epsilon \rightarrow 0.$

Vibert
  • 437
  • 1
    Nice, +1. An alternative calculation of $\zeta(0)=-1/2$ appears early in this article: http://motls.blogspot.com/2014/01/a-recursive-evaluation-of-zeta-of.html?m=1 - See my comment under the OP's question for 4 more articles about the topic if you wish... – Luboš Motl Jan 23 '14 at 18:49
  • 2
    @LubošMotl Here you claim the values of $\zeta(2m+1)$ are transcendental... when was this proven? – Pedro Jan 24 '14 at 05:45
  • Nope, it's just believed by me and almost everyone that they're transcendental. The much "easier" proof that zeta(3) is irrational came just recently, in 1978, see https://en.wikipedia.org/wiki/Ap%C3%A9ry's_theorem – Luboš Motl Jan 24 '14 at 06:43
  • @Vibert I know Taylor expansion and I've just learned about analytic continuation but I don't get why does $S_s$ turn out to be $-\frac{1}{2}-\frac{1}{2}\ln(2\pi)s+\cdots$? Could you please give me a detail explanation about that? Thanks... – Tunk-Fey Jan 24 '14 at 10:29
  • @PedroTamaroff Not really. It is not even known whether all of them are irrational. Though there have been a little work on them and the best result so far AFAIK is that infinitely many such zeta constants are irrational. – Balarka Sen Jan 24 '14 at 10:32
  • 5
    @LubošMotl Right, thatis my (indirect point). Why not make it clear they are believed to be transcendental rather than claim it? It is surely misleading for the casual reader, and slightly irresponsible coming from someone that knows about it, don't you think? – Pedro Jan 24 '14 at 10:48
  • Dear @Pedro, I am saying that $\zeta(2k+1)$ are transcendental numbers because I am ready to bet all my wealth on the validity of the claim. After all, even $\pi^n$ are transcendental numbers for rational nonzero values of $n$ so even zeta of even positive integers are (provably) transcendental. But the zetas of odd positive integers are (I claim without a rigorous proof) much more transcendental than the powers of $\pi$. I never claimed that I had a rigorous proof of the transcendentality but if you want to claim that I said something incorrect, you should have a proof which you don't have. – Luboš Motl Jan 24 '14 at 10:58
  • @Tunk-Fey: there's no easy explanation. If you know the story about analytically continuing through different patches, it's not difficult to write a computer program to estimate these coefficients. But the precise numbers only follow from serious computations; typically it involves using an integral representation for the $\zeta$ function, contour integrals etc. Another starting point can be the reflection formula for the gamma function. – Vibert Jan 24 '14 at 12:20
  • @Vibert So why does $\zeta(s)=-\frac{1}{2}-\frac{1}{2}\ln(2\pi)s+\cdots$? What is the analytic function that you used for this series? – Tunk-Fey Jan 24 '14 at 12:24
  • @Tunk-Fey: I didn't calculate these numbers myself, Mathematica (or any other serious CAS) does it for you. – Vibert Jan 24 '14 at 12:27
3

The result you obtain when calculating sums like $$S=\sum_{n=1}^\infty T_n$$ depends on how you define them. Here $T_n$ denotes anything that we may want to insert there.

The most intuitive way to define an infinite sum is by using partial sums. The idea is to introduce a sequence of sums $$S_N=\sum_{n=1}^N T_n$$ and then define the infinite sum $S$ as the following limit $$S=\lim_{N\to \infty}S_N.$$ Obviously, each partial sum $S_N$ is finite, however the problem is in this limit, that may diverge. For your example, evidently, this limit diverges and doesn't give anything useful.

To deal with this kind of sums people invented another approach called analytical continuation, that was described in the answer by Vibert. Not to repeat it I'll just say, that intuitively the idea is to consider a convergent sum instead of our divergent one. Then replace this sum by an analytical function (say Riemann zeta function). Finally, we take a limit of this analytical function in that region, where the initial sum diverges.

An example of analytical continuation is the well-known gamma function $\Gamma(n)$, that coincides with the function $(n-1)!$ when $n\in \mathbb{Z}$. However, $\Gamma(z)$ is defined for any complex $z\in\mathbb{C}$.

Edvard
  • 196
  • I know Taylor expansion and I've just learned about analytic continuation but in solution by Vibert, I don't get why does $S_s$ turn out to be $-\frac{1}{2}-\frac{1}{2}\ln(2\pi)s+\cdots$? If you know, could you please give me a detail explanation about that? Thanks... – Tunk-Fey Jan 24 '14 at 10:31
  • For convergent sums of the form $S_s=\sum_{n=1}^{\infty}1/n^s$, i.e. when $s>1$ we define zera function as $\zeta(s)=S_s$. To make it work for $s\leq1$ we use analytical continuation and obtain zeta function defined for any $s$. Then we put by hand, that this zeta function represents sums $S_s$ for $s\leq1$ as well. Basically, we replace the formal expression $\sum_{n=1}^{\infty}n^{s}$ with $s\leq1$ by an analytical function. The precise form $\zeta(s)=-1/2+...$ follows from its properties, I do not know the details. – Edvard Jan 24 '14 at 11:10
  • So what is the analytic function of this series? – Tunk-Fey Jan 24 '14 at 11:16
  • Riemann zeta function. – Edvard Jan 24 '14 at 12:15
2

Use following functional equation: [which is not trivial to get]

$$\pi^{-\frac{1}{2}s}\Gamma\left(\frac{1}{2}s\right)\zeta(s)=\pi^{-\frac{1}{2}(1-s)}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)$$

PS: Page 43 of the following "paper" http://www.math.ethz.ch/~gruppe5/group5/lectures/mmp/hs13/Files/Lecture%20notes%20(November%2029).pdf

b00n heT
  • 16,360
  • 1
  • 36
  • 46
  • Can you use simple explanation as is shown in this link to prove my question? – Tunk-Fey Jan 23 '14 at 13:11
  • I've taken a look at the video and it's good fun, yet in your case unfortunately you need to work with the analitic continuation of the $\zeta$ function, which is quite a mess but leads to astoundishing results as for the one given here – b00n heT Jan 23 '14 at 13:17
  • OK, let we ignore the part about using analytic continuation to answer this question. So, is there simple method to prove that $\zeta(0)=-\frac{1}{2}$? – Tunk-Fey Jan 23 '14 at 13:33
  • No. $\zeta(s)$ as you intend is is only defined for $\Re(s)>1$ – b00n heT Jan 23 '14 at 13:36
  • How about this one? As you can see in equation (4), (5), and (6) in Prof Tao's blog, $\zeta(s)$ is also defined for $\hbox{Re}(s) \le 1$. – Tunk-Fey Jan 23 '14 at 13:47
  • Quoting: "For $\Re(s) \le 1$, the series on the right-hand side of $(1)$ is no longer absolutely convergent, or even conditionally convergent. Nevertheless, the $\zeta$ function can be extended to this region (with a pole at $s=1$) by analytic continuation. For instance, it can be shown that after analytic continuation, one has..." – b00n heT Jan 23 '14 at 13:53
  • Fine then. It looks like I just accept that $\zeta(0)=-\frac{1}{2}$ without arguing anything. – Tunk-Fey Jan 23 '14 at 14:23