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?

  • 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: – Siva Jan 23 '14 at 16:41
  • 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.:…………… – Luboš Motl Jan 23 '14 at 18:31
  • 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 8:25
up vote 22 down vote accepted

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 $\epsilon \rightarrow 0.$

  • 1
    Nice, +1. An alternative calculation of $\zeta(0)=-1/2$ appears early in this article:… - 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 Tamaroff Jan 24 '14 at 5:45
  • 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's_theorem – Luboš Motl Jan 24 '14 at 6:43
  • @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
  • 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 Tamaroff Jan 24 '14 at 10:48

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 $z\in\mathbb{C}$.

  • 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

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


PS: Page 43 of the following "paper"

  • 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

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.