The Riemann zeta function is defined in the region by the absolutely convergent series

Thus, for instance, it is known that , and thus

For , the series on the right-hand side of (1) is no longer absolutely convergent, or even conditionally convergent. Nevertheless, the function can be extended to this region (with a pole at ) by analytic continuation. For instance, it can be shown that after analytic continuation, one has , , and , and more generally

for , where are the Bernoulli numbers. If one *formally* applies (1) at these values of , one obtains the somewhat bizarre formulae

Clearly, these formulae do not make sense if one stays within the traditional way to evaluate infinite series, and so it seems that one is forced to use the somewhat unintuitive analytic continuation interpretation of such sums to make these formulae rigorous. But as it stands, the formulae look “wrong” for several reasons. Most obviously, the summands on the left are all positive, but the right-hand sides can be zero or negative. A little more subtly, the identities do not appear to be consistent with each other. For instance, if one adds (4) to (5), one obtains

whereas if one subtracts from (5) one obtains instead

and the two equations seem inconsistent with each other.

However, it is possible to interpret (4), (5), (6) by purely real-variable methods, without recourse to complex analysis methods such as analytic continuation, thus giving an “elementary” interpretation of these sums that only requires undergraduate calculus; we will later also explain how this interpretation deals with the apparent inconsistencies pointed out above.

To see this, let us first consider a convergent sum such as (2). The classical interpretation of this formula is the assertion that the partial sums

converge to as , or in other words that

where denotes a quantity that goes to zero as . Actually, by using the integral test estimate

we have the sharper result

Thus we can view as the leading coefficient of the asymptotic expansion of the partial sums of .

One can then try to inspect the partial sums of the expressions in (4), (5), (6), but the coefficients bear no obvious relationship to the right-hand sides:

For (7), the classical Faulhaber formula (or *Bernoulli formula*) gives

for , which has a vague resemblance to (7), but again the connection is not particularly clear.

The problem here is the discrete nature of the partial sum

which (if is viewed as a real number) has jump discontinuities at each positive integer value of . These discontinuities yield various artefacts when trying to approximate this sum by a polynomial in . (These artefacts also occur in (2), but happen in that case to be obscured in the error term ; but for the divergent sums (4), (5), (6), (7), they are large enough to cause real trouble.)

However, these issues can be resolved by replacing the abruptly truncated partial sums with smoothed sums , where is a *cutoff function*, or more precisely a compactly supported bounded function that equals at . The case when is the indicator function then corresponds to the traditional partial sums, with all the attendant discretisation artefacts; but if one chooses a smoother cutoff, then these artefacts begin to disappear (or at least become lower order), and the true asymptotic expansion becomes more manifest.

Note that smoothing does not affect the asymptotic value of sums that were already absolutely convergent, thanks to the dominated convergence theorem. For instance, we have

whenever is a cutoff function (since pointwise as and is uniformly bounded). If is equal to on a neighbourhood of the origin, then the integral test argument then recovers the decay rate:

However, smoothing can greatly improve the convergence properties of a divergent sum. The simplest example is Grandi’s series

The partial sums

oscillate between and , and so this series is not conditionally convergent (and certainly not absolutely convergent). However, if one performs analytic continuation on the series

and sets , one obtains a formal value of for this series. This value can also be obtained by smooth summation. Indeed, for any cutoff function , we can regroup

If is twice continuously differentiable (i.e. ), then from Taylor expansion we see that the summand has size , and also (from the compact support of ) is only non-zero when . This leads to the asymptotic

and so we recover the value of as the leading term of the asymptotic expansion.

Exercise 1Show that if is merely once continuously differentiable (i.e. ), then we have a similar asymptotic, but with an error term of instead of . This is an instance of a more general principle that smoother cutoffs lead to better error terms, though the improvement sometimes stops after some degree of regularity.

Remark 1The most famous instance of smoothed summation is Cesáro summation, which corresponds to the cutoff function . Unsurprisingly, when Cesáro summation is applied to Grandi’s series, one again recovers the value of .

If we now revisit the divergent series (4), (5), (6), (7) with smooth summation in mind, we finally begin to see the origin of the right-hand sides. Indeed, for any fixed smooth cutoff function , we will shortly show that

for any fixed where is the Archimedean factor

(which is also essentially the Mellin transform of ). Thus we see that the values (4), (5), (6), (7) obtained by analytic continuation are nothing more than the constant terms of the asymptotic expansion of the *smoothed* partial sums. This is not a coincidence; we will explain the equivalence of these two interpretations of such sums (in the model case when the analytic continuation has only finitely many poles and does not grow too fast at infinity) below the fold.

This interpretation clears up the apparent inconsistencies alluded to earlier. For instance, the sum consists only of non-negative terms, as does its smoothed partial sums (if is non-negative). Comparing this with (13), we see that this forces the highest-order term to be non-negative (as indeed it is), but does not prohibit the *lower-order* constant term from being negative (which of course it is).

Similarly, if we add together (12) and (11) we obtain

while if we subtract from (12) we obtain

These two asymptotics are not inconsistent with each other; indeed, if we shift the index of summation in (17), we can write

and so we now see that the discrepancy between the two sums in (8), (9) come from the shifting of the cutoff , which is invisible in the formal expressions in (8), (9) but become manifestly present in the smoothed sum formulation.

Exercise 2By Taylor expanding and using (11), (18) show that (16) and (17) are indeed consistent with each other, and in particular one can deduce the latter from the former.

** — 1. Smoothed asymptotics — **

We now prove (11), (12), (13), (14). We will prove the first few asymptotics by *ad hoc* methods, but then switch to the systematic method of the Euler-Maclaurin formula to establish the general case.

For sake of argument we shall assume that the smooth cutoff is supported in the interval (the general case is similar, and can also be deduced from this case by redefining the parameter). Thus the sum is now only non-trivial in the range .

To establish (11), we shall exploit the trapezoidal rule. For any smooth function , and on an interval , we see from Taylor expansion that

for any , . In particular we have

and

eliminating , we conclude that

Summing in , we conclude the trapezoidal rule

We apply this with , which has a norm of from the chain rule, and conclude that

But from (15) and a change of variables, the left-hand side is just . This gives (11).

The same argument does not quite work with (12); one would like to now set , but the norm is now too large ( instead of ). To get around this we have to refine the trapezoidal rule by performing the more precise Taylor expansion

where . Now we have

and

We cannot simultaneously eliminate both and . However, using the additional Taylor expansion

one obtains

and thus on summing in , and assuming that vanishes to second order at , one has (by telescoping series)

We apply this with . After a few applications of the chain rule and product rule, we see that ; also, , , and . This gives (12).

The proof of (13) is similar. With a fourth order Taylor expansion, the above arguments give

and

Here we have a minor miracle (equivalent to the vanishing of the third Bernoulli number ) that the term is automatically eliminated when we eliminate the term, yielding

and thus

With , the left-hand side is , the first two terms on the right-hand side vanish, and the norm is , giving (13).

Now we do the general case (14). We define the Bernoulli numbers recursively by the formula

The first few values of can then be computed:

From (19) we see that

for any polynomial (with being the -fold derivative of ); indeed, (19) is precisely this identity with , and the general case then follows by linearity.

As (20) holds for all polynomials, it also holds for all formal power series (if we ignore convergence issues). If we then replace by the formal power series

we conclude the formal power series (in ) identity

leading to the familiar generating function

for the Bernoulli numbers.

If we apply (20) with equal to the antiderivative of another polynomial , we conclude that

which we rearrange as the identity

which can be viewed as a precise version of the trapezoidal rule in the polynomial case. Note that if has degree , the only the summands with can be non-vanishing.

Now let be a smooth function. We have a Taylor expansion

for and some polynomial of degree at most ; also

for and . We conclude that

Translating this by an arbitrary integer (which does not affect the norm), we obtain

Summing the telescoping series, and assuming that vanishes to a sufficiently high order at , we conclude the Euler-Maclaurin formula

We apply this with . The left-hand side is . All the terms in the sum vanish except for the term, which is . Finally, from many applications of the product rule and chain rule (or by viewing where is the smooth function ) we see that , and the claim (14) follows.

Remark 2By using a higher regularity norm than the norm, we see that the error term can in fact be improved to for any fixed , if is sufficiently smooth.

Exercise 3Use (21) to derive Faulhaber’s formula (10). Note how the presence of boundary terms at cause the right-hand side of (10) to be quite different from the right-hand side of (14); thus we see how non-smooth partial summation creates artefacts that can completely obscure the smoothed asymptotics.

** — 2. Connection with analytic continuation — **

Now we connect the interpretation of divergent series as the constant term of smoothed partial sum asymptotics, with the more traditional interpretation via analytic continuation. For sake of concreteness we shall just discuss the situation with the Riemann zeta function series , though the connection extends to far more general series than just this one.

In the previous section, we have computed asymptotics for the partial sums

when is a negative integer. A key point (which was somewhat glossed over in the above analysis) was that the function was smooth, even at the origin; this was implicitly used to bound various norms in the error terms.

Now suppose that is a complex number with , which is not necessarily a negative integer. Then becomes singular at the origin, and the above asymptotic analysis is not directly applicable. However, if one instead considers the telescoped partial sum

with equal to near the origin, then by applying (22) to the function (which vanishes near the origin, and is now smooth everywhere), we soon obtain the asymptotic

Applying this with equal to a power of two and summing the telescoping series, one concludes that

for some complex number which is basically the sum of the various terms appearing in (23). By modifying the above arguments, it is not difficult to extend this asymptotic to other numbers than powers of two, and to show that is independent of the choice of cutoff .

From (24) we have

which can be viewed as a definition of in the region . For instance, from (14), we have now proven (3) with this definition of . However it is difficult to compute exactly for most other values of .

For each fixed , it is not hard to see that the expression is complex analytic in . Also, by a closer inspection of the error terms in the Euler-Maclaurin formula analysis, it is not difficult to show that for in any compact region of , these expressions converge uniformly as . Applying Morera’s theorem, we conclude that our definition of is complex analytic in the region .

We still have to connect this definition with the traditional definition (1) of the zeta function on the other half of the complex plane. To do this, we observe that

for large enough. Thus we have

for . The point of doing this is that this definition also makes sense in the region (due to the absolute convergence of the sum and integral . By using the trapezoidal rule, one also sees that this definition makes sense in the region , with locally uniform convergence there also. So we in fact have a globally complex analytic definition of , and thus a meromorphic definition of on the complex plane. Note also that this definition gives the asymptotic

near , where is Euler’s constant.

We have thus seen that asymptotics on smoothed partial sums of gives rise to the familiar meromorphic properties of the Riemann zeta function . It turns out that by combining the tools of Fourier analysis and complex analysis, one can reverse this procedure and deduce the asymptotics of from the meromorphic properties of the zeta function.

Let’s see how. Fix a complex number with , and a smooth cutoff function which equals one near the origin, and consider the expression

where is a large number. We let be a large number, and rewrite this as

where

The function is in the Schwartz class. By the Fourier inversion formula, it has a Fourier representation

where

and so (26) can be rewritten as

The function is also Schwartz. If is large enough, we may then interchange the integral and sum and use (1) to rewrite (26) as

Now we have

integrating by parts (which is justified when is large enough) we have

where

We can thus write (26) as a contour integral

Note that is compactly supported away from zero, which makes an entire function of , which is uniformly bounded whenever is bounded. Furthermore, from repeated integration by parts we see that is rapidly decreasing as , uniformly for in a compact set. Meanwhile, standard estimates show that is of polynomial growth in for in a compact set. Finally, the meromorphic function has a simple pole at (with residue ) and at (with residue ). Applying the residue theorem, we can write (26) as

for any . Using the various bounds on and , we see that the integral is . From integration by parts we have and

and thus we have

for any , which is (14) (with the refined error term indicated in Remark 2).

The above argument reveals that the simple pole of at is directly connected to the term in the asymptotics of the smoothed partial sums. More generally, if a Dirichlet series

has a meromorphic continuation to the entire complex plane, and does not grow too fast at infinity, then one (heuristically at least) has the asymptotic

where ranges over the poles of , and are the residues at those poles. For instance, one has the famous *explicit formula*

where is the von Mangoldt function, are the non-trivial zeroes of the Riemann zeta function (counting multiplicity, if any), and is an error term (basically arising from the trivial zeroes of zeta); this ultimately reflects the fact that the Dirichlet series

has a simple pole at (with residue ) and simple poles at every zero of the zeta function with residue (weighted again by multiplicity, though it is not believed that multiple zeroes actually exist).

The link between poles of the zeta function (and its relatives) and asymptotics of (smoothed) partial sums of arithmetical functions can be used to compare elementary methods in analytic number theory with complex methods. Roughly speaking, elementary methods are based on leading term asymptotics of partial sums of arithmetical functions, and are mostly based on exploiting the simple pole of at (and the lack of a simple zero of Dirichlet -functions at ); in contrast, complex methods also take full advantage of the zeroes of and Dirichlet -functions (or the lack thereof) in the entire complex plane, as well as the functional equation (which, in terms of smoothed partial sums, manifests itself through the Poisson summation formula). Indeed, using the above correspondences it is not hard to see that the prime number theorem (for instance) is equivalent to the lack of zeroes of the Riemann zeta function on the line .

With this dictionary between elementary methods and complex methods, the Dirichlet hyperbola method in elementary analytic number theory corresponds to analysing the behaviour of poles and residues when multiplying together two Dirichlet series. For instance, by using the formula (11) and the hyperbola method, together with the asymptotic

which can be obtained from the trapezoidal rule and the definition of , one can obtain the asymptotic

where is the divisor function (and in fact one can improve the bound substantially by being more careful); this corresponds to the fact that the Dirichlet series

has a double pole at with expansion

and no other poles, which of course follows by multiplying (25) with itself.

Remark 3In the literature, elementary methods in analytic number theorem often use sharply truncated sums rather than smoothed sums. However, as indicated earlier, the error terms tend to be slightly better when working with smoothed sums (although not much gain is obtained in this manner when dealing with sums of functions that are sensitive to the primes, such as , as the terms arising from the zeroes of the zeta function tend to dominate any saving in this regard).

## 88 comments

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10 April, 2010 at 5:01 pm

Allen KnutsonDo you understand how Planck’s black-body radiation formula “should” be related to Euler-Maclaurin? I don’t think I’m even asking the question correctly, but confident you can deal with that part.

10 April, 2010 at 8:44 pm

Terence TaoAh, I remember discussing these sorts of questions back in grad school :-)

Well, ostensibly most visible connection between the two is that the expression appears in both. OK, let’s try deconstructing this, starting with Euler-Maclaurin. One way to formally derive it (which I didn’t emphasise above, as I wanted to keep things fairly elementary, and also rigorous) is to start with the Taylor expansion, which one can write as

where is the derivative operator. In particular

.

Using the power series expansion

mentioned in the post and using Taylor expansion and the fundamental theorem of calculus (to invert ) one can then quickly obtain a formal derivation of the Euler-Maclaurin formula.

Now for black-body radiation. The key calculation here involves the expected particle number of a boson gas at frequency . Ignoring vacuum energy issues, a n-particle state here should have energy . At temperature , the proportion of the state with particle number n is thus something like , where is the partition function at this frequency (ignoring state multiplicity arising from spin). Summing the geometric-like series, we see that the expected particle number

is then something like , which then appears in the black-body radiation law. (In three dimensions, one then gets two extra powers of coming from polar coordinates as ranges over or a discretised version thereof.)

In one dimension, the frequency is associated to the operator (this is just reflecting the relationship ). So the energy is associated to (this reflects the relation ) and so the partition function formally resembles something like . So I guess this is a sort of Wick rotation of the summation that appears in Euler-Maclaurin. The expected particle number can be viewed as a variant of the partition function. I guess if one is working in the low frequency regime then one can then Taylor expand in D and see a series with Bernoulli number coefficients appear.

To summarise: the low-frequency expansion of the black body radiation formula can be derived from a variant of a higher-dimensional Wick-rotated version of Euler-Maclaurin. It’s sort of a six-degrees-of-separation thing, but I guess there is some slight connection.

10 April, 2010 at 11:43 pm

AnonymousShould there be an expository tag?

[Added, thanks - T.]11 April, 2010 at 12:04 am

Américo TavaresSmall typo between (2) and (3)

[Corrected, thanks - T.]13 April, 2010 at 10:53 pm

Bryan Jacobstypo is back!

[Corrected (again), thanks. - T.]12 April, 2010 at 6:31 am

Mark MeckesTypo: after “integral test estimate” the first integral sign should be a summation.

[Corrected, thanks - T.]12 April, 2010 at 8:57 pm

JamalYOU ARE GREAT! How can I learn mathematics by hard byhard

13 April, 2010 at 12:17 am

Bo JacobyThanks!

Shouldn’t formula (8)

read

to avoid confusion with

[Corrected, thanks - T.]The expression

should be understood as

Otherwise the expression depend on N.

Likewise the upper summation limit in formulas 11-12-13-14-16-17-18 should be N rather than \infty.

The formula

\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \eta(n/N) = \frac{1}{2} + \sum_{m=1}^\infty \frac{\eta((2m-1)/N) – 2\eta(2m/N) + \eta((2m+1)/N)}{2}.

must be split into two lines in order to display correctly.

13 April, 2010 at 12:43 am

Bo JacobyI regret my second comment above. Now I got it.

15 April, 2010 at 11:14 am

JavierPf. Tao,

I hope you do not shoot messengers bearing bad news!

I noticed equation 19 is incorrect. The series in equation (19) is equal to zero (I find it incredible no one has made a comment about this). Therefore, the recursion formula you suggest is incorrect. (I tested the recursion for the first few Bernoulli numbers, and I also checked in wolfram website in http://mathworld.wolfram.com/BernoulliNumber.html eq. #34). Therefore, the basis of your development in this blog seems suspect to me since equation 20 has no basis for polynomials unless you replace the first derivative evaluated at one to evaluation at zero for P(x;s) = x^s. It seem interesting to me that yet equation 20 is valid for P(x;t) = exp(xt). I think this has to do with the fact that P(0;t) is identical to one.

Sincerely,

Javier

15 April, 2010 at 1:04 pm

Terence TaoDear Javier,

I believe you are using the alternate definition of the Bernoulli numbers, in which is set to -1/2 rather than 1/2. There does not appear to be a consensus on which one is the canonical definition; both definitions have their own (minor) advantages. For instance, with the convention, the formula (7) can be extended to the case s=0.

15 April, 2010 at 2:11 pm

JavierDear Pf. Tao,

I must admit do not have a math degree, but I love mathematics because of it’s beauty. I am a test engineer by trade and therefore like to test the mathematical ideas I read. I thought B_1 = -1/2 was the “original” Bernoulli number ;-) Do you have a reference for the “extra crispy” definition of the Bernoulli, B_1 = 1/2?

I think I understand your article’s motivation: you are trying to create meaning to non-convergent series by applying a smoothing function to give a reasonable values to functions with integral representations (which can be analytically continued in the complex domain), thereby using the “asymptotic series” as a quick way to calculate values of these functions, such as the Riemann Zeta function (RZ). Is this correct? If this is the case, could this “approximation” method give an advantage in calculating the zeros of RZ ? I do not know, but my intuition tells me no.

Again thanks for the clarification and not shooting the messenger :-)

Javier

28 May, 2010 at 10:00 am

Divergent sums and the class number formula « Secret Blogging Seminar[...] why Orde’s argument works. Specifically, I am going to use an idea which I learned from Terry Tao’s blog: Arguments about divergent sums are often really arguments about the constant term in asymptotic [...]

29 May, 2010 at 12:03 am

Jeremy WilliamsBefore learning of Bernoulli Numbers, I myself have derived a closed formula for the summation of consecutive integers raised to the i power. Within this fomula I derived another closed formula for computing Bernoulli Numbers.

After this, I derived a much more efficient formula for generating the summation of consecutive integers riased to the i power. If you are interest in looking at my work (in pdf format) please reply to this post, and I will email it to you.

29 May, 2010 at 12:05 am

Jeremy Williamsby the way, all of my computations and derivations give B(1) = +1/2

18 August, 2010 at 5:25 pm

Dan ChristensenIn the discussion of Grandi’s series, in the displayed equation after

“Indeed, for any cutoff function , we can regroup”

shouldn’t the leading term actually be ?

[Corrected, thanks - T.]3 September, 2010 at 2:29 pm

teretaI found my question,thanks”

19 September, 2010 at 4:54 am

Max AtkinDear Prof. Tao,

I found this article a very interesting and fun read. I was wondering if there exists a more precise statement of the following (which appears just before you mention the “explicit formula” );

“More generally, if a Dirichlet series

has a meromorphic continuation to the entire complex plane, and does not grow too fast at infinity, then one (heuristically at least) has the asymptotic

where ranges over the poles of , and are the residues at those poles.”

In particular, I would be interested to know what qualifications are hiding in “heuristically at least” and “does not grow too fast at infinity” – is there a theorem that states this somewhere?

Thanks,

Max

19 September, 2010 at 8:17 am

Terence TaoWell, I do not know of a precise reference, but most texts in analytic number theory would have some results along these lines, perhaps specialised to specific Dirichlet series such as the one for 1/zeta(s) or zeta'(s)/zeta(s). But the basic idea is simply to mimic the discussion given previously for the case to whatever Dirichlet series one is studying.

21 September, 2010 at 1:55 pm

Max AtkinThank you for taking the time to reply. It sounds as if this is something that is approached on a case by case basis rather than proving a set of properties that a_n must satisfy in order for an “explicit formula” to exist. Is this because the latter approach is too hard?

21 September, 2010 at 2:50 pm

Terence TaoIt’s more an issue of non-uniqueness. There are many different choices for what hypotheses to place here, and various conclusions one could reach (involving different hypotheses on the cutoff , different bounds on the error term, etc.). [This is common to many principles in analysis; they are not easily formalised by a "one-size-fits-all" theorem, in contrast to the more algebraic portions of mathematics, but instead need to be tailored to each separate application.]

Also, in most number-theoretic applications, the Dirichlet series in question obeys a functional equation, which can be used to strengthen the asymptotic to a significantly greater extent than can be done for a generic Dirichlet series. So it isn’t all that worthwhile to write down a specific instantiation of the heuristic formula until it is actually needed, other than as a useful exercise to test if one understands how such formulae would be derived.

6 October, 2010 at 8:49 am

Yet Another Article to Read During Break | What's Up[...] From Terry Tao’s blog. [...]

23 July, 2011 at 7:40 pm

Erdos’ divisor bound « What’s new[...] sharper bounds available by using tools such as the Euler-Maclaurin formula (see this blog post). Exponentiating such asymptotics, incidentally, leads to one of the standard proofs of [...]

28 March, 2012 at 8:01 pm

marchDr.Tao ,at s=0 the value should be -1/2 if i’m not mistaking ,anyway my question is this the only way this value obtainable ?

30 March, 2012 at 9:20 pm

marchI got it,the series is 1/2 but the value of zeta at s=0 is -1/2

4 June, 2012 at 10:36 am

What happens when a physicist does math :) | Room 196, Hilbert's Hotel[...] good reference to that is here. Wanna share?TwitterFacebookLinkedInEmailPrintLike this:LikeOne blogger likes this [...]

24 April, 2013 at 1:18 am

jose garciathanks for your lesson Professor Tao :)

my question is instead of smoothed sums could we use regulators ? for exapmle for the series {1+2+3+4+5+6+7+….} we introduce a parameter or regulator ‘s’ so { \sum_{n=1}^{\infty}n^{1-s}} so for big ‘s’ the sum conveges ?

30 June, 2013 at 12:50 pm

AnonymousZETA REGULARIZATION can also be extended to inlcude the renormalizatio of integrals http://vixra.org/pdf/1305.0171v2.pdf however no one pays me attention since i am not famous..

26 October, 2013 at 5:55 am

Jose Javier Garcia Moretait should be worth mentioning that ZETA REGULARIZATION can be also be applied to divergent integrals in Quantum Field theory and not only for divergent series look http://vixra.org/abs/1305.0171

21 January, 2014 at 12:04 pm

MrCactu5 (@MonsieurCactus)it is been my experience information travels faster by word of mouth than first-hand experience.

a discussion of zeta-regularization – stripped of any physics – can be found on math.StackExchange:

http://math.stackexchange.com/questions/39802/why-does-123-dots-1-over-12

I personally have doubts… you can pick any collection of weights and have them approach 1 and take the constant term. Standard procedure in QFT.

Nonetheless, -1/12 seems to be agreed as the value of the sum of the positive integers.

I like Dr. Tao’s explanation which uses numerical analysis rather than complex analysis, which feels contrived here. And he points out some holds I hadn’t observed before.

18 January, 2014 at 8:40 pm

Rahul RajDon’t underestimate the Zeros!

This has a mix of Good Maths and Bad Maths which muddles the mind of the less-than-mathematically-inclined, including yours truly. However, the equations [4], [5], [6], [7], [8] and [9] (at least) make the same bad deductions as the numberphile video. You can see the fallacy here http://rahulraj-says.blogspot.in/2014/01/the-power-of-zero.html.

If the car runs at a constant speed for an hour, it *seriously* impacts the averages, and it does not actually start running 4 times faster if it does a 4x speed only the other hour.

19 January, 2014 at 10:45 pm

AnonymousBut you did not take your example to infinity. When you do the result can be negative. This is a well observed occurrence in several branches of physics starting with thermodynamics through string theory. It has even been experimentally demonstrated (in atomic physics). It is certainly unintuitive that summing positive numbers could result in a negative value, but the piece missing is that infinity is not intuitive — any introductory course on number theory or complex analysis will inform you of this. I recommend you do a bit more exploratory learning rather than claiming fallacies where they do not exist. The mathematics here is sound: Professor Tao even demonstrated several different methods at arriving with the same conclusion.

20 January, 2014 at 9:42 am

jjonly because it’s not intuitive doesn’t mean a nonsense result isn’t a nonsense result. “the result isn’t intuitive, but infinity isn’t, so this must be correct then” is no valid method of proof.

the classical sum over the natural numbers doesn’t exist (nor is it negative).

and terence tao says that in this very blog post, look above. and the wikipedia page on it says it right in the beginning. but many people tend to ignore that. because it’d be cooler to sum infinitely many positive numbers to get a negative number. that’s more psychology than mathematics.

what is being done here is not calculating the classical sum of those divergent series by “some new method”, but a method of assigning a value to a divergent sum, which isn’t really free of inconsistencies.

we need to be careful and not mix up the theory of sequences and infinite sums with what might be called ramanujan summation.

similarly we shouldn’t mix up the (cauchy) principal value of some divergent integral with the proper integral. for example

int 1/x over [-1,1] doesn’t exist, but the cauchy principal value would be 1. they are 2 different things, and a thing like the principal value exists, because it’s useful, but it’s not necessarily the “better concept”.

18 January, 2014 at 11:07 pm

TennisonSum of all the positive numbers is equal to a negative number ???????

How is it possible ?????????

19 January, 2014 at 5:09 am

AnonymousInfinity behaves in misterious ways. Why are you surprised that positive numbers can add up to a negative number in infinity, but not be touched by the counter-intuitive result of equation (2) ? The sum of rational numbers gives an irrational number as a result, in infinity… if you accept this as a well-known result, I think you can accept just as easily that adding up positive integers may lead to a negative fraction in infinity.

20 January, 2014 at 9:47 am

jjyou should only accept correct results.

while there are results that might seem counter-intuitive but are correct, there are also counter-intuitive results that are simply wrong. =)

if you actually calculate partial sums (while doing that you should [in the long run] get closer to the limit of the series, if it exists!) you will realise that if you sum (2) you really get close to 1/6 pi^2. while if you calculate partial sums of the sum of natural numbers you will realise that the number gets bigger, and the series diverges. and you can show that in under 3 lines.

calculating partial sums is no proof but maybe it will help you to visualize. convergence or divergence of the respective series can be proven without big effort.

19 January, 2014 at 11:00 am

The Sum Of All Positive Numbers[…] Numberphile and physicists Tony Padilla and Ed Copeland share one of the most mind-blowing concepts we’ve ever heard. Apparently, the sum of all positive numbers is not infinity. It’s not even a positive number. Mathlethes report here. […]

19 January, 2014 at 1:21 pm

Matt's Homepage » Blog Archive[…] http://periodicvideos.blogspot.co.uk/2014/01/thanks.html But how does it really work? TerryTao goes into tremendous detail of how the summation works and why the Grandi series can sum […]

19 January, 2014 at 1:25 pm

ericI know I’m late to this column, but I’ve never seen these connections before. Can someone point me to some references?

19 January, 2014 at 4:56 pm

More Infinite Series Madness | Of Prime Interest[…] The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic contin… […]

20 January, 2014 at 1:49 pm

The 1+2+3+…=-1/12 bonanza | Moses Supposes[…] are two ways to do this: Lamb explains analytic continuation and links to Terry Tao’s great post proving the same result using calculus. Analytic continuation by imperfect analogy: I have a car […]

20 January, 2014 at 5:23 pm

Le scandale des séries divergentes ! (ou le retour de 1+2+3+4+5+… = -1/12) | Science étonnante[…] Un excellent billet de Terence Tao qui donne des réponses à plusieurs questions qui se posent à la fin de ce billet : pourquoi la méthode zeta n’est pas stable, quelle est la connection avec Euler-McLaurin, etc. Ca picote un peu, mais il est fort le bougre ! (merci Hervé !) […]

21 January, 2014 at 11:48 am

Benjamin SprungWhen you did regroup Grandi’s series together with smooth summation (shortly before exercise 1) it should be , right?

[Corrected, thanks - T.]21 January, 2014 at 6:09 pm

Naturalness and Infinity | The Furloff[…] to our emotional selves to think that stuff can cancel out or that nature abhors infinities, yet mathematicians study these things all the time and have been for nearly three […]

22 January, 2014 at 6:31 am

Here's what you get if you add up every whole number from 1 to infinity[…] to -1/12. Watch the video above for an explanation (a technical demonstration is here; there is more than one way of getting to the same […]

24 January, 2014 at 6:08 am

An infinite series of blog posts which sums to -1/12 | The Aperiodical[…] An old post by Terry Tao on the topic: The Euler-Maclaurin formula, Bernoulli numbers, the zeta function, and real-variable analytic contin… […]

24 January, 2014 at 8:55 am

How (Not) to Sum the Natural Numbers: Zeta Function Regularization | 4 gravitons and a grad student[…] This discussion is not going to be mathematically rigorous, but it should give an authentic and accurate view of where these results come from. If you’re interested in the full mathematical details, a later discussion by Numberphile should help, and the mathematically confident should read Terence Tao’s treatment from back in 2010. […]

26 January, 2014 at 11:26 am

Yes, 1+2+3+… = -1/12. Sort of. | Michael Lauer's Weblog[…] much more satisfying. One fortuitous consequence of this brouhaha has been a lot of links to a post of Terry Tao’s of a few years ago, which I have found enormously helpful. It’s pretty […]

26 January, 2014 at 4:29 pm

Dick GrossTerry,

This is an excellent post. If you want a historical reference to the evaluation of the zeta function at negative integers, you should look at Euler’s great paper of 1749 “Remarques sur un beau rapport…”. In this paper, he is very clear that the sums he is evaluating are not convergent in the traditional sense. Euler also discovered the functional equation of the zeta function (at positive and negative integers).

Dick Gross

27 January, 2014 at 5:00 am

David ColquhounI am merely an amateur mathematician, The result 1 + 2 + 3 + . . . = -1/12 appears to me to be ‘obviously’ absurd, if only because of the sign.

I would have presumed that the result is wrong because it is based on the Grandi series which does not have any well-defined sum. Yet I see that string theorists take it literally. Do you think that they are justified in doing so?

27 January, 2014 at 5:39 pm

DougNo. As “jj” pointed out above, this is more like “pop” math. It’s cool because it’s extremely counter-intuitive. In the numberphile vid on this, when pressed for answers on how unbelievable it was, the guy filmed just cutely smiles and says “infinity” as if now it’s magically believable. Because infinity is *magic*! lols

30 January, 2014 at 1:39 pm

Kevin VanOrdThat’s an extreme mischaracterization of the Numberphile video, which did not go into nearly as much mathematical detail as Terry does, but certainly does not throw its figurative hands up in the air and say “infinity is magic lolz.” In fact, those professors go into some detail and provide multiple proofs of this particular summation. Infinity is not magic, but it is also not sensible. Your dismissal is no different from the dismissal you perceived from Drs Padilla and Copeland. Rather than make semantic points and accusations of these useful maths (I would remind you that these equations are used in physics and describe observation to essentially 100% accuracy), I’d rather see your proof showing that the sum cannot equal -1/12.

14 February, 2014 at 1:27 am

JocYou dismiss Doug’s comment, but your dismissal is of the exact sort you claim his is. In fact, your asking for a proof showing that the sum cannot be equal to -1/12 merely shows that you do not know what a correct proof is nor how to check whether a proof is correct, otherwise you would know better than to accept the fake proof given in the Numberphile video. Any mathematician will be able to tell you that the “-1/12″ is not a rational number, not to say a negative one! Anyone who does not make clear the distinction between convergent series and formal series is just deliberately misleading or misled. See Terry’s comments at https://plus.google.com/114134834346472219368/posts/ZuJDv3daT9n, which show two things. Firstly, that he also thinks it should not be written as just “-1/12″ because it certainly doesn’t mean that. Secondly, that some people choose to ignore the infinities in the asymptotic expansion, for questionable reasons. Remember that for a theory to have explanatory power it must not have unbounded arbitrarily chosen parameters. Discarding infinity is equivalent to saying that the theory works because we have managed to fit it to the data we have now, and if future data doesn’t agree we can easily create new cutoff functions to fit to it. In other words, we can never ever be wrong. Now that is foolish.

27 January, 2014 at 8:49 pm

Apparently 1+2+3+... = -1/12[…] to do this is by something called analytic continuation. Some further reading if you are curious: http://terrytao.wordpress.com/2010/0…-continuation/. Might have to do a bit of wiki hyperlink chasing […]

28 January, 2014 at 7:42 pm

JamesIs there a formal reason for why the following naive manipulation is not problematic?

S = 1+1+1+… = 1+(1+1+…) = 1+S => 0=1

3 February, 2014 at 11:25 am

AndreYes, exactly here: …)

This is where this manipulation fails.

14 February, 2014 at 1:34 am

JocIndeed it is problematic, and it is also problematic that the Numberphile video committed exactly the same mistake when he shifted one series by adding an extra zero term in front and claimed that it had the same value.

29 January, 2014 at 4:14 pm

Open Thread « Econstudentlog[…] I felt pretty much that way after watching the video below and skimming parts of Terence Tao’s related blogpost. […]

29 January, 2014 at 6:37 pm

The Sum of Positive Integers | Quantitative Scientific Solutions Blog[…] technically misleading as well). For a much more correct (and technical discussion), see this blog post by Terence […]

4 February, 2014 at 12:09 am

Din ecosistemul științei: Cum folosesc matematica fizicienii | Isarlâk[…] februarie. De asemenea (credit Steven Strogatz, pe Twitter) blogul Scientific American. Vezi și un mesaj mai vechi al lui Terry Tao in care se discută background-ul […]

4 February, 2014 at 9:19 am

One number to rule them all: -1/12 « Why Evolution Is True[…] Before you all start chipping in below pointing out basic mistakes in the maths (I don’t think there are any), read this response by Tony Padilla (I got lost when the Greek letters began) or, for the truly mathematically endowed, watch this extra and more elaborate proof which takes you through Euler’s calculation and read this mind-bogglingly difficult piece. […]

4 February, 2014 at 1:29 pm

AllenIn regards to the proof of why 1+2+3+4+….=-1/12 simpler proof,

When they move over the bottom S2 and say it is the same as the top S2. I believe that the bottom number if given the same number of terms as the top number would be equal to S2 plus infinity or S2 minus infinity. The lower moved over version of S2 is S2-1 after 1 term, then S2 +1 after 2 terms then S2-2 then S2 +2 then -3 then +3 then -4 then +4 eventually going to – infinity and +infinity greater to make the result they want by adding the 2 S2’s together. As they did with S1, one could argue that that averages to 0. That may be true mathematically in which case I wholeheartedly accept their result. But if you just go with this idea for a second, the result of what they call 2 * S2 would be S2 + S2 + or (and?) – infinity. Working through the problem that equals S1 so in the end, the series 1+2+3…… =-1/12 + or(and?) – infinity over 6 (which is still + or(and?) – infinity). I find that won’t damage the significance of -1/12 to string theory but I find this result, which I think is correct, much more philosophically rewarding. The principle of + and – which could be thought of as + and – infinity or + or – infinity is a fundamental concept in a number of religions (buddhism, shintoism, taoism) and since mathematically + and – infinity equals 0 it does nothing to the original claim that 1 + 2 + 3 +……=-1/12.

5 February, 2014 at 1:44 am

fintanDear Dr Tao, thank you for your nice presentation. You’ve accounted for the inconsistencies between (4)(5) and (6), but not for the more obvious problem: “Most obviously, the summands on the left are all positive, but the right-hand sides can be zero or negative.” This problem means that either there is a logical flaw in the statement S: “summing numbers that are all positive cannot give zero or a negative result”, or there is a logical flaw in the proofs for (4)(5) and (6). I guess statement S follows from Peano’s axioms. Can you say something about where the logical flaw lies?

5 February, 2014 at 8:03 am

Terence TaoSee the paragraph before equation (16), or my more recent discussion at https://plus.google.com/114134834346472219368/posts/ZuJDv3daT9n

6 February, 2014 at 2:22 pm

Philipp ScholzIts a question on the extra footage video:

So i showed this to my dad who is a physician and he had the problem that you simply set s=-1.

In the series 1+2x+3x^2+4x^3+…=1/(1-x)^2 my father sad that the |x| has to be <1 |x|<1 because if you set |x|=1/-1 you would get a problem. So how do you justify that you just break that rule/insert s=-1?

7 February, 2014 at 5:15 am

¿Es cierto que 1+2+3+4+…=-1/12? | Blog del Departamento de Álgebra[…] Para leer más sobre el tema, recomiendo los excelentes posts del maestro Terence Tao en Google+ y en su blog. […]

7 February, 2014 at 5:39 am

Linkage | An Ergodic Walk[…] some press now. It’s part of the Numberphile series. Terry Tao (as usual) has a pretty definitive post on […]

9 February, 2014 at 7:19 am

Nature already patched it | The Gauge Connection[…] could say. Of course, smarter mathematicians are well aware of this as you can read from Terry Tao’s blog. Indeed, Terry Tao is one of the smartest living mathematicians. One of his latest successes is […]

9 February, 2014 at 4:51 pm

MrCactu5 (@MonsieurCactus)Does (11) just say that the integeral of your cutoff function and the riemann sum of the cutoff function differ by 1/2 in the large N limit?

10 February, 2014 at 2:38 am

Series convergentes y métodos de sumación | Adsu's Blog[…] este post, Tao intenta dar una interpretación consistente de estos valores, y lo hace basandose en […]

10 February, 2014 at 9:07 am

The sum of the natural numbers is -1/12? | Scientific Clearing House[…] Terence Tao has a nice blog post on evaluating such sums. In a “smoothed” version of the sum, it can be thought of as […]

15 February, 2014 at 2:28 pm

Сумма всех натуральных чисел: 1 + 2 + 3 + 4 +… » CreativLabs[…] два раза не ходить, ещё пара интересных примеров с […]

15 February, 2014 at 7:28 pm

Сумма всех натуральных чисел: 1 + 2 + 3 + 4 +… | Вести3.ру — Информационный журнал[…] два раза не ходить, ещё пара интересных примеров с […]

20 March, 2014 at 12:07 am

JamilasadekI disagree completely with this proof because it can never work you need to set a point where to stop to make this proof. For example:

S1= 1-1+1-1+1-1+1…..

2S1=0+1-1+1-1+1-1+1…..

now here is where we go wrong. You need to have the same number of ones written wherever u stop so the above means that 2S1=2 which means S1=1 not 1/2. So it all depends on where you stop and wherever you stop there is a different answer given. So this previous cannot work as you are not supposed to stop. So my whole point is that this proof cannot work as you have to set a limit to it for it to work.

CASE CLOSED!

22 March, 2014 at 4:57 am

Robert AldridgeGrandi’s Series: In my opinion the answer is indeterminate. It’s either 0 or 1 but we don’t know which.

Every proof I’ve seen so far proves that the average of 0 and 1 is a half.

I agree that the average is a half but this isn’t the question – we want the sum.

1-S = S seems to prove the sum is a half but the S’s are NOT equal.

Both S’s are indeterminate but one is 0 while the other is 1 so we can’t rearrange to get 2S.

22 March, 2014 at 4:57 pm

Fabi@Grandi’s series:

how do you recover from the term above?

I only get after Taylor expansion.

Or do you assume, that equals 1 near the origin?

22 March, 2014 at 5:05 pm

Terence TaoYes; see the second full paragraph after (10).

23 March, 2014 at 3:31 pm

FabiThanks for the quick answer.

I intended to do your exercises since it seems fun, but I’m already stuck at exercise 1. I think the proof is similar to the one above the exercise. Maybe I didn’t get it right:

Let be twice differentiable. From taylor expansion we see:

with

And

with

If we put that in the smoothed Grandi’s series after the regroup we obtain

Since is compact supported the summand is Ok.

But you claimed that 1. the sum is only non-zero if and 2. thus the sum is .

I don’t get neither 1. and 2. …

(Yesterday I thought I had it, but I misread again…)

25 March, 2014 at 10:37 am

AnonymousDear Prof. Tao,

Can your approach be used to show

where the numbers are given in terms of the Bernoulli numbers?

The equations after (20) of this blog post look very similar.

Can this be used to shift the sum of the Riemann -function and its odd derivatives from the critical line in the Levinson-Conrey method as claimed here?

They also claim to construct a Dirichlet series which obeys a functional equation coming from this shift in addition to properties coming from the Riemann functional equation . Can this really be used to show that almost all of the zeros of the zeta function are on the critical line?

26 March, 2014 at 10:32 am

GeorgeI can’t say I can comprehend everything written here. However, this I know as fact. I believe that Terry would also agree with me (I think).

The series 4+8+12+16+… is not the same as the series 0+4+0+8+… If we could describe “1+2+3+4+…” as a number, then it is a different number than “0+1+0+2+…” In other words, if you are going to say that c=1+2+3+4+…, then keep it consistent and don’t change the definition of “c” for usage in the same equation. Using it inconsistently is like having c=5 and c=4 in the same equation. In other words, it would be having c equal to two different numbers, which defies laws of mathematical rigor that allow us to only assign one definition to a term.

26 March, 2014 at 2:53 pm

GeorgeWhile it would take work for me to learn all the stuff needed for me to get all of those integrals, the fact that 0+4+0+8+… is not the same as 4+8+12+16… demonstrates that 1+2+3+4+… is definitely equal to an infinitely large number rather than -1/12. In fact, if you notice, [1-(1+2)/4]*(1-4)=-1/12, explaining how -1/12 is achieved in the first place. 1-4 is from c-4c.

Pretending as if 0+4+0+8+… is the same as 4+8+12+16+… and then substituting the two series so easily is sure to get an answer that is false. No matter how amazing the math is afterwards, if you screw up in the beginning then you will not find the truth. If anyone considers 0+4+0+8+… to be the same as 4+8+12+16+…, then they are simply deluding themselves (which many smart people are keen on doing). Complicating the math is not going to change the result. The wiki page makes the same error, even with the zeta function. It STILL puts a space (in other words +0) between each number in the series, leading to a number that hides the truth.

27 March, 2014 at 8:11 am

GeorgeI believe though that you stated that traditional methods of treating infinity do not allow for this. The reason why traditional methods do not allow for this is because 0+4+0+8+… is a different value from 4+8+12+16+…, and thus, having c=1+2+3+4+… and calculating 4c to be 0+4+0+8+… instead of having c=4+8+12+16+… is wrong, ruining all the mathematics done afterwards.

27 March, 2014 at 10:32 pm

ghmathHello Professor Tao,

An interesting article, thanks. This article “whispered” other idea about polylogarithms and divisors sum series estimation. This method can be extended for them also.

Thanks,

Gevorg.

31 March, 2014 at 9:35 am

The Sum of Positive Integers | Quantitative Scientific Solutions | QS-2[…] technically misleading as well). For a much more correct (and technical discussion), see this blog post by Terence […]

4 May, 2014 at 2:44 am

Aake Roffo[…] From http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-func… […]

8 May, 2014 at 3:25 pm

KimHello Prof. Tao,

in paragraph 1 (smoothed asymptotics) you didn’t define

I assumed it is

In paragraph 1, there are no problems with that definition.

However, in paragraph 2 (at the beginning) you use the Euler-Maclaurin formula in paragraph 1 with where .

What is for where ?

Confused,

Kim

8 May, 2014 at 7:18 pm

Terence TaoThe complex parameter in Section 2 is unrelated to the regularity parameter in Section 1.

18 June, 2014 at 12:50 am

Grandi’s series and taking sums of infinite series by averaging | Scorkle's Media[…] was also made to Terry Tao’s blog post explaining some different ways to overcome non-converging series with alternative outcomes (i.e. […]