Why Does 1+2+3+... = -1/12? The Honest Answer

On 27 February 1913, writing from Madras to G. H. Hardy in Cambridge, Srinivasa Ramanujan set down the claim that the sum of an infinite number of terms of the series $1 + 2 + 3 + 4 + \cdots$ equals $-1/12$ under his theory, and then added, in the same breath, that if he told Hardy this, Hardy would at once point out the lunatic asylum as his goal. Ramanujan understood exactly how the statement reads to someone who has not been told what it means. A century later, after a Numberphile video in January 2014 pushed the identity into mainstream news coverage, most people encountering it are in Hardy’s position, minus the context.

So why does $1+2+3+\cdots = -1/12$ ? The honest answer is that, read as an ordinary sum, it does not. The series diverges. Its partial sums are the triangular numbers $S_n = n(n+1)/2$ , which increase without bound, and by the definition of convergence used everywhere in analysis the series has no sum at all: not $-1/12$ , not infinity as a number, nothing. What is true is a different statement about a different object. The Riemann zeta function $\zeta(s)$ agrees with the formula $\sum_{n=1}^{\infty} n^{-s}$ only on the half-plane $\mathrm{Re}(s) > 1$ . It is defined on almost the whole complex plane by analytic continuation, and at $s = -1$ , which is nowhere near that half-plane, it takes the value $\zeta(-1) = -1/12$ .

The equation is therefore an abbreviation, and a misleading one. It writes down the series that the zeta function would formally be at $s = -1$ , and sets it equal to the value of the continued function at that point. Those are two different mathematical objects, and the abbreviation hides the switch. Everything else in this article is an elaboration of that one sentence: what analytic continuation is, why the popular derivations get the right number by illegitimate means, what Ramanujan actually had, and why physicists use the result without believing the series literally sums to a negative fraction.

What the Series Actually Does

There is no subtlety in the divergence. The partial sums are

$S_n = 1 + 2 + \cdots + n = \frac{n(n+1)}{2},$

so $S_{10} = 55$ , $S_{100} = 5050$ , and $S_{1{,}000{,}000} = 500{,}000{,}500{,}000$ . For any bound $M$ you name, $S_n$ exceeds it once $n$ is large enough. A series converges to a limit $L$ when its partial sums eventually stay arbitrarily close to $L$ ; these stay arbitrarily far from every real number. The sequence $S_n$ does not approach $-1/12$ , does not oscillate near it, and never takes a negative value.

This matters because a common defence of the identity is that infinity is strange and intuition fails. Intuition does not fail here. Divergence to infinity is one of the least strange things in analysis, and the series $1+2+3+\cdots$ is about as unambiguously divergent as a series can be. Whatever $-1/12$ is, it is not the limit of those partial sums.

The Same Trick in Miniature

Before touching the zeta function, look at a case small enough to see through completely. For $|x| < 1$ ,

$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}.$

Both sides are perfectly respectable on the disc $|x| < 1$ . At $x = 1/2$ the left side sums to 2 and the right side gives $1/(1 - 1/2) = 2$ . Agreement.

Now put $x = 2$ . The left side is $1 + 2 + 4 + 8 + \cdots$ , which diverges. The right side is $\frac{1}{1-2} = -1$ , a perfectly ordinary number. And there is a popular-style derivation that produces it:

$S = 1 + 2 + 4 + 8 + \cdots, \qquad 2S = 2 + 4 + 8 + \cdots = S - 1, \qquad \text{so } S = -1.$

This is exactly the shape of the $-1/12$ argument, in a setting where nobody feels obliged to defend it. The manipulation assumed that $S$ is a number obeying the usual rules of arithmetic, which is precisely what is in question. What the calculation really shows is that if the divergent series is assigned a value by any method that is linear and respects shifting, that value must be $-1$ . It does not show the series sums to $-1$ , because it does not sum to anything.

The relation between $-1$ and $1+2+4+\cdots$ is the same relation as between $-1/12$ and $1+2+3+\cdots$ : a formula valid in one region, evaluated outside it. Nothing about infinity is being revealed. A function is being extended.

Analytic Continuation, Without the Machinery

The extension is not arbitrary, and this is the point that gets lost. Suppose you have a function defined by a formula that only works on some region, and you want to extend it. In general there are infinitely many ways to extend a function. But if you insist that the extension be analytic, meaning complex-differentiable, the freedom evaporates.

The identity theorem says: if two analytic functions defined on a connected open set agree on any subset with an accumulation point, they agree everywhere on that set. So an analytic function is rigid. Knowing it on a tiny disc determines it on the entire connected domain to which it can be analytically extended. If an extension exists, it is unique.

That uniqueness is what makes $\zeta(-1) = -1/12$ a fact rather than a convention. There is exactly one analytic function on $\mathbb{C} \setminus \{1\}$ that agrees with $\sum n^{-s}$ where that series converges. It is not a matter of taste which value to assign at $s = -1$ ; the value is forced by the behaviour of the function on the half-plane where the series does converge. In the geometric example, $1/(1-x)$ is the unique analytic extension of $\sum x^n$ beyond the unit disc, which is why $-1$ and not some other number is the natural value at $x = 2$ .

What analytic continuation does not do is make the divergent series converge. The extended function is a genuinely different object from the series formula. They coincide on an overlap region and elsewhere only one of them exists.

Computing the Value Honestly

Riemann’s functional equation relates $\zeta(s)$ to $\zeta(1-s)$ :

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

Set $s = -1$ . Then $2^{-1} = \tfrac12$ , $\pi^{s-1} = \pi^{-2}$ , $\sin(-\pi/2) = -1$ , $\Gamma(2) = 1$ , and $\zeta(2) = \pi^2/6$ , the value Euler found in 1735. Multiplying:

$\zeta(-1) = \frac{1}{2} \cdot \frac{1}{\pi^2} \cdot (-1) \cdot 1 \cdot \frac{\pi^2}{6} = -\frac{1}{12}.$

The $\pi^2$ cancels and out comes $-1/12$ . Notice what went into the computation: the convergent value $\zeta(2) = \pi^2/6 \approx 1.6449$ , the gamma function, and a sine. At no point did anyone add up the natural numbers.

The same machinery gives the other values people quote. Since $\zeta(-n) = -B_{n+1}/(n+1)$ in terms of Bernoulli numbers, one gets $\zeta(0) = -1/2$ , which is the source of the formal statement that $1 + 1 + 1 + \cdots = -1/2$ , and $\zeta(-3) = 1/120$ , which is the one the Casimir calculation actually uses in three dimensions.

Why the Shifting Argument Is Not a Proof

The derivation that spread through the internet runs in three steps. First, Grandi’s series:

$S_1 = 1 - 1 + 1 - 1 + \cdots = \tfrac12.$

Second, $S_2 = 1 - 2 + 3 - 4 + \cdots$ , added to a shifted copy of itself to give $2S_2 = S_1 = \tfrac12$ , hence $S_2 = \tfrac14$ . Third,

$S - S_2 = (1+2+3+4+\cdots) - (1-2+3-4+\cdots) = 4 + 8 + 12 + \cdots = 4S,$

so $-3S = S_2 = \tfrac14$ and $S = -1/12$ .

Every step treats divergent series as if they were numbers that can be added, shifted and subtracted. They are not. The first line already asserts a value for a series whose partial sums are $1, 0, 1, 0, \ldots$ and which therefore has no limit. The third line subtracts two divergent series term by term, and the rearrangement into $4 + 8 + 12 + \cdots$ silently reorders terms.

Riemann’s rearrangement theorem is the standard demonstration of why this reasoning is untrustworthy even for series that do converge. The alternating harmonic series

$1 - \tfrac12 + \tfrac13 - \tfrac14 + \cdots = \ln 2 \approx 0.6931$

converges, but only conditionally: the sum of the absolute values diverges. Riemann showed that such a series can be rearranged to converge to any real number you choose, or to diverge. Concretely, taking two positive terms for each negative one,

$1 + \tfrac13 - \tfrac12 + \tfrac15 + \tfrac17 - \tfrac14 + \cdots = \tfrac32 \ln 2 \approx 1.0397,$

using exactly the same terms in a different order. If reordering can change the answer for a convergent series, then reordering and shifting a divergent one carries no authority at all. An argument of that form can be made to produce whatever you want.

To see how little authority it carries, apply the same moves to Grandi’s series without the summation theory behind them. Group the terms as $(1-1) + (1-1) + \cdots$ and you get 0. Group them as $1 - (1-1) - (1-1) - \cdots$ and you get 1. Write $S_1 = 1 - (1 - 1 + 1 - \cdots) = 1 - S_1$ and you get $1/2$ . Three groupings, three answers, all obtained by manipulations that look identical to the ones in the popular derivation. The difference between the last of these and the first two is not visible in the algebra. It becomes visible only when you specify a summation method and check that the method is linear and stable, which is exactly the step the video skips.

This is why the response from mathematicians in 2014 was not that the number was wrong but that the presentation was. The video, which featured the physicists Tony Padilla and Ed Copeland, alluded to the deeper theory only in passing while performing the arithmetic as though it were ordinary algebra, and the coverage that followed, including a piece in the New York Times, largely reproduced the framing. Evelyn Lamb, writing in Scientific American’s Roots of Unity blog, put the objection squarely: the manipulations shown are not what justifies the result, and presenting them as if they were teaches the wrong lesson about infinite series. The correction is not pedantry. Someone who takes the derivation at face value has learned a false rule that will produce nonsense the next time they use it.

Why the Wrong Method Gets the Right Answer

And yet the derivation lands on $-1/12$ . That is not luck. The individual steps, taken separately, correspond to genuine summation methods that are known to be consistent with analytic continuation.

Grandi’s series is Cesàro summable to $1/2$ . Cesàro summation replaces the partial sums by their running averages. Here the partial sums are $1, 0, 1, 0, \ldots$ and the averages are

$1, \ \tfrac12, \ \tfrac23, \ \tfrac12, \ \tfrac35, \ \tfrac12, \ \tfrac47, \ \ldots$

which converge to $1/2$ . That is a rigorous statement about a well-defined procedure, not a fudge.

The series $1 - 2 + 3 - 4 + \cdots$ is Abel summable to $1/4$ . Abel summation forms the power series $\sum a_n x^n$ and takes the limit as $x \to 1^-$ . Here

$\sum_{n=1}^{\infty} (-1)^{n-1} n x^{n-1} = \frac{1}{(1+x)^2} \longrightarrow \frac14 .$

Again rigorous. And it agrees with the zeta function, because the Dirichlet eta function satisfies $\eta(s) = (1 - 2^{1-s})\zeta(s)$ , so

$\eta(-1) = (1 - 2^{2})\,\zeta(-1) = -3 \cdot \left(-\tfrac{1}{12}\right) = \tfrac14 .$

The consistency is real: different regularisation schemes, applied where they apply, give the same answers, which is what makes them trustworthy tools.

The crucial honest caveat is this. Neither Cesàro nor Abel summation assigns any value to $1 + 2 + 3 + \cdots$ itself. Abel summation gives $\sum n x^n = x/(1-x)^2$ , which blows up as $x \to 1^-$ . Every summation method that is linear, stable and regular (meaning it agrees with the ordinary sum on convergent series) must fail on a series of increasing positive terms. The $-1/12$ does not come from any of the gentle methods. It comes from zeta regularisation or Ramanujan summation, which are more aggressive constructions, and it is the popular derivation’s mixing of levels that makes it look like arithmetic.

There is a further way to see where the number lives. Insert a smooth cutoff: instead of the bare sum, compute $\sum_{n=1}^{\infty} n \, \eta(n/N)$ for a smooth function $\eta$ with $\eta(0) = 1$ that decays to zero. The result has an asymptotic expansion consisting of a term proportional to $N^2$ , whose coefficient depends on the cutoff, plus a constant term, plus terms that vanish as $N \to \infty$ . The constant term is always $-\frac{1}{12}$ , whatever cutoff you chose. Terence Tao has given a detailed real-variable account of this. It is the cleanest statement of what $-1/12$ is: the cutoff-independent constant hiding underneath a divergence, not the value of a sum.

What Ramanujan Actually Had

Ramanujan’s technique was not the shifting argument. Ramanujan summation attaches to a series $\sum f(n)$ a constant extracted from the Euler-Maclaurin formula, which relates a sum to an integral plus correction terms:

$\sum_{n=a}^{b} f(n) \approx \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k \ge 1} \frac{B_{2k}}{(2k)!}\left[f^{(2k-1)}(b) - f^{(2k-1)}(a)\right].$

Roughly, one discards the divergent integral and reads off the remaining constant. It is a defined operation with stated properties, and it reproduces $-1/12$ for $f(n) = n$ and $-1/2$ for $f(n) = 1$ . It is emphatically not a claim that the partial sums converge, and Ramanujan, who was self-taught but not naive, flagged the point himself in the letter to Hardy by joking about the asylum. In the same letter he told Hardy that Hardy would not be able to follow his methods of proof if he indicated them in a single letter, which was a fair warning rather than a boast.

The historically accurate version is therefore: Ramanujan did write this, in his second letter to Hardy dated 27 February 1913, he had a real method behind it, and he did not claim it was an ordinary sum.

Where the Number Shows Up in Physics

The reason $-1/12$ is not merely a curiosity is that it appears in calculations whose predictions are measured.

The Casimir effect. In 1948 Hendrik Casimir predicted that two uncharged parallel conducting plates in vacuum attract each other, because the boundary conditions restrict which electromagnetic modes can exist between them. The zero-point energy is a sum over modes, $\tfrac12 \sum \hbar \omega_n$ , and it diverges. In the simplified one-dimensional version with $\omega_n = n\pi c / a$ , the energy is

$E = \frac{\pi \hbar c}{2a} \sum_{n=1}^{\infty} n \ \longrightarrow \ \frac{\pi \hbar c}{2a}\,\zeta(-1) = -\frac{\pi \hbar c}{24 a}.$

The realistic three-dimensional electromagnetic calculation uses $\zeta(-3) = 1/120$ and gives a force per unit area

$\frac{F}{A} = -\frac{\pi^2 \hbar c}{240\, a^4}.$

Put in numbers. At a plate separation of $a = 100$ nm, with $\hbar c \approx 3.16 \times 10^{-26}$ J·m, this gives a pressure of about 13 pascals, which is small but far from negligible at that scale and rises as the fourth power of the reciprocal separation. Casimir’s prediction resisted measurement for decades; the first firm experimental confirmation came in 1997, when Steve Lamoreaux measured the force between a gold-coated sphere and a plate using a torsion pendulum and found agreement with theory at the level of a few per cent. Later experiments have tightened this considerably.

That is the real argument for taking the regularisation seriously. A divergent quantity was replaced by a finite one via a rule, the rule produced a number, and the number was found in the laboratory.

Bosonic string theory. In light-cone quantisation, the mass-shell condition for the string contains a normal-ordering constant built from the zero-point energies of the transverse oscillator modes, of the form $\frac{D-2}{2}\sum_{n \ge 1} n$ for $D$ spacetime dimensions. Substituting $\zeta(-1)$ turns this into $-\frac{D-2}{24}$ , and consistency of the quantised theory, specifically the closure of the Lorentz algebra, forces $\frac{D-2}{24} = 1$ . Hence $D - 2 = 24$ and $D = 26$ : the critical dimension of the bosonic string.

It should be said clearly that this is a shortcut rather than the foundation. The modern covariant derivation computes the conformal anomaly, finds it proportional to $D - 26$ , and requires it to vanish. No divergent sum appears. That the quick zeta argument and the careful anomaly argument agree is the sort of cross-check that justifies the shortcut, and it is a reason to describe $-1/12$ as reliable bookkeeping rather than as a hidden truth about addition.

The Verdict, Stated Precisely

Statement	Status
$1+2+3+\cdots$ converges to $-1/12$	False
$1+2+3+\cdots$ diverges	True
$\zeta(-1) = -1/12$	True
$\sum n^{-s} = \zeta(s)$ at $s=-1$	False; the formula holds only for $\mathrm{Re}(s)>1$
$1+2+3+\cdots$ is Abel or Cesàro summable to $-1/12$	False; neither method applies
Zeta-regularised value of $\sum n$ is $-1/12$	True, by definition of the regularisation
Ramanujan summation of $\sum n$ gives $-1/12$	True

Read down that column and the confusion dissolves. The popular headline compresses a true statement about the analytic continuation of a function into a false statement about a sum, and the compression is the whole of the problem. The right instinct on first hearing it, which is that you cannot add positive numbers and get a negative one, is correct. The right conclusion is not that mathematics is playing tricks, but that the equals sign in $1+2+3+\cdots = -1/12$ is not the equals sign you learned in school. It is shorthand for a specific and well-defined regularisation, and once you know which one, the mystery is gone and something more interesting takes its place: the fact that a function defined by a series in one part of the plane carries, in its analytic continuation, information that the series itself cannot express.

Frequently asked

Is 1+2+3+... = -1/12 actually true?

Not as an ordinary sum. The partial sums are n(n+1)/2, which grow without bound, so the series diverges and has no sum in the standard sense. The true statement is that the Riemann zeta function, which equals the sum of n to the power minus s only when the real part of s exceeds 1, takes the value -1/12 at s equal to -1. That is a fact about the zeta function, not about adding up the natural numbers.

Did Ramanujan prove that the sum of all natural numbers is -1/12?

He wrote it in his second letter to G. H. Hardy, dated 27 February 1913, and immediately added that if he told Hardy this, Hardy would at once point out the lunatic asylum as his goal. Ramanujan was aware the statement looked absurd out of context. What he had was a genuine and well-defined technique, now called Ramanujan summation, which assigns a finite constant to certain divergent series via the Euler-Maclaurin formula. It is not a proof that the series converges.

Is the Numberphile video wrong?

The final number is right and the reasoning is not a proof. The video manipulates divergent series by shifting and subtracting them, operations that are not valid for series without a sum. The same style of manipulation can be used to produce contradictions. It lands on the correct value because the steps implicitly compute Abel-type summations that happen to be consistent with analytic continuation, but the video presents them as ordinary arithmetic, which they are not.

How can adding positive numbers give a negative answer?

It cannot. No process of adding finitely or infinitely many positive numbers produces a negative result. What happens is that a formula which agrees with the sum in one region is evaluated in a region where the sum does not exist. The geometric series is the same phenomenon in miniature: the sum of powers of x equals 1 over 1 minus x only when the absolute value of x is below 1, yet the right-hand side is perfectly well defined at x equal to 2, where it gives -1.

How is -1/12 used in string theory?

In light-cone quantisation of the bosonic string, the normal-ordering constant involves a divergent sum over oscillator modes. Replacing that sum by the zeta value -1/12 gives a constant equal to D minus 2, all divided by 24, and consistency of the quantised theory forces this to equal 1, hence D equals 26. The same regularisation gives the Casimir energy its finite value, and the Casimir force has been measured, which is why physicists trust the bookkeeping.