# $1+2+3+4+\cdots=-\frac{1}{12}$?

No, folks! I am not drunk nor I am pot-headed. Yet, I am about to discuss the crazy identity
$$1+2+3+4+\cdots=-\frac{1}{12}.$$
No, I am not joking either. This is actually pretty serious mathematics and is also pretty serious stuff even to physicists. I promise you that by the time you finish reading this blog article, it will all make sense to you (I hope). So, please bear with me.

The very first thing we learned about numbers in elementary school was how to add two numbers, and that was not bad. But when it came to adding three numbers, things were confusing. Because we didn’t know what $1+2+3$ meant. It could mean $(1+2)+3$ i.e. add 1 and 2 first, and then add the resulting number (which is 3) to 3. Or, it could mean $1+(2+3)$ i.e. add 2 and 3 first and then add 1 to the resulting number (which is 5). It turns out that whichever you do it does not matter. They will all turn out to be the same number 6. In fact, for any real numbers $a$, $b$ and $c$, the following property holds:
$$(a+b)+c=a+(b+c).$$
This property, as we recall, is called the associative law. By the induction process, we know that what is true for three numbers is true for any finitely many numbers. For instance, knowing that the associative law holds for three numbers, we also get
$$(a+b+c)+d=a+(b+c+d)$$
for any real numbers $a,b,c,d$. This means that the sum $a+b+c+d$  can be obtained by any of the following four ways
$$[(a+b)+c]+d,\ [a+(b+c)]+d,\ a+[(b+c)+d],\ a+[b+(c+d)].$$
For multiplication, while complex numbers (2-dimensional numbers) and quaternions (4-dimensional numbers) satisfy the associative law, octonions (8-dimensional numbers) do not. The associative law, on the other hand, does not hold in general when we add infinitely many numbers. For instance, let us consider the Grandi’s series
$$1+(-1)+1+(-1)+\cdots.$$
If we assume that the associative law still holds for this case, we can prove something interesting. First, by the associative law we obtain
\begin{align*}
1+(-1)+1+(-1)+\cdots&=(1-1)+(1-1)+(1-1)+(1-1)+\cdots\\
&=0+0+0+0+\cdots\\
&=0.
\end{align*}
But then again by the associative law we also obtain
\begin{align*}
1+(-1)+1+(-1)+\cdots&=1+(-1+1)+(-1+1)+(-1+1)+\cdots\\
&=1+0+0+0+\cdots\\
&=1.
\end{align*}
Therefore, we just proved that $0=1$. Of course this is bullshit! Now that we know the associative law does not hold in general for infinite sums, our question is what do we mean by adding an infinitely many numbers? That is what do we mean by
$$\sum_{k=1}^\infty a_k=a_1+a_2+a_3+\cdots+a_k+\cdots?$$
Although we think that we can perceive the notion of infinity, our brain can actually process only finitely many things just like computers do. So this is the way we perceive natural numbers. We don’t actually perceive the entire infinitely many natural numbers. We can only count finitely many of them but our mind can convince us that the process continues indefinitely as 2 comes after 1, 3 comes after 2,$\cdots$,  1 million 1 comes after 1 million, and so on so forth. This is the way we perceive infinity. We do not perceive actual infinity but only potential infinity through finite processes. From intuitionism point of view, the actual infinity such as the set $\mathbb{N}$ of all natural numbers is an illusion and it should not be considered as a mathematical object. Finitism even rejects the notion of potential infinity and says that “a mathematical object does not exist unless it can be constructed from the natural numbers in a finite number of steps.” (By the way I am not an intuitionist but a
Platonist.) Back to our previous question. What we really can do is the finite sum
$$s_n=\sum_{k=1}^n a_k=a_1+a_2+a_3+\cdots+a_n,$$
which is called the $n$-th partial sum, but we can define the infinite sum $\displaystyle\sum_{k=1}^\infty a_k$ as the limit of the $n$-th partial sum $s_n$,  $\displaystyle\lim_{n\to\infty}s_n$. If this limit exists as a finite number, we say the infinite sum $\displaystyle\sum_{k=1}^\infty a_k$ exists. If the limit does not exist or becomes $\infty$ or $-\infty$, we say the infinite sum does not exist. The $n$-th partial sum $s_n$ for the Grandi’s series is $1,0,1,0,1,0,\cdots$ so the sequence of partial sums $\{s_n\}$ does not converge, and hence the Grandi’s series does not converge in ordinary sense i.e. the way we learned in calculus. Although this definition of infinite sums appears to be most natural and intuitive, there may be other legitimate ways to define infinite sums. In fact, there are. One of them is Cesàro’s sum. Cesàro’s sum of an infinite series is defined by
$$\lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^ns_k}{n}$$
i.e. the limit of the arithmetic mean of the first $n$ partial sums as $n\to \infty$. If an infinite series is summable, it is Cesàro summable. But the converse need not be true. A counterexample for the converse is Grandi’s series. The partial sums of Grandi’s series are
$$\frac{1}{1},\frac{1}{2},\frac{2}{3},\frac{2}{4},\frac{3}{5},\frac{3}{6},\frac{4}{7},\frac{4}{8},\cdots$$
and the limit of this sequence is $\frac{1}{2}$ as seen in the following picture.

Cesaro’s sum of Grandi’s series

There is another summation method called Abel summation which is more powerful than Cesàro’s summation. It uses a different mean called the abelian mean. Let $\{\lambda_n\}_{n=0}^\infty$ be a strictly increasing sequence such that $\lambda_0\geq 0$ and $\displaystyle\lim_{n\to\infty}\lambda_n=\infty$. Let $f(x)=\displaystyle\sum_{n=0}^\infty a_ne^{-\lambda_nx}$. Suppose that $f(x)$ converges for all real numbers $x>0$. Then the abelian mean $A_\lambda$ is defined as
$$A_\lambda(s)=\lim_{x\to 0+}f(x).$$
Now let $\lambda_n=n$. Then we obtain
$$f(x)=\sum_{n=0}^\infty a_ne^{-nx}=\sum_{n=0}^\infty a_nz^n\ (z=e^{-x})$$
and
$$\lim_{x\to 0+}f(x)=\lim_{z\to 1-}\sum_{n=0}^\infty a_nz^n.$$
This limit is called Abel summation. Let us consider the infinite sequence
$$\{a_n\}=\{(-1)^n(n+1):n=0,1,2,\cdots\}=\{1,-2,3,-4,\cdots\}.$$
The series $\displaystyle\sum_{n=0}^\infty a_n$ is not summable nor is Cesàro summable as seen in the following pictures.

Cesaro’s sum of 1-2+3-4+…

But it is Abel summable. To see this,
\begin{align*}
\sum_{n=0}^\infty a_nz^n&=1-2z+3z^2-4z^3+\cdots\\
&=\frac{1}{(1+z)^2}
\end{align*}
for $|z|<1$. Thus,
$$\lim_{z\to 1-}\sum_{n=0}^\infty(-1)^n(n+1)z^n=\lim_{z\to 1-}\frac{1}{(1+z)^2}=\frac{1}{4}.$$
This means than
$$1-2+3-4+\cdots=\frac{1}{4}$$
as Abel summation. Abel summation also can calculate Grandi’s series. Let $a_n=(-1)^n$, $n=0,1,2,\cdots$. Then
\begin{align*}
\sum_{n=0}^\infty a_nz^n&=1-z+z^2-z^3+\cdots\\
&=\frac{1}{1+z}
\end{align*}
for $|z|<1$. Hence, we obtain
$$\lim_{z\to 1-}\sum_{n=0}^\infty a_nz^n=\lim_{z\to 1-}\frac{1}{1+z}=\frac{1}{2}$$
as we have seen earlier. From this we see that
$$(1-1+1-1+\cdots)^2=1-2+3-4+\cdots.$$
This identity can be also obtained by the Cauchy product. Cesàro summation or Abel summation can be used to calculate oscillating divergent series to possibly produce a finite answer. However, they cannot produce a finite answer for a series that diverges to $\infty$. For example, $1+2+3+4+\cdots$ is neither Cesàro summable nor Abel summable.

Interestingly Srinivasa Ramanujan (1877-1920) showed in his notebook that
$$1+2+3+4+\cdots=-\frac{1}{12}$$
and his proof is pretty elementary. Let $c=1+2+3+4+\cdots$. Then
\begin{aligned} c&= 1&+&2+3&+&4+5&+&6&+&\cdots\\ 4c&= &+&4 &+&8 &+&12&+&\cdots \end{aligned}
Subtracting the second identity from the first results
$$-3c=1-2+3-4+5-6+\cdots=\frac{1}{4}$$
and hence we obtain
$$c=-\frac{1}{12}\ \mbox{i.e.}\ 1+2+3+4+\cdots=-\frac{1}{12}.$$
This appears to be a way too elementary and simple proof for the outrageous claim. The problem with this proof is that it is carried out by assuming that $c$ is a finite number. But we don’t know that, do we? In ordinary sense, $c=\infty$ so subtracting $4c$ from $c$ would result $\infty-\infty$ on the left hand side which is undefined, while it would result $1-2+3-4+5-6+\cdots$ on the right hand side. You need to be very careful when you deal with infinite sums and treating them like finite numbers is dangerous as it may result an inconsistent result. While I am not happy with Ramanujan’s proof, what he claimed may still be true. And yes, I am still sober. In fact, there is a much more sophisticated way to show that $1+2+3+4+\cdots=-\frac{1}{12}$. It is called zeta function regularization. Let
$$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\cdots$$
where $s=\sigma+it$ is a complex variable. $\zeta(s)$ converges for all complex numbers $s$ with $\sigma=\mathrm{Re}(s)>1$. $\zeta(s)$ diverges when $\sigma=\mathrm{Re}(s)\leq 1$, in particular when $s=-1$ in which case we obtain
$$\zeta(-1)=1+2+3+4+\cdots.$$
Bernhard Riemann (1826-1866) showed that $\zeta(s)$ can be continued analytically to the punctured plane $\mathbb{C}\setminus\{1\}$. The analytic continuation of $\zeta(s)$ is called the Riemann zeta function. If you are not familiar with analytic continuation, I explained the idea of analytic continuation using a simple example here. The Reimann zeta function is a meromorphic function on $\mathbb{C}$, which is holomorphic everywhere except for a simple pole at $s=1$.
\begin{aligned} \zeta(s)&=1^{-s}&+&2^{-s}&+&3^{-s}&+&4^{-s}&+&5^{-s}&+&6^{-s}&+&\cdots\\ 2\cdot 2^{-s}\zeta(s)&=0&+&2\cdot 2^{-s}&+&0&+&2\cdot 4^{-s}&+&0&+&2\cdot 6^{-s}&+&\cdots \end{aligned}
Subtracting the second identity from the first results
$$(1-2^{1-s})\zeta(s)=1^{-s}-2^{-s}+3^{-s}-4^{-s}+5^{-s}-6^{-s}+\cdots=\eta(s).$$
The Dirichlet series
$$\eta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^s}=\frac{1}{1^s}-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\cdots$$
converges only for complex numbers $s$ with $\mathrm{Re}(s)>0$. However, $\eta(s)$ is Abel summable for any complex number $s$. Hence, it can be analytically continued to the entire complex plane $\mathbb{C}$. The analytic continuation of $\eta(s)$ which is an entire function is called the Dirichlet eta function. The identity $(1-2^{1-s})\zeta(s)=\eta(s)$ still holds when both functions are continued analytically to the punctured plane $\mathbb{C}\setminus\{1\}$. Substituting $s=-1$, we obtain $-3\zeta(-1)=\eta(-1)$. $\eta(-1)=1-2+3-4+5-6+\cdots$ diverges to $\infty$ but it is Abel summable and the Abel sum of the series is $\frac{1}{4}$
$$\eta(-1)=\lim_{z\to 1-}(1-2z+3z^2-4z^3+\cdots)=\lim_{z\to 1-}\frac{1}{(1+z)^2}=\frac{1}{4}$$
as we have seen earlier. $\eta(-1)$ as the analytic continuation of $\eta(s)$ evaluated at $s=-1$ is the Abel sum $\frac{1}{4}$. Therefore, we obtain $\zeta(-1)=-\frac{1}{12}$ i.e. $1+2+3+4+\cdots=-\frac{1}{12}$. So, have we actually proved $1+2+3+4+\cdots=-\frac{1}{12}$ now? The answer is yes and no. No, no, I am not playing with you. Let me explain. What we actually have proved here is $\zeta(-1)=-\frac{1}{12}$ and in fact you also remember that $\zeta(-1)$ is not defined because it diverges. So here $\zeta(-1)$ is not really $1+2+3+4+\cdots$ but the Riemann zeta function i.e. the analytic continuation of $\zeta(s)$ evaluated at $s=-1$. Precisely speaking, it is not that $1+2+3+4+\cdots=-\frac{1}{12}$ but that we can assign the infinite series $1+2+3+4+\cdots$ a unique finite number $-\frac{1}{12}$, which coincides with Ramanujan’s calculation, using the analytic continuation of $\zeta(s)$. As an analogy, in my notes here, $\displaystyle f_1(s)=\sum_{n=0}^\infty(-1)^ns^n$ is not defined at $s=\frac{3}{2}i$ because $f_1\left(\frac{3}{2}i\right)$ diverges to $\infty$, however the analytic continuation $F(s)$ is defined at $s=\frac{3}{2}i$ and that $F\left(\frac{3}{2}i\right)=\frac{4}{13}-\frac{6}{13}i$. Hence, we may assign the divergent series $f_1\left(\frac{3}{2}i\right)$ a unique finite number $\frac{4}{13}-\frac{6}{13}i$.

Giving a finite value to a divergent quantity is not an unusual practice in mathematics. For instance, we can turn the infinite number line $\mathbb{R}=(-\infty,\infty)$ into a unit circle $S^1$ by assigning a finite point to the boundary $\{\pm\infty\}$ of $\mathbb{R}$. Here is one way to do it. The infinite number line $(-\infty,\infty)$ is homeomorphic to the open interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ via the map $f(t)=\arctan t$. This means that $f(t)$ is a homeomorphism i.e it is one-to-one and onto, is continuous and its inverse $f^{-1}(t)$ is also continuous ($f^{-1}(t)=\tan t$). In the eyes of topologists two homeomorphic spaces are the same i.e. they do not distinguish $(-\infty,\infty)$ and $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. The open interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is then homeomorphic to $(0,2\pi)$ via the homeomorphism $g(s)=2s+\pi$. The open interval $(0,2\pi)$ is then homeomorphic to the unit circle $S^1$ with one point $(1,0)$ removed via the homeomorphism $h(\theta)=(\cos\theta,\sin\theta)$. The composition $(h\circ g\circ f)(t)$ is then a homeomorphism from $(-\infty,\infty)$ to $S^1\setminus\{(1,0)\}$. By adding a point $(1,0)$ to $S^1\setminus\{(1,0)\}$, we obtain a unit circle $S^1$. This process is called one-point compactification in topology. Similarly adding a point $\infty$ (this is called the ideal point) to the infinite plane $\mathbb{R}^2$ and identifying the boundary of $\mathbb{R}^2$ with the ideal point $\infty$ results a unit sphere $S^2$. This whole process can be done via the stereographic projection from the north pole of $S^2$ as seen in the following picture. In fact, the stereographic projection can be also used to show that we can obtain a unit circle $S^1$ by adding a point to the real line $\mathbb{R}$ (and this point is identified with the boundaries $\{\pm\infty\}$ of $\mathbb{R}$).

Stereographic projection from the north pole

In the study of instantons in physics, an infrared cut-off, the finite-energy condition imposed on Yang-Mills action with $|x|\to\infty$ where $x\in\mathbb{R}^4$, geometrically amounts to the one-point compactification of $\mathbb{R}^4$ which is the 4-sphere $S^4$.

In the strip $0<\mathrm{Re}(s)<1$, the zeta function satisfies the functional equation
$$\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s).$$
One can immediately see that $\zeta(s)=0$ for $s=-2,-4,-6,\cdots$, negative even integers. These are called trivial zeros of the Riemann zeta function and there are indeed nontrivial zeros of the Riemann zeta function as well. There is a famous conjecture, called the Riemann Hypothesis, regarding the nontrivial zeros of the Riemann zeta function, namely the nontrivial zeros $s$ of the Riemann zeta function all have real part $\mathrm{Re}(s)=\frac{1}{2}$. This conjecture still has not been resolved (proved or disproved). It is part of Hilbert’s eighth problem along with the Golbach conjecture in the Hilbert’s 23 unsolved problems and is also one of the Clay Mathematics Institute Millennium Prize Problems. Using the Riemman zeta function we can calculate a finite answer for another divergent series $1+1+1+\cdots$.
\begin{align*}
\zeta(0)&=\lim_{s\to 0}2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s)\\
&=\frac{1}{\pi}\lim_{s\to 0}\sin\left(\frac{\pi s}{2}\right)\zeta(1-s)\\
&=\frac{1}{\pi}\lim_{s\to 0}\left(\frac{\pi s}{2}-\frac{\pi^3s^3}{48}+\cdots\right)\left(-\frac{1}{s}+\cdots\right)=-\frac{1}{2}.
\end{align*}
That is, we obtain
$$1+1+1+\cdots=-\frac{1}{2}.$$

This sort of regularization i.e. giving a finite value to a divergent quantity is particularly important to physicists along with renormalization as the divergence of path integrals often appears in quantum field theory even when physically you expect finite values for those integrals. Many physicists are trying to find the resolution for the divergences from mathematics such as regularization. However, I don’t think that the answer is in mathematics. I suspect that the occurrence of divergences in quantum field theory may have originated from its foundation and I believe that the answer can be found by carefully and thoroughly reexamining the way the current quantum field theory is formulated. For one, in quantum field theory particles are treated as mathematical points i.e. there are no inner structures of particles contrary to the nature of actual particles. Second, I also believe that the way the path integral was formulated is incorrect. (I am not saying that the idea of path integral formulation is wrong. I do believe that the idea itself is correct. I just believe that the use of complex numbers in the formulation of path integrals is incorrect. For this, see my blog article here.) I will delve into this issue at some other time when I have a better understanding.

Back to mathematics, so what now? Well, we have two conflicting types of arithmetic. One type of arithmetic, which coincides with our perception of numbers, says that $1+2+3+4+\cdots=\infty$ and the other type of arithmetic says that
$1+2+3+4+\cdots=-\frac{1}{12}$. While this second type of arithmetic appears to be mathematically consistent, it is also against everything we experience about numbers. For instance, we know that if we add any finitely many positive numbers, the result would still be positive. Now the infinite sum $1+2+3+4+\cdots$ is not only finite but also negative! The big question is how these two conflicting types of arithmetic can be consistent with each other and how do we cope with this trouble? The situation may be parallel to what happened in geometry about a couple centuries ago. The parallel postulate, also called Euclid’s fifth postulate (because it is the fifth postulate in Euclid’s Elements) states that, in two-dimensional geometry,

If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.” In 1795, John Playfair (1748-1819) showed that Euclid’s parallel postulate can be replaced by the following axiom

At most one line can be drawn through any point not on a given line parallel to the given line in a plane.”

Many mathematicians thought Euclid’s parallel postulate might be proved from the other four postulates of Euclid and attempted to prove it and failed. Among them includes Adrien-Marie Legendre (1752-1833). In the beginning of 19th century, mathematicians realized the possibility of other geometries by negating the parallel postulate. Carl Friedrich Gauß (1777-1855) knew the possibility of non-Euclidean geometry but never publicized his finding, perhaps because he was afraid of criticisms from other mathematicians. There are two different cases with negating the Euclid’s parallel postulate. One is “In a plane, given a point $p$ and a line $\ell$ not passing through $p$, there exist two lines through $p$ which do not meet $\ell$.” Replacing the Euclid’s parallel postulate by this postulate, János Bolyai (1802-1860) and Nikolai Ivanovich Lobachevsky (1792-1856) independently discovered a non-Euclidean geometry, called hyperbolic geometry. In fact, in plane hyperbolic geometry given a point $p$ and a line $\ell$ not passing through $p$, there exist infinitely many lines through $p$ which do not meet $\ell$, i.e. there are infinitely many parallel lines. The other case is “In a plane, given a point $p$ and a line $\ell$ not passing through $p$, all the lines through $p$ meet $\ell$,” i.e. there are no parallel lines. Replacing the Euclid’s parallel postulate by this one, Bernhard Riemann discovered a non-Euclidean geometry called elliptic geometry which is the simplest case of Riemannian geometry. The existence of plane geometries with three different parallel postulates may appear to be a contradiction. Due to Arthur Cayley (1821-95), Eugenio Beltrami (1835-1912), Felix Klein (1849-1925), and Henri Poincaré (1854-1912), et. al. it turns out that non-Euclidean geometries can be modelled within Euclidean geometry, and hence there is no contradiction with having both Euclidean and non-Euclidean geometries and they can all be consistent with each other.

Having a lesson from geometry, one must wonder whether we are in the same situation with the two different types of arithmetic we have now, which appear to be conflicting with each other. We obtained $1+2+3+4+\cdots=-\frac{1}{12}$ within the conventional arithmetic through analytic continuation. To me this appears to be a reminiscence of obtaining models of non-Euclidean geometries within Euclidean geometry. But a suitable mathematical or physical interpretation of this new arithmetic still lacks. I cannot shake off the feeling that we are missing something big here with a chain of questions. What is the meaning of this new arithmetic? What are the implications of this new arithmetic regarding its possible impact on mathematics? Does it indicate that there may be a new type of mathematics we don’t yet know about? If so, what could be possible impact of such new mathematics on physics? I will get to these questions in due course and hopefully by then I will have answers for them.

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### 4 Responses to $1+2+3+4+\cdots=-\frac{1}{12}$?

1. Khin Maung Maung says:

Sung,

I am glad to see that you are back to mathematics. Wonderful!!! Let’s get back to our problem.

TTYL,

Khin

• sunglee says:

Thanks! Khin

Yes, let us get back to our problem.

TTYS,
Sung

2. sunglee says:

To clarify, my criticism above on Ramanujan’s proof is solely based on our usual perspective of the infinite sum $1+2+3+\cdots$ which is $\infty$. Logically, there is nothing wrong with assuming that $1+2+3+\cdots$ is finite if one does not accept the usual definition of infinite sums.

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