How does the sum of the series “$1 + 2 + 3 + 4 + 5 + 6\ldots$” to infinity = “$-1/12$”? [duplicate]

(I was requested to edit the question to explain why it is different that a proposed duplicate question. This seems counterproductive to do here, inside the question it self, but that is what I have been asked by the site and moderators. There is no way for me to vote against their votes. So, here I go: Please stop voting this as a duplicate so quickly, which will eventually lead to this question being closed off. Yes, the other question linked to asks the same math, but any newcomer to the problem who was exposed to it via physics, as I was, will prefer this question instead of the one that is purely mathematically. I beg the moderators to not be pedantic on this one. This question spills into physics, which is why I did the cross post to the physics forum as well.)

How does the sum of the series “1 + 2 + 3 + 4 + 5 + 6…” to infinity = “-1/12”, in the context of physics?

I heard Lawrence Krauss say this once during a debate with Hamza Tzortzis (http://youtu.be/uSwJuOPG4FI). I found a transcript of another debate between Krauss and William Lane Craig which has the same sum. Here is the paragraph in full:

Let’s go to some of the things Dr. Craig talked about. In fact, the existence of infinity, which he talked about which is self-contradictory, is not self-contradictory at all. Mathematicians know precisely how to deal with infinity; so do physicists. We rely on infinities. In fact, there’s a field of mathematics called “Complex Variables” which is the basis of much of modern physics, from electro-magnetism to quantum mechanics and beyond, where in fact we learn to deal with infinity; without the infinities we couldn’t do the physics. We know how to sum infinite series because we can do complex analysis. Mathematicians have taught us how. It’s strange and very unappetizing, and in fact you can sum things that look ridiculous. For example, if you sum the series, “1 + 2 + 3 + 4 + 5 + 6…” to infinity, what’s the answer? “-1/12.” You don’t like it? Too bad! The mathematics is consistent if we assign that. The world is the way it is whether we like it or not.

-- Lawrence Krauss, debating William Lane Craig, March 30, 2011

Source: http://www.reasonablefaith.org/the-craig-krauss-debate-at-north-carolina-state-university

CROSS POST: I'm not sure if I should post this in mathematics or physics, so I posted it in both. Cross post: https://physics.stackexchange.com/questions/92739/how-does-the-sum-of-the-series-1-2-3-4-5-6-to-infinity-1-12

EDIT: I did not mean to begin a debate on why Krauss said this. I only wished to understand this interesting math. He was likely trying to showcase Craig's lack of understanding of mathematics or logic or physics or something. Whatever his purpose can be determined from the context of the full script that I linked to above. Anyone who is interested, please do. Please do not judge him out of context. Since I have watched one of these debates, I understand the context and do not hold the lack of a full breakdown as being ignorant. Keep in mind the debate I heard this in was different from the debate above.

Solution 1:

As a Physicist, I'll admit that the above is one of the many abuses of mathematics that occurs in theoretical physics like in string theory and in quantum field theory. In fact, that pseudo "analytic continuation" of $1+2+3+\dots$ being $-1/12$ using the Riemann's zeta function is done as a drop in for that infinite series which actually does occur in the development of string theory to sort of do away with it. Many other such upsetting things happen in physics like that.

Call it shameless, but physicists are that way. Kind of like how we shriek when chemists think that electronic configurations are actually accurate above hydrogen, the "lesser field" is always less rigorous than the antecedent...

Solution 2:

That is either a joke or a proof of ignorance: we have Riemann's zeta funcion

$$\zeta(s):=\sum_{n=1}^\infty\frac1{n^s}\;\;,\;\;\text{which converges for}\;\;\text{Re}(s)>1$$

If we put $s=-1$, we indeed get

$$\sum_{n=1}^\infty\frac1{n^{-1}}=1+2+3+\ldots$$

and we know that the zeta function's value at $\;z=-1\;$ is $\;-\frac1{12}\;$ ....but ...the Riemann zeta function at that value is not the above infinite series but way another thing (google" functional equation for zeta function), and from it, we get that value of $\;-\frac1{12}\;$ .

To put $\;1+2+3+\ldots=-\frac1{12}\;$ is like a mathematical joke (don't debate the humour...), and whoever who actually claims this is probably an ignorant of the above facts.

Solution 3:

He probably meant Ramanujan summation, which assigns divergent series a number. This is still not "The Answer" to the sum of the series, as this is clearly infinity, it's just a number assigned to a series.

If that is indeed what he meant, it rather ruins his argument, but we'll assume it was said in the heat of the debate.

Solution 4:

As pointed out in another answer, the "result" is related to the Riemann zeta function, $\zeta(z)$, which can be defined by the series $\zeta(z)=\sum_{n=1}^{\infty}n^{-z}$ when $\text{Re}(z)>1$. This can be extended by analytic continuation to all $z$; the unique result is the zeta function, which has a simple pole at $z=1$ and is holomorphic on the rest of $\mathbb{C}$.

The process of analytic continuation is interesting in its own right, and you may want to know how it works in this case. One way of describing it is that you can use the value of a function and its first $n$ derivatives at a point $z_0$ to estimate the value of the function and its first $n-1$ derivatives at a second point $z_1$ (and if $|z_1-z_0|$ is small enough, this process will converge as $n\rightarrow \infty$). You can then estimate the value of the function and its first $n-2$ derivatives at a third point $z_2$ (close enough to $z_1$), and so on. By tracing out a path that avoids singularities (such that $z_{i+1}$ is always within the radius of convergence of the Taylor series expansion at $z_{i}$), you can define the value of the function outside the original region, purely in terms of the function's value and derivatives ("germ") at a single point. In this case, for instance, you might choose $z=2$ to start from; the germ then depends only on the sums $$ \frac{d^k}{dz^k}\sum_{n=1}^{\infty}n^{-z}\bigg\vert_{z=2}=\sum_{n=1}^{\infty}\frac{(-1)^k\log^k n}{n^2}. $$