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This question already has an answer here:

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://math.stackexchange.com/questions/630490/how-does-the-sum-of-the-series-1-2-3-4-5-6-ldots-to-infinity

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.

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marked as duplicate by Emilio Pisanty, John Rennie, Kyle Kanos, Brandon Enright, Qmechanic Jan 7 '14 at 19:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ From a mathematical point of view, the answer is here terrytao.wordpress.com/2010/04/10/… $\endgroup$ – joshphysics Jan 7 '14 at 18:44
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    $\begingroup$ This question appears to be off-topic because it is about mathematics. $\endgroup$ – Emilio Pisanty Jan 7 '14 at 18:45
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    $\begingroup$ On the other hand, only a physicist would say something like this. ;-) $\endgroup$ – ShreevatsaR Jan 7 '14 at 18:46
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    $\begingroup$ We all love a proof that concludes with some absurd result, but this is definately maths not physics. $\endgroup$ – John Rennie Jan 7 '14 at 18:49
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    $\begingroup$ This particular sum is also discussed here, here and here, and on Math.SE here. See also this Phys.SE post. Also related Phys.SE post here. $\endgroup$ – Qmechanic Jan 7 '14 at 19:03
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This is a statement about taking improper limits while doing analysis. In particular, this case is one of the Riemann Zeta function:

$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}}$$

it can be shown that $\lim_{s\rightarrow 1} \zeta(s) = -\frac{1}{12}$. This regularization is often used in string theory to handle infinities of the type you describe.

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    $\begingroup$ Well, actually the definition is $\zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^{s}}$ for $Re s >1$. Next this function is analytically (uniquely) extended in the whole complex plane, and the extended function turns out to be everywhere analytic except for $s=1$, where it has a simple pole. With the given definition, $\zeta(-1) = -\frac{1}{12}$. Obviously, for $s=-1$ the initial series does not converge! $\endgroup$ – Valter Moretti Jan 7 '14 at 19:15
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I think the point is that you have some object, which when you naively try to calculate, happily gives you it's perturbative (series) expansion $1+2+3\dots$. The mathematician then sees the mistake and recognizes the object for what it is, and tells you the answer.

Further reading: https://www.google.com/search?btnG=1&pws=0&q=when+does+a+function+equal+to+its+taylor+series

Incidently, one of the places you encounter this sum in physics is in bosonic string theory, where each number summed represents the contribution of a mode on the string. Something similar also arises in the casimir force calculation

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