First of all, I'm aware lots of very similar questions have been asked and answered here and on other sites. However, after browsing literally tens of explanations of this fact, I couldn't find a single one that satisfactorily convinced we can really do this.

In fact, for some reason, people seem to be looking sideways and not giving straight answers whenever this is asked, which makes me think this might be a big inside joke at the expense of spreading misinformation.

If you consider this question to be a duplicate, please point out where I can find an explanation which sheds light to the points below:

I'm going to list a few premises. Please bear in mind I've only taken a Real Analysis course. Since I'm talking about a sum of real numbers, I'd expect one could explain it without making reference to complex analysis.

-Equality between real numbers means double inclusion between the sets they represent

-An infinite sum is the limit of the partial sums, from the $\epsilon-\delta$ defition of limit

-Divergent series are not equal to any number, since, by virtue of their divergence and the archimedean property, we can show they are different to any given x

Now, every explanation of why $1+2+3+4+5+6+...=-1/12$ seems to violate one of the above, either by manipulating infinite series as something other than the limit of partial sums, or by violating the radius of convergence, or by claiming equality means something else rather than equality. To me those explanations (particularly the ones referring to analytical continuations) seem akin to saying:

$f(x)=x^2$ behaves like $y=0$ near the origin. Hence, $f(7)=7^2=0$.

With that in mind, I ask the following questions:

  1. What does it mean, precisely, to take infinite sums, if we are to accept $1+2+3+...=-1/12$?

  2. What does $=$ means, precisely, in this context?

  3. (bonus) I know this is relevant to string theory. Has it ever been used to make demonstrable predictions about the real world?

  • 1
    Different notions of summability of divergent series usually violate «An infinite sum is the limit of the partial sums». One easy to comprehend example is Cesàro summation. – Grigory M Jan 9 '14 at 20:16
  • See also Ramanujan summation – Grigory M Jan 9 '14 at 20:22
  • Related: math.stackexchange.com/q/39802 (esp., IMHO, answers of Matt E. and Luboš Motl). – Grigory M Jan 9 '14 at 20:29
  • 1
    I am also curious as to what all this actually means. From what I gather (I may be wrong) it seems that this 1+2+3+... is just "loaded notation" coming from a series whose domain is being extended to have an answer when the series "looks like" 1+2+3+.... If that is the case it seems like the answers that have been discussing this recently should be more upfront about that. – Paul Plummer Jan 9 '14 at 21:13
  • Asymptotic series and Carlson's theorem could also be potentially related. – lcv Oct 5 '14 at 1:14
up vote 7 down vote accepted

Certainly $1+2+3+\ldots$ is a divergent series. However, there are (many!) methods for attaching a value to (some) divergent series, which deserve the name "summation" in that they assign the usual value (the limit of the partial sums) to convergent series. Many of these methods additionally have nice properties like linearity, finite reindexability, etc.

Formally, one can define a regular, linear series-summation method $\Lambda$ to be a partial function from $\mathbb{R}^{\mathbb{N}}$ to ${\mathbb{R}}$ such that $$ \Lambda(a_1,a_2,\ldots)=\sum_{i=1}^{\infty}a_i $$ whenever the latter sum converges, and such that $\Lambda(\mu a + b)=\mu \Lambda(a) + \Lambda(b)$ for any $\mu\in\mathbb{R}$ and whenever $\Lambda(a)$ and $\Lambda(b)$ are defined. A stronger constraint is stability, such that $$ \Lambda(a_1,a_2,\ldots)=\Lambda(a_{n+1},a_{n+2},\ldots) + \sum_{i=1}^{n}a_i $$ for any $n\in\mathbb{N}$. A number of different methods are described in detail in the Wikipedia entry on divergent series. One method that gives the result $-1/12$ for the series $1+2+3+\ldots$ is $\zeta$-function regularization.

The equality is a mere definition. For example, consider a power series that converges to a function $S(x)$ for $|x|<1$ (and diverges for $|x|\geq1$). If there exists $\lim_{x\to1^-}(aS(x^2)-S(x))$ for a certain $a\neq1$, the sum of the divergent series at $x=1$ can simply be defined by $$\frac{\lim_{x\to1^-}(aS(x^2)-S(x))}{a-1}.$$

Example 1.

$$x+x^2+x^3+\cdots=\frac{x}{1-x}$$ and $$2\frac{x^2}{1-x^2}-\frac{x}{1-x}=-\frac{x}{1+x}.$$ By definition, $1+1+1+\cdots=-\frac{1}{2}$.

Example 2.

$$x+2x^2+3x^3+\cdots=\frac{x}{(1-x)^2}$$ and $$4\frac{x^2}{(1-x^2)^2}-\frac{x}{(1-x)^2}=-\frac{x}{(1+x)^2},$$ By definition, $1+2+3+\cdots=-\frac{1}{12}$.

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