Is $1+2+3+4+\cdots=-\frac{1}{12}$ the unique ''value'' of this series?

Solution 1:

Our problem is: Given a sequence $(a_n)_{n=0}^\infty$ of numbers, assign a "sum" to it, i..e, give a meaning to the pixel pattern $\sum_{n=0}^\infty a_n$. This is easy the sequence of partial sums converges as we can simply assign the limit of these. This does not define $\sum_n a_n$ for all given sequences $(a_n)_n$, but only to a certain subset $S$ of the set of all sequences. But we have a collection of nice rules:

• $S$ is a vector space and $\sum$ is linear, that is
  • If $(a_n)_n,(b_n)_n\in S$, then $(a_n+b_n)_n\in S$ and $\sum(a_n+b_n)=\sum a_n+\sum b_n$
  • If $(a_n)\in S$ and $c\in\Bbb C$, then $(ca_n)_n\in S$ and $\sum ca_n=c\sum a_n$.
• $S$ is closed under adding/dropping/modifiying finitely many terms, which boils down to
  • $(a_n)_n\in S$ if and only if $(a_{n+1})_n\in S$. In this case $\sum a_{n}=a_0+\sum a_{n+1}$

Regularization is the attempt to enlarge $S$ in a useful way. There are many ways to do so, and in particular if one ignores the above nice rules, one can do it almost arbitrarily (using the Axiom of Choice, perhaps). On the other hand, if one wants to have permanence of the above rules (or perhaps others), then quite often the extension is uniquely determined.

For example, if we want to enlarge $S$ in such a way that it contains $(2^n)_n=(1,2,4,8,\ldots)$ and we assign (by any computational method at all) a value $\sum 2^n=c$, then $S$ must also contain $(2\cdot 2^n)_n=(2,4,8,16,\ldots)$ and $\sum 2^{n+1}=2c$, and $S$ must contiain this prepended with $1$ - which is again the original sequence - and so we obtain the equality $2c+1=c$. Therefore, if we want to assign a value to $\sum 2^n$ and want permanence of the above rule, we must agree to set $\sum 2^n=-1$. Similarly, it follows that $(1)_n$ cannot be $\in S$ because $c+1=c$ has no solution.

But can we extend $S$ to contain $(n)_n$? If so, then also $(n+1)_n\in S$ and their difference $(1)_n\in S$ - which we have just seen is impossible. Thus any attempt to consistently assign a value to $1+2+3+4+\ldots$ must drop one of the very reasonable rules listed above. But with these rules as goalposts removed, one would first have to agree what constitutes a "valid" extension of summation before one can make statements about whether a certain value (if it is assigned at all) is necessarily correct (as we did above for $\sum 2^n$).

Solution 2:

I'll go a bit further than Hagen Von Eitzen's (very good) answer.

I had considered the general case as he does, to see whether $1+2+...= -1/12$ had any "deep" content.

So I'll give here the results I found (which do not, if I remember correctly, require the axiom of choice), and if you are interested I can provide proofs.

Call admissible space a sub-vector space of $K^{\Bbb{N}}$ for $K=\Bbb{R}, \Bbb{C}$ that contains all summable sequences and is closed under $(a_n)_n \to (0,a_0,a_1,....)$.

If $H$ is an admissible space, and if $T$ is a linear form on $H$, say that $T$ is a supersummation if and only if :

1. it extends the sum for summable sequences, and
2. $T((0,a_0,a_1,\ldots)) = T((a_n)_n)$, ie if appending $0$ to the front of a sequence does not change the ‘sum’.

Say that a supersummation $T$ on the admissible space $H$ is proper if and only if $T$ is the unique supersummation on $H$.

Now one has the following results :

• Assume $H$ is an admissible space, with a proper supersummation $T$ and $u$ is a given fixed sequence. Then $H$ is contained in an admissible space containing $u$ with a proper supersummation on it, if and only if $u$ satisfies a linear induction relation, up to an $H$-term. The second condition reads, more precisely: there exist $k\in\Bbb{N}$, $a_0,\ldots,a_{k-1}$ and $h\in H$ such that $u_{n+k} = \displaystyle\sum_{i=0}^{k-1} a_i u_{n+i} + h_n$ for all $n\in\Bbb{N}$.

• There exists a (necessarily unique) admissible space $H$ satisfying the following properties: 1. $H$ has a proper supersummation; 2. no admissible space properly containing $H$ has a proper supersummation; moreover 3. every admissible space with a proper supersummation, $(H’,T’)$, is contained in $(H,T)$, ie, $H’\subseteq H$ und $T’=T\mid H'$.

This second result is very interesting as it can be read ‘If you want to define a generalization of infinite sums, there is only one way to do it right, and the way of doing it is, in a sense, absolute’. And, as Hagen Von Eitzen shows, $H$ cannot contain $(n)_n$ but it does contain $(2^n)_n$, and the value of the supersummation of $(2^n)$ is $-1$.

EDIT: I'm adding to this answer a short "paper" that I had written about this topic. A few remarks : 1. This is in french (because I'm french), so until I trnalsate it (if I translate it), only those who read french can understand it. 2. I wrote it at the beginning of my second year post-highschool, so there are some things that I could have done differently and really more easily -don't be surprised if there are calculations that could be avoided, or trivial arguments that are developed etc. 3. At some point I use Zorn's lemma, but as those who read it can see, it's not necessary. I simply use it to get the existence of a maximal element, but I prove its existence in another way later: I'm only using it to show what the maximal element might look like, so that I can better pursue the search for said element. With this in mind, here's the "paper"