What $+1+1+\cdots$ really equals

Question

$1+1+1+\cdots$ is clearly a divergent series, so you'd say that it tends towards infinity? Through analytic continuation of the zetafunction the value $-1/2$ could be assigned the sum, right? But if we consider the sum of all integers: $\cdots-2+-1+0+1+2+\cdots$ and group them up pairwise as follows: $0+(-1+1)+(-2+2)+(-3+3)+\cdots=0+0+0+\cdots=0$. You could also group them up like this: $(0+1)+(-1+2)+(-2+3)+\cdots=1+1+1+1+\cdots$ which is the series we started with, so you could argue that $1+1+1+\cdots=0$ as well? So what's really going on? Could the sum be given any value what so ever? And if so; Could every divergent series be given any value through steps like those I used? And in that case; how would things like the analytic continuation of the zeta function be useful at all?

Answer 1

The analytic continuation assigns a formal value to that sum. You cannot do summation rearrangements with these kinds of sums because they don't converge absolutely (they don't even converge) and expect to get the same answer. Hence why $$\sum_{n=0}^{\infty} (-1)^n=(1-1)+(1-1)+(1-\cdots)=0=1+(-1+1)+(-1+1)+\cdots=1$$

Edit: A sum can be given "any value" I suppose. You could declare that $\sum (-1)^n=15$ and it would make sense as long as it is understood that the sign "=" is just a convention (say your definition), just as when we say $\sum \frac{1}{2^n}=1$, we really mean that the partial sums converge to this value. The point of things like Cesaro summation and other techniques is that they assign the proper value to convergent sums (namely the value they converge to) in addition to assigning values to various non-convergent sums.

Edit 2: Why are these things useful at all, such as analytic continuation of the zeta function to conclude $\sum n=\frac{-1}{12}$? There are a variety of reasons. In the case I mentioned, the analytic continuation gives you an interpretation of this sum because of the way the zeta function is defined. It is a reasonable definition based on this fact (although the zeta function is not literally defined to give you this value). People sometimes want to do things with nonconvergent series. The series I mentioned is physically relevant, and so it makes sense to assign a finite value to it.

Answer 2

For the most part, rearranging terms allows any answer to occur, which is why we usually try to avoid rearranging of terms.

So I suppose you could get 'any' answer from this, but in essence, what happens is that you are no longer searching for the answer of the original summation, but rather you are searching for a summation that 'looks' similar to the original in order to obtain a different value then expected.