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First thing I want to mention is that this is not a topic about why $1+2+3+... = -1/12$ but rather the connection between this summation and $\zeta$.

I perfectly understand that the definition using the summation $\sum_{k=1}^\infty k^{-s}$ of the zeta function is only valid for $Re(s) > 1$ and that the function is then extrapolated through analytic continuation in the whole complex plan.

However some details bother me : Why can we manipulate the sum and still obtain correct final answer. $$ S_1 = 1-1+1-1+1-1+... = 1-(1-1+1-1+1-...)= 1-S_1 \implies S_1 = \frac{1}{2} \\ S_2 = 1-2+3-4+5-... \implies S_2 - S_1 = 0-1+2-3+4-5... = -S_2 \implies S_2 = \frac{1}{4} \\ S = 1+2+3+4+5+... \implies S-S_2 = 4(1+2+3+4+...) = 4S \implies S = -\frac{1}{12} \\ S "=" \zeta(-1) $$ Clearly these manipulations are not legal since we're dealing with infinite non-converging sums. But it works ! Why ? Is there a real connection between the analytic continuation which yields the "true" value $\zeta(-1) = -1/12$ and these "forbidden manipulations" ? Could we somehow consider these manipulations as "continuation of non-converging sums" ? If so, is there a well-defined framework with defined rules because it is clear that we must be careful when playing with non-converging sums if we don't want to break the mathematics ! (For example Riemann rearrangement theorem)

And since it seems that these illegal operations can be used to compute some value of zeta in the extended domain $Re(s) < 1$, are there other examples of such derivations, for example $0 = \zeta(-2) "=" 1^2 + 2^2 + 3^2 + 4^2 + ...$ ?

Hopefully this is not an umpteenth vague question about zeta and $1+2+3+4...$ I did some research about it but couldn't find any satisfying answer. Thanks !

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  • $\begingroup$ Unfortunately, this is the umpteenth time I've seen this question alone. Regardless, you might get an answer, but I would recommend deeper research. $\endgroup$ – Brevan Ellefsen Dec 15 '16 at 17:29
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    $\begingroup$ Moreover, you do ask for justifications of the operations themselves, which is a good thing. Would answer myself it not for lack of time. Best of luck getting an answer! $\endgroup$ – Brevan Ellefsen Dec 15 '16 at 17:30
  • $\begingroup$ @BrevanEllrfsen I look forward to some good answers for this interesting question. $\endgroup$ – Simply Beautiful Art Dec 15 '16 at 17:49
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    $\begingroup$ This may or may not answer your question, but I think a lot of insight and intuition about the nature of this function and analytic continuation in general can be found here: youtube.com/watch?v=sD0NjbwqlYw $\endgroup$ – AlexanderJ93 Dec 15 '16 at 17:49
  • $\begingroup$ @AlexanderJ93 haven't watched that video yet, watching it later. $\endgroup$ – Simply Beautiful Art Dec 15 '16 at 17:53
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They "work" because the manipulations you present work where the original definition was valid. Notice that:

$$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}$$

This function can be made to converge for all $s:$

$$\eta(s)=\lim_{r\to1^{-1}}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}r^n$$

Whereupon we find that for $s=0$,

$$\begin{align}S_{1r}&=r-r^2+r^3-\dots\\&=r-r(r-r-r^2+r^3-\dots)\\&=r-rS_{1r}\\\implies S_{1r}&=\frac r{1+r}\end{align}$$

Then take $r\to1$ to get $S_1=1/2$. Notice the similarities and differences between this and your method.

In the same manner,

$$\begin{align}S_{2r}&=r-2r^2+3r^3-\dots\\&=(r-r^2+r^3-\dots)-r(r-2r^2+3r^3-\dots)\\&=S_{1r}-rS_{2r}\\\implies S_{2r}&=\frac r{(1+r)^2}\end{align}$$

Letting $r\to1$, we get $S_2=1/4$.

Notice that in each of the steps above, if we replace $r$ with $1$, we get the methods you present, despite that they don't really make sense in that way.


Also notice that, by manipulation of the original definitions,

$$\begin{align}\zeta(s)-\eta(s)&=2\left(\frac1{2^s}+\frac1{4^s}+\frac1{6^s}+\dots\right)\\&=2^{1-s}\left(\frac1{1^s}+\frac1{2^s}+\frac1{3^s}+\dots\right)\\&=2^{1-s}\zeta(s)\end{align}$$

$$\zeta(s)=\frac1{1-2^{1-s}}\eta(s)=\frac1{1-2^{1-s}}\lim_{r\to1^{-1}}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}r^n$$

In this way, everything you have presented makes sense in the context of $\Re(s)>1$, so it holds for all $s$ by analytic continuation.

Notice things like the Riemann rearrangement theorem does not come in to play due to absolute convergence for $\Re(s)>1$, where we derive these formulas. Also, $S_r$ converges absolutely for any $|r|<1$. Also notice that what determines where a certain set of parentheses are allowed comes from these convergent scenarios.

From all of the above, I note that instead of using algebra, derivatives may be used to give

$$\zeta(-s)=\frac1{1-2^{1+s}}\lim_{r\to1}\left[\underbrace{r\frac r{dr}r\frac r{dr}\dots r\frac r{dr}}_s\frac{-1}{1+r}\right]$$

For whole numbers $s>0$.


The rules that allow this to work is to work only with convergent sums that are analytic continuations of the original sums. Then, everything becomes normal and there are not so many worries.

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  • $\begingroup$ Thank you for your answer ! $\endgroup$ – Zubzub Dec 16 '16 at 16:20
  • $\begingroup$ =D you are welcomed! $\endgroup$ – Simply Beautiful Art Dec 16 '16 at 17:07
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The question I would have is "why should these rearrangements you've made work, but not all possible ones?" In short, what makes that particular parentheses combination so special? If I throw a ball into a tiny cup at random, surely every once in a while I'll get lucky and it will go in, but that doesn't mean I should but stock in the fact the idea that any given guess will work. In particular, it's not useful building forward: how can I use this technique on general things if I don't know it will always work? What use is it to building newer mathematics if we rely on attractive psychological eye-candy that happens to work in some special cases?

There are some handy notions of convergence which are in some ways much better than traditional convergence, eg Cesaro summation or abelian summation. If you work in a topology where convergence means convergence in such a mode, that's fine--you need to specify it, but that's just bookkeeping--the danger is in saying that one sequence which diverges in one topology--in this case the typical metric topology on $\Bbb C$--which has a lot of very hard-worked theorems on what algebraic manipulations are compatible with the analytic structure of convergence is somehow not doing what it's doing because a separate topology demonstrates convergence.

So I suppose the easiest way to say it is: we as humans are used to the traditional topology to the point we are almost willing to say things like those sums actually converge to those values they do not converge to. The real solution here is to embrace the fact that there are alternative topologies which can give us insights on possible values (in these cases correct values!) because they happen to coincide in all the cases where the regular notion of convergence holds. In short, to not be so caught up in making everything we want be "correct" in the usual topology.

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