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I'm not sure whether $\sum\limits_{x = -\infty}^{\infty}x$ is equal to $0$ or undefined. For example,

$$\sum_{x = -\infty}^{\infty}x = \displaystyle \sum_{x = -\infty}^{-1}x + \displaystyle \sum_{x = 1}^{\infty}x = -\infty + \infty$$

So with that approach it is undefined. However, clearly all the negative elements in the summation cancel with the positive elements, so that makes it seem like it should be zero. So which is it?

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    $\begingroup$ Using $$...$$ already puts you in displaystyle hence using \displaystyle inside $$...$$ is useless. $\endgroup$ – Did Feb 12 '17 at 10:57
  • $\begingroup$ This thread may help too. $\endgroup$ – Raymond Manzoni Feb 12 '17 at 12:05
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You can see that the sum does not converge because the limit $$\lim_{(m, n) \to (-\infty, \infty)} \sum_{x=m} ^{n} x\tag{1}$$ does not exist. On the other hand note that the limit $$\lim_{n\to\infty} \sum_{x=-n} ^{n} x\tag{2}$$ exists and is equal to $0$. The sum in your question is defined via $(1)$ and not via $(2)$.

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  • $\begingroup$ Why do you say the sum in my case is defined via (1) and not (2)? It seems like there is a choice we can make here of which to use? This seems like a very strange situation because we can clearly identify the positive and negative elements so it seems they should always cancel, as they do in your case (2). $\endgroup$ – csss Feb 13 '17 at 20:42
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    $\begingroup$ Note that $\infty$ and $-\infty$ are two different limiting cases and the better approach is to handle them separately / independently like equation $(1)$. Defining in this manner allows the usual properties of limits to hold. $\endgroup$ – Paramanand Singh Feb 14 '17 at 5:01
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The rearanging of the coefficitiens is not allowed. To ensure the rearrangement you need to know about the absolute convegence of the series which obviously does not hold.

A typical example is the nonconvergent series $$\sum_{n\in \mathbf N} (-1)^n.$$

The Riemann series theorem might be interesting for you in this context.

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  • $\begingroup$ So can an infinite double summation ever converge? Can you give me an example of an infinite double summation that does converge so I can compare and contrast it with the one in my post? $\endgroup$ – csss Feb 13 '17 at 20:45
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The answer really depends entirely on what you mean by "the sum $\sum_{n=-\infty}^{n=\infty}x$".

The standard definition for the sum of an infinite series given in most books only holds when the sum is done from $n=0$ to $n=\infty$. (A "one-tailed" series). However, your series is two-tailed and the definition doesn't apply.

So you need to extend your definition first, and that alone will dictate what your answer is.

That being said, a common situation where two-tailed series comes up is in Laurant series (compare with Taylor series). In this case, a Laurant series will have an annulus of convergence (compare to the radius of convergence). Inside the annulus of convergence, the natural way to define the sum is the naive way: add up partial sums starting at $0$ and going out, interleaving positive and negative values of $n$. The sum will converge to a finite value and rearrangement is immaterial.

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    $\begingroup$ Hmmm... there is a canonical definition, even in this case. $\endgroup$ – Did Feb 12 '17 at 10:35
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    $\begingroup$ Of course, you can take any definition as "canonical". And there are certainly sensible defaults if you are confronted with such a series. But my point stands. $\endgroup$ – Tac-Tics Feb 12 '17 at 10:38
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    $\begingroup$ Yes. That is what I am saying. In the more familiar case of exponents, many people seem to think the definition of x^y is somehow God-given, and it causes much confusion when you investigate quantities like (-1)^(1/2) or later, i^i when learning complex numbers. But definitions are not God-given. They are driven by heuristics. And simply put, most basic calculus books don't cover two-tailed series. And I would argue that they are much more niche, save in the absolute convergent case. So I don't see why you take such issue with my reply, even if it differs in style from how you would reply. $\endgroup$ – Tac-Tics Feb 12 '17 at 17:51
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    $\begingroup$ "why you take such issue with my reply, even if it differs in style from how you would reply" Not a question of style. Simply I am mentioning to the interested reader (and to the OP if they are not stuck in "defending" their answer) that the assertion that the convergence of "two-tailed" series has no canonical definition, is wrong. By comparison, $i^i$ has no canonical definition as a complex number so your comparison is moot (even if, yes, some people are trying to save the notion pretending they mean it as a set, but they soon treat it as a number, predictably running into problems) $\endgroup$ – Did Feb 12 '17 at 18:05
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    $\begingroup$ You are arguing over what definitions are 'canonical'. I am not wrong here. I am being conservative. On the other hand, I could take an exaggeration of your stance and stubbornly claim $i^i$ does have a canonical value. (And of course, it should coincide with Mathematica). $\endgroup$ – Tac-Tics Feb 12 '17 at 18:14

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