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What else can the definition of the ellipsis symbol, "$\dots$", mean in this context?

$$S = x_1 + x_2 + x_3 + x_4 + \dots$$

All I can see is that you have an infinite sum of $x$s, where the first one is $x_1$, the next is $x_2$, then $x_3$, and so on forever, for as many natural numbers as exist and in order. But, for some reason, I am being told that such a definition is ambiguous and meaningless compared to formal mathematics.

Blue
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Ivan Hieno
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  • What if the next term is $x_5$, for example? I think that's what the person who told you that means. It's not as precise as writing $S=\sum_{n=1}^\infty x_n$ – Jakobian Oct 04 '18 at 18:13
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    The expression $x_1+x_2+x_3+x_4+\dots$ generally is taken to mean $\lim\limits_{n\to\infty}\sum\limits_{i=1}^n x_i$. On the other hand, $x_1+x_2+x_3+\dots+x_n$ is generally taken to mean $\sum\limits_{i=1}^n x_i$. – JMoravitz Oct 04 '18 at 18:14
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    @ Jakobian, sure the next term could ba anything, even $p_1g$, but a person would really have to desire to misunderstand to make such an objection. – Ivan Hieno Oct 04 '18 at 18:21
  • The only scenario that usually comes up that I would say is ambiguous (only in that people don't follow standard convention) is that $\lim\limits_{n\to\infty}\sum\limits_{i=1}^n i = \infty$ while some people will write something like "$1+2+3+\dots = -\frac{1}{12}$" such as here. It is important to note that people who write that are going against standard convention. Most mathematicians will agree that the only valid answer to evaluating $1+2+3+\dots$ is $\infty$. – JMoravitz Oct 04 '18 at 18:22
  • What they intend with the $-\frac{1}{12}$ interpretation is that $1+2+3+\dots$ is in effect shorthand for the analytic continuation of the zeta function. – JMoravitz Oct 04 '18 at 18:23
  • Usually it implies an infinite number of terms follow, but one needs additional machinery (such as a norm) to ascribe a meaning to the term. – copper.hat Oct 04 '18 at 18:23
  • Arrrrgh, why does this $-{1 \over 12}$ keep popping up? – copper.hat Oct 04 '18 at 18:26
  • I don't see why the switch of notations in comprehension or extension would interfere with the convergence issue. These are two independent topics. –  Oct 04 '18 at 18:45
  • " for as many natural numbers as exist and in order" Where is that stated in the "definition"? $S= x_1 + x_2 + x_3 + x_4+....$ Nowhere in that definition is is stated the that $1,2,3,4$ are the first four natural numbers and that our intent is to represent every natural number in order. Sure, it's IMPLIED and you'd have to be pretty brain dead to miss, it be never is is statement. We can't have definitions where the reader has to second guess the author's intent. – fleablood Oct 04 '18 at 18:51
  • With subscripted elements, as in the example you cite, I believe the expression is perfectly meaningful (and less intimidating to casual readers than sigma notation). If a pattern in the summands were unclear, then ellipses would, of course, be problematic. (Just look around here and see how many "What's the next term in the sequence?" questions get the "Could be anything!" response.) A way to alleviate confusion is to provide a formula for the $k$-th term: $$S=\frac{1}{4}-\frac{2}{9}+\frac{3}{16}-\cdots+(-1)^{k+1}\frac{k}{(k+1)^2}+\cdots$$ (Not a great example, but I trust you get my point.) – Blue Oct 04 '18 at 19:07
  • @Blue: for clarity one can also write $$S=\frac1{2^2}-\frac2{3^2}+\frac3{4^2}-\cdots$$ –  Oct 04 '18 at 19:12
  • @YvesDaoust: That's actually why my example wasn't so great! :) (I ran out of comment space, so couldn't explain myself or show the alternative form.) One can even be more explicit, with $(1+1)^2$, $(2+1)^2$, $(3+1)^2$ in the denominators. At some point, that kind of thing gets cumbersome, so there's a balance to be struck by the author. In any case, it should be clear that ellipsis notation can be just as unambiguous as sigma notation. – Blue Oct 04 '18 at 19:31
  • $S=-1.$ $-S=1.$ $(1-2)S=1.$ $S-2S=1.$ $S=1+2S.$ $S=1+2(1+2S).$ $S=1+2+4+8+\cdots+2^n+2^{n+1}S.$ $(-1)=1+2+4+8+\cdots+2^n+2^{n+1}(-1).$

    There is nothing fishy about this. It's not even infinite. It's recursion. $S=-1$ can be expanded like $1+2+4+8+\cdots$ as many times as you like. $S=-1$ doesn't mean what people think it means.

     – Rob Sep 18 '23 at 20:20

3 Answers3

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Stating ambiguity of this summation is bad faith. The pattern is obvious and the ellipsis clearly indicates an unlimited sequence of terms.

I would be more critical towards a sum like

$$1+2+4+\cdots\ ?$$

In common practice, if the first few terms/indexes (as little as $3$) follow an arithmetic progression, it can be considered implied.

  • actually, it is ambiguous, because the tail is chopped off. $S=1+2S.$ then $S=1+2+4+8+\cdots+2^n+2^{n+1}S$. Note that $S$ can be solved for -1. But $S = Sum(S) + Tail(S)$ cant be ignored. $Sum(S) = S - Tail(S)$. The fact that $S=-1$ is why you can solve for $Sum(S) = 2^{n+1}-1$. – Rob Sep 18 '23 at 20:34
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To give you an idea of what can go wrong with arbitrary entries in the summation the following example shows these complications. Begin with $S=1 + 2 + 4 + 8 + ...$ and so $S=1+2(1+2+4+8+...)$ and so $S=1+2S$ giving us finally that $S=-1$ and we now have a sum of positive numbers equal to a negative number, which is absurd. These complications must be addressed in order to deal with sums of this kind meaningfully.

CyclotomicField
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    This "problem" has nothing to do with the ellipsis notation. –  Oct 01 '19 at 21:04
  • Actually, it tells you that $S$ is not the sum. In a recursive definition, $S=1+2S$, it means $S = Sum(S) + Tail(S)$, so you have to solve for the sum $Sum(S) = S - Tail(S)$, which is what everybody misunderstands. $(-1)=1+2(-1)=1+2+4(-1)$. $S$ isn't just a name. It's what lets you solve for the full sum. $(-1)=1+2+4+8+\cdots+128+256(-1)$ ... it's not that $n$ is infinite, but that it's true no matter how many times you expand it. – Rob Sep 18 '23 at 20:42
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Be careful with recursively defined sums. $S=1+2S$ doesn't solve for -1 by accident. ie:

$S=1+2S. (1-2)S=1. S=-1.$

The thing is that $S$ is not the sum! $S = Sum(S) + Tail(S)$ in a recursive sequence. So, $S - Tail(S) = Sum(S)$. That's where the negative values come from. $S = 1 + 2 + 4 + 8 + \cdots + 2^n + 2^{n+1}S$. It's important to not stop at "$\cdots$" and throw the tail away.

$(-1) = 1 + 2 + 4 + 8 + 16(-1).$

Notice this:

$S - 2^{n+1}S = 1 + 2 + 4 + \cdots + 2^n.$

There's no paradox at all. The $S=-1$ is why you can now solve for $Sum(S) = 2^{n+1}-1$, and this is a general phenomenon. S is not the sum! $Sum(S)$ is actually $(S - Tail(S))$.

Rob
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