1

I stumbled upon "the God proof" which goes:

$0 = 0 + 0 + 0...$

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

$= 1 - 1 + 1 - 1 + 1 - 1 + ...$

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

$= 1$

Even though this result is obviously wrong, I can't quite pinpoint exactly what the 'illegal' operation here is? I know that it's possible to make a conditionally convergent infinite series converge to any value, so that can't be the issue (?).

It makes me think there's something dangerous about representing numbers by infinite series that aren't absolutely convergent, but I haven't been able to find something that specifically addresses this, if someone could tell me where to look for more information, I'd be grateful.

[background: undergrad student]

Jan
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user261581
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3 Answers3

4

Infinite sums are not "associative" like that. You can only rearrange the order of summation if the sum converges "absolutely".

Gregory Grant
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  • I saw this comment on quora: "Associativity holds if the sum is convergent. Commutativity holds if the sum is absolutely convergent." I guess the first statement is wrong then? Is the second statement right? – user261581 Aug 12 '15 at 15:00
  • @user261581 No that's true, if it converges than you can associate terms and add the associated ones first, but you cannot rearrange the order. But the above proof that $0=1$ does not start with a convergent series, so you cannot do either. – Gregory Grant Aug 12 '15 at 15:02
  • Thanks. When you say start, do you mean the line "0+0+0+.." or "(1-1)+(1-1)+..."? I was under the impression (from another answer) that with the parentheses intact, the series is convergent, and that the flaw is removing the parentheses? (and adding new ones) – user261581 Aug 12 '15 at 15:07
  • @user261581 The series has to be convergent with no parentheses, and then you can put in parentheses wherever you want. But you can't just require it converges with some parentheses, that is not a strong enough criteria. – Gregory Grant Aug 12 '15 at 15:22
2

The equality $$(1-1) + (1-1) + (1-1)...=1-1+1-1+1-1+...$$ is wrong, as the first series converges, while the second doesn't.

N. S.
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  • Actually, written as $\sum_{n=1}^\infty 1-1$, it converges. The issue is when we split it into $\sum_{n=0}^\infty (-1)^n$. – Ian Aug 12 '15 at 14:28
  • Why doesn't $\sum_{n=1}^\infty (1-1)$ converge? I certainly understand why the next line $\sum_{n=1}^\infty (-1)^{n+1}$ doesn't converge, but I'm uncertain about why you pick on this first equality. – Simon S Aug 12 '15 at 14:28
  • @Ian you are right, fixed it. – N. S. Aug 12 '15 at 14:30
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    @SimonS you are right, fixed it. – N. S. Aug 12 '15 at 14:31
  • Is this because the partial sums no longer converge when the parentheses are removed? – user261581 Aug 12 '15 at 14:51
  • @user261581 Yes. Actually the entire argument reduces to the following: when the aranthesys are removed, then $T_{2n}$ converges to $0$ while $T_{2n+1}$ converges to $1$. The "proof" concludes that this means that $0=1$. – N. S. Aug 12 '15 at 15:05
0

See Grandi's series. Answer comes out to be $0.5$

Clement C.
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joe1984
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  • I didn't downvote it but I'm sure whoever did, did it because your link is broken. Instead of just pasting the link, try this: put a description between square brackets, like [Grandi's series], then immediately following the closing bracket, put the link surrounded by parentheses. I tried to edit your answer but that fix is less than 6 characters long and any edit by someone other than the author must have more than that in changes. It should look like Grandi's series after you've fixed it. – wltrup Aug 12 '15 at 14:29
  • @wltrup More likely, whoever did, did it because it's not true. Not "really", anyway. – Vincenzo Oliva Aug 12 '15 at 14:59
  • @VincenzoOliva Well, that too, I suppose. TBH, I clicked on the link and it didn't work, so I assumed that that was the reason for the down-votes. Didn't get to read the link. Still haven't. Not enough hands, fingers, eyes, and brain to do all the things I'm trying to do at once today! lol – wltrup Aug 12 '15 at 15:02
  • I see. That seems awfully pedantic, as a simple copy/paste of the link would have led to the correct information. :/ My response has been the only one so far to address this topic with the most relevant information, and it gets voted down. This is quite discouraging, because I like mathematics, and I don't feel much like contributing if the good information is going to get lost because of pedantry. – joe1984 Aug 12 '15 at 15:02
  • @joe1984 Who are you addressing, me or Vincenzo? The link copy and paste didn't work because of the underscore. It breaks the link parser. Yes, not your fault, I know. – wltrup Aug 12 '15 at 15:04
  • @wltrup I could see why clicking on the link would break it, because the parser won't include anything in the html link at the underscore. I get that. I'm talking about highlighting the link text with your cursor to copy and pasting it into the location bar in the browser. That definitely should have worked. But oh well... – joe1984 Aug 12 '15 at 15:07
  • @wltrup That's our curse! :D – Vincenzo Oliva Aug 12 '15 at 15:08
  • @joe1984 I sincerely hope that this experience didn't turn you away from math.se. This is a great place to learn and share. Yes, occasionally there are mildly upsetting things that happen, but for the most part it's a really cool place. I hope you'll decide to stay and contribute. – wltrup Aug 12 '15 at 15:12
  • @joe1984 We have all the math cookies, don't leave our side. – Vincenzo Oliva Aug 12 '15 at 15:14
  • We'll see what happens. I'll definitely not be running into the erroneous link trap again. Thanks – joe1984 Aug 12 '15 at 15:25