# 1+2+3+4+... = -1/12. In what sense? **URL:** \ **Category:** Factual Questions\ **Created:** [May 9, 2007, 12:12am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329 "2007-05-09T00:12:46Z")\ **Posts on this page:** 20\ **Page:** 6
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 22, 2010, 4:54am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/101 "2010-07-22T04:54:09Z")
> [@tim314](#): > > Anyhow, if we say 1 + 2 + 3 + … = infinity, essentially what we[sup]\*[/sup] _mean_ is that if you pick some really big number, and then you add up the numbers in that series, eventually the sum of the series will surpass that really big number, and in fact this is true for _any_ number you could have picked, no matter how big. That’s what “equals infinity” really means. Whereas if we say 1 + 2 + (-3) = 0 what we really mean is “you can add up all those numbers and you get zero”. Which is quite a different thing. Ah, but what if we say -1 + 1/2 + 1/4 + 1/8 + 1/16 + … = 0? This is rather more like 1 + 2 + 3 + … = infinity than 1 + 2 + (-3) = 0. I wouldn’t say this illustrates a difference in “numberyness” between infinity and 0, even if one feels there is such a difference; it just illustrates a difference in the meaning of finitary and infinitary addition. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 22, 2010, 7:29am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/102 "2010-07-22T07:29:35Z")
> [@septimus](#): > > I think this is a key point. Similarly, the “value” of 0^0 depends on whether you’re seeking the limit of 0^x, x^0 or x^x. > > I will now demonstrate that the positive integers sum instead to -1/10 ! This is probably just a silly mathematical fallacy, like 10 = 12, but it still might be interesting to hear mathematicians’ comments. I like your example! I’ll note for now that a simpler example along the same lines is as follows: A = 1 + 2 + 3 + 4 + … 3A = 3 + 6 + 9 + 12 + … 6A = (2 + 4) + (5 + 7) + (8 + 10) + (11 + 13) + … So we might conclude that A = 1 + 6A + 3A and thus A = -1/8. As noted above, we actually can, through these kinds of manipulations, get A to equal anything we like, though it is the specific manipulation which was shown to produce -1/12 which is relevant to the zeta function, since that is the only one which works more abstractly on 1[sup]s[/sup] + 2[sup]s[/sup] + 3[sup]s[/sup] + … (I’ll give a more detailed response to this tomorrow, maybe.) ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [July 22, 2010, 3:45pm UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/103 "2010-07-22T15:45:50Z")
I’ll try to clear up some of the confusion I caused. Much less interesting and important than the analytic continuation of zeta, but still possibly fun! > [@Indistinguishable](#): > > I like your example! I’ll note for now that a simpler example along the same lines is as follows: > > A = 1 + 2 + 3 + 4 + … > 3A = 3 + 6 + 9 + 12 + … > 6A = (2 + 4) + (5 + 7) + (8 + 10) + (11 + 13) + … > > So we might conclude that A = 1 + 6A + 3A and thus A = -1/8. Your example is based, in a sense, on 1/3 + 1/3 + 1/3 = 1, with the denominaters summing to 9, and the resulting value “-1/(9-1).” Similarly my examples were based on 1/2 + 1/3 + 1/6 with the result “-1/(11-1)”, and 1/3 + 1/4 + 1/6 + 1/6 + 1/12 with the result “-3/(31-1).” From 1/5+1/5+1/5+1/5+1/5 we get -3/(25-1) = -1/8 From 1/7+1/7+1/7+1/7+1/7+1/7+1/7 we get -6/(49-1) = -1/8 From 1/2+1/8+1/8+1/8+1/8= 1 we can get -6/(34-1) = -2/11 as seen next: 2A = 2 + 4 + 6 + … 8A+8A = (3+13) + (11+21) + (19+29) + … 8A+8A = (7+9) + (15+17) + (23+25) + … Sum to get 34A + 6 = A or A = -2/11 So, -1/8, -1/10, and -2/11 are among the solutions derivable in this way. Complete list of such “solutions” is left as an exercise. 😃 ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 22, 2010, 5:17pm UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/104 "2010-07-22T17:17:19Z")
Well, the complete list is everything; A = kA + (1-k)A. Plug in one derived value for the A on the left and another different derived value for the A on the right, then vary k till the result is whatever you want. 🙂 ---
**Author:** ![Darth\_Panda](https://avatars.discourse-cdn.com/v4/letter/d/ee7513/32.png) [@Darth\_Panda](https://boards.straightdope.com/u/Darth_Panda)\ **Post date:** [July 22, 2010, 7:23pm UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/105 "2010-07-22T19:23:07Z")
I think the real question is this: If I owed you -1/12 dollars, would you accept my paying you back by having you give me 1 dollar, then two dollars, then three dollars, then four dollars, then five dollars, etc. If so, I owe you -1/12 dollars. Now, put your money where your math is. 🙂 ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 22, 2010, 7:32pm UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/106 "2010-07-22T19:32:06Z")
If you owe me -1/12 dollars, then _you_ should be giving _me_ the 1 + 2 + 3 + 4 + … dollars. 🙂 ---
**Author:** ![Colophon](https://avatars.discourse-cdn.com/v4/letter/c/f05b48/32.png) [@Colophon](https://boards.straightdope.com/u/Colophon)\ **Post date:** [July 22, 2010, 7:50pm UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/107 "2010-07-22T19:50:07Z")
> [@Indistinguishable](#): > > If you owe me -1/12 dollars, then _you_ should be giving _me_ the 1 + 2 + 3 + 4 + … dollars. 🙂 Yeah. But owing -1/12 dollars is the same as being owed 1/12 dollars. Could you lend me 8 cents? (I’ll let you off the third of a cent) 😃 ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 22, 2010, 8:22pm UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/108 "2010-07-22T20:22:01Z")
Heh, sure: for me to lend you 8 cents is for me to give you 1/12 of a dollar minus 1/3 of a cent. That’s the same as for you to give me -1/12 of a dollar plus 1/3 of a cent. So the situation you’re asking for, as I choose to see it, is for _you_ to give _me_ 1 + 2 + 3 + 4 + … dollars, plus 1/3 of a cent. Pay up! 🙂 ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [July 22, 2010, 9:47pm UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/109 "2010-07-22T21:47:42Z")
> [@Indistinguishable](#): > > If you owe me -1/12 dollars, then _you_ should be giving _me_ the 1 + 2 + 3 + 4 + … dollars. 🙂 But in order to settle your 8.33 cent debt that way, he’d have to **complete the infinity**. At some point he’ll have scoured the reachable universe to come up with an installment of 50 gazillion tons of rhodium and will be unable to locate the required 100 gazillion tons for the next installment. Would you settle for a pint of beer instead? ---
**Author:** ![Darth\_Panda](https://avatars.discourse-cdn.com/v4/letter/d/ee7513/32.png) [@Darth\_Panda](https://boards.straightdope.com/u/Darth_Panda)\ **Post date:** [July 22, 2010, 9:55pm UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/110 "2010-07-22T21:55:30Z")
> [@septimus](#): > > But in order to settle your 8.33 cent debt that way, he’d have to **complete the infinity**. At some point he’ll have scoured the reachable universe to come up with an installment of 50 gazillion tons of rhodium and will be unable to locate the required 100 gazillion tons for the next installment. > > Would you settle for a pint of beer instead? I’ll settle for a pint and a little of that there rhodium. ---
**Author:** ![Nicolas\_Bourbaki](https://avatars.discourse-cdn.com/v4/letter/n/ecc23a/32.png) [@Nicolas\_Bourbaki](https://boards.straightdope.com/u/Nicolas_Bourbaki)\ **Post date:** [July 23, 2010, 5:47am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/111 "2010-07-23T05:47:51Z")
> [@septimus](#): > > But in order to settle your 8.33 cent debt that way, he’d have to **complete the infinity**. At some point he’ll have scoured the reachable universe to come up with an installment of 50 gazillion tons of rhodium and will be unable to locate the required 100 gazillion tons for the next installment. In a sense, this. A lot of the counter-arguments here rely on what works for sums of finitely many terms, or sums of _very many_ finite terms, or for a finite number of terms for every finite number, but none of them actually addressed what would happen for infinitely many terms. The problem is we can only give an answer for the most well-behaved circumstances, and if any aspect of the question is even the slightest bit not kosher, we have to resort to indirect methods, and put restrictions in place to guarantee uniqueness. As an example, riddle me this: suppose I have a chute and infinitely many balls to throw in it. At 11 o’clock, one ball is thrown in the chute, and is let out. At 11:30, two balls are thrown in, and one is let out. At 11:45, 4 balls are thrown in, and another ball is let out. At 11:52:30, 11:56:45, 11:58:47.5, . . . . the pattern continues, with twice as many balls being thrown in each time, and one being let out. How many balls are in the chute at noon? (Admittedly, the answer is not “-1/12”) ---
**Author:** ![tim314](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tim314/32/3468_2.png) [@tim314](https://boards.straightdope.com/u/tim314)\ **Post date:** [July 24, 2010, 4:10am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/112 "2010-07-24T04:10:34Z")
> [@Indistinguishable](#): > > Ah, but what if we say -1 + 1/2 + 1/4 + 1/8 + 1/16 + … = 0? This is rather more like 1 + 2 + 3 + … = infinity than 1 + 2 + (-3) = 0. I wouldn’t say this illustrates a difference in “numberyness” between infinity and 0, even if one feels there is such a difference; it just illustrates a difference in the meaning of finitary and infinitary addition. Eh, that’s a fair point. But I guess what I’m getting at is that infinity is only a limit, whereas zero, while it _can_ be a limit, is also a plain old number that you can do regular arithmetic with. (Well, as long as it’s not the divisor.) Except I balked a bit at saying “infinity is only a limit”, since if you ask me “How many integers are there?” I’d have to say “infinity”, and I’m not sure that such usage is really invoking a limit. So maybe I’d say "infinity is the limit of some sequences, and it’s the number of elements in some sets, and it’s probably some other things too… but zero is all that stuff plus it’s something that you can do grade school arithmetic with. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 24, 2010, 4:23am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/113 "2010-07-24T04:23:02Z")
> [@tim314](#): > > Eh, that’s a fair point. But I guess what I’m getting at is that infinity is only a limit, whereas zero, while it _can_ be a limit, is also a plain old number that you can do regular arithmetic with. (Well, as long as it’s not the divisor.) In the same sense, you can do grade school arithmetic with ∞, as long as you don’t subtract it from itself, you don’t divide it by itself (just as you can’t divide 0 by itself), and you don’t multiply it by 0 (a gap for which 0 and ∞ are equally “blameworthy”). The quite simple rules being a + ∞ = a, a \* ∞ = ∞ (for non-zero a), a/0 = ∞ (for non-zero a), and a/∞ = 0 (for non-infinite a). There’s no fixed sense of what “number” means. Various abstract systems are called systems of “numbers” when they have (subjectively) enough resemblances of enough kinds to cause us to consider them of a kind with the familiar existing archetypes. I certainly think the system outlined above meets this criterion (part of what’s often called [“the extended real numbers”](http://en.wikipedia.org/wiki/Extended_real_numbers)). Any objection raised against ∞ “being a number” could just as well be matched by similar objections raised against more familiar objects blessed only by a longer history of acceptance. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 24, 2010, 4:45am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/114 "2010-07-24T04:45:36Z")
Er, sorry, I made a mismash of combining the projectively extended and affinely extended real numbers. (Also, I made an obvious typo in the addition rule). Anyway, you can have the following systems: Affinely extended reals: ∞ and -∞ are distinct. You can’t add ∞ to -∞, you can’t divide an infinite quantity by an infinite quantity, you can’t multiply an infinite quantity by 0, and you still can’t divide anything by 0. The rules beyond that are that an infinite quantity plus anything appropriate equals the same infinite quantity, an infinite quantity times anything appropriate equals the infinite quantity of appropriate sign, and any finite quantity divided by an infinite quantity is zero. Projectively extended reals: ∞ and -∞ are the same. You can’t add ∞ to -∞ (which means you can’t add ∞ to ∞), you can’t divide an infinite quantity by an infinite quantity, and you can’t multiply an infinite quantity by 0. However, you _can_ divide a non-zero quantity by 0. The rules are that ∞ + a = ∞, ∞ \* a = ∞, a / 0 = ∞, and a / ∞ = 0, subject to the previous restrictions on when these are defined. Both of these define perfectly reasonable arithmetic systems along the lines of grade school arithmetic; ∞ is missing some properties in each, sure, but no worse than we all accept 0 to be missing in more familiar arithmetic systems. (And the projectively extended reals partially remedy the gap of division by 0). Plus, there are myriad arithmetic systems beyond these of interest; for instance, in many contexts, one deals with quantities for which ordering and addition make sense but multiplication is meaningless, and which cannot go negative but can turn infinite. Well, here it makes sense to speak of “numbers” as either reals \>= 0 or ∞, which can be added to one’s heart’s content with the same rules as those of the extended reals. And so on, in any particular case one picking or devising a system of “numbers” appropriate to whatever one intends to do with them… ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 24, 2010, 4:57am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/115 "2010-07-24T04:57:30Z")
(Of course, you can make up whatever systems you want; the point of the systems illustrated above is that they’re actually very commonly useful to mathematicians, enough so that they’ve managed to develop standard names, just like the integers, the real numbers (god, I hate that name), the complex numbers, and so on) ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 24, 2010, 6:18am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/116 "2010-07-24T06:18:12Z")
Quoth **Nicolas** : > [@](#): > > As an example, riddle me this: suppose I have a chute and infinitely many balls to throw in it. At 11 o’clock, one ball is thrown in the chute, and is let out. At 11:30, two balls are thrown in, and one is let out. At 11:45, 4 balls are thrown in, and another ball is let out. At 11:52:30, 11:56:45, 11:58:47.5, . . . . the pattern continues, with twice as many balls being thrown in each time, and one being let out. How many balls are in the chute at noon? (Admittedly, the answer is not “-1/12”) I don’t know; you haven’t specified the behavior of the system at noon, only at times previous. You were probably aiming for an answer of “zero”, but there’s insufficient information to support that. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 24, 2010, 6:27am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/117 "2010-07-24T06:27:39Z")
Yes, that’s the response I would give as well. Though I think maybe **Nicolas** wanted us to adopt the principle “A ball is only present at time T if it is introduced at or prior to time T without being removed in the meanwhile” and “No balls are added or removed at any other times than the stated ones”, and then to note the still-present ambiguity: if, for example, the ball removed at each step is the one which was introduced earliest out of all present, then all balls will eventually be removed, leaving none at noon. On the other hand, if the ball removed at each step is the one which was last introduced, then there would be infinitely many balls left (the only balls removed would be the n-th ones to be introduced, where n is one less than a power of two). And, of course, other ways of picking which ball to remove at each step would leave any particular subset of the balls remaining at the end which you like. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 24, 2010, 6:35am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/118 "2010-07-24T06:35:27Z")
(In the above descriptions, I in turn intend for no ball to ever be re-introduced upon removal. Lots of variables here…) ---
**Author:** ![BigT](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bigt/32/12044_2.png) [@BigT](https://boards.straightdope.com/u/BigT)\ **Post date:** [July 24, 2010, 11:02am UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/119 "2010-07-24T11:02:19Z")
It being a chute, the one ball being removed would be the earliest one put in. Yet I fail to see how it would ever reduce to zero. It seems to me that it would be subtracting two different sets of infinity, one that includes all whole numbers, and one that only includes 2[sup]x[/sup] - 1 . Does subtracting all the even numbers from the whole numbers produce zero? No. You still have {1, 3, 5, 7, 9, …} left. I thought that was the whole point of discovering there were different sets of infinity. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [July 24, 2010, 2:07pm UTC](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329/120 "2010-07-24T14:07:50Z")
Good point about the mechanics of chutes; I suppose then the intention wasn’t to point out the ambiguity given by choice of ball to remove. Given the assumptions above including that the earliest ball is always removed: 11:00: In goes ball #1. Out goes ball #1. 11:30: In go balls #2 and 3. Out goes ball #2. 11:45: In go balls #4, 5, 6, 7. Out goes ball #3. 11:52:30: In go balls #8 through 15. Out goes ball #4. … Note that, even though at any moment prior to noon there are many, many unremoved balls, every ball that ever goes in eventually comes out, the n-th ball to be introduced being also the n-th ball to be removed. Thus, with the mechanics that a ball is present at noon if and only if it has previously been introduced without being removed in the meanwhile, no ball will be left present at noon, even though the number of balls grows larger and larger at an exponential rate up until noon. [Previous page](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329.md?page=5) [Next page](https://boards.straightdope.com/t/1-2-3-4-1-12-in-what-sense/403329.md?page=7)