# 1 - 1 + 1 - 1 + 1 - 1 ... = ??? **URL:** \ **Category:** Factual Questions\ **Created:** [October 16, 2003, 2:58am UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719 "2003-10-16T02:58:48Z")\ **Posts on this page:** 10\ **Page:** 2
**Author:** ![The\_Ryan](https://avatars.discourse-cdn.com/v4/letter/t/7feea3/32.png) [@The\_Ryan](https://boards.straightdope.com/u/The_Ryan)\ **Post date:** [October 17, 2003, 12:35am UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/21 "2003-10-17T00:35:19Z")
> [@](#): > > \*Originally posted by Derleth \* > \*\*It only proves that the Universe is running on a two’s-compliment machine. > > Now the only question is… big-endian or little-endian? \*\* Yes, the universe is constantly telling me what nice twos I have. 🙂 Moe > [@](#): > > I really don’t get well over half of what’s been said here, but I still don’t see how it can be anything but 0, +1, or -1. No, 1-1+1-1…=1+1-1-1…=1+1-1+1-1-1…=1+1-1+1-1…=2+0=2 so the sum is two. Seriously though, the possibility that you’re missing is that it’s not equal to anything. ---
**Author:** ![Moe](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/moe/32/2972_2.png) [@Moe](https://boards.straightdope.com/u/Moe)\ **Post date:** [October 17, 2003, 12:56am UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/22 "2003-10-17T00:56:48Z")
OK, my little voice is telling me I should definitely be letting this one go, but I have much less fun when I listen to him. If the series is infinite doesn’t that mean that there are an infinite number of both +1’s and -1’s? If you have two positive terms in a row, and then continue the series in the usual alternating manner, aren’t you assuming that there will always be one more positive term than negative one? Which there can’t be since there are an infinite number of both. If you take the series to infinity why can’t you conclude that for every +1 there will be a -1 to cancel it out? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [October 17, 2003, 3:02am UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/23 "2003-10-17T03:02:15Z")
> [@](#): > > \*Originally posted by Moe \* > \*\*OK, my little voice is telling me I should definitely be letting this one go, but I have much less fun when I listen to him. > > If the series is infinite doesn’t that mean that there are an infinite number of both +1’s and -1’s? If you have two positive terms in a row, and then continue the series in the usual alternating manner, aren’t you assuming that there will always be one more positive term than negative one? Which there can’t be since there are an infinite number of both. If you take the series to infinity why can’t you conclude that for every +1 there will be a -1 to cancel it out? \*\* The value of a series is defined to be the limit of the sequence of partial sums. The partial sums on this one go 1, 0, 1, 0, … That’s not a convergent sequence, so this series has no value. ---
**Author:** ![Thaumaturge](https://avatars.discourse-cdn.com/v4/letter/t/858c86/32.png) [@Thaumaturge](https://boards.straightdope.com/u/Thaumaturge)\ **Post date:** [October 17, 2003, 3:37am UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/24 "2003-10-17T03:37:43Z")
Of course, the _real_ solution is that there is no such thing in the universe as an infinite series of numbers, and the answer depends upon where you decide to finish. But try to tell a mathemetician that. 😃 ---
**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:** [October 17, 2003, 4:50am UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/25 "2003-10-17T04:50:26Z")
> [@](#): > > If the series is infinite doesn’t that mean that there are an infinite number of both +1’s and -1’s? If you have two positive terms in a row, and then continue the series in the usual alternating manner, aren’t you assuming that there will always be one more positive term than negative one? Which there can’t be since there are an infinite number of both. If you take the series to infinity why can’t you conclude that for every +1 there will be a -1 to cancel it out? Answer me this: After you’ve used up three “ones” and only one “negative one”, how many of each are left? Using methods like this, one can get that series to sum to any integer at all, and by using some other mathematical tricks which really don’t apply here, you can get any number at all as your answer. There are some places in mathematics where it makes sense to say that a problem has more than one solution, or even an infinite number of solutions. But when you try to say that _everything_ is a solution, you might as well just say that the problem isn’t well-defined. By the way, in case anyone isn’t familiar with the “logic” for the twos-complement problem: x = 1 + 2 + 4 + 8 + 16 + … 2x = 2 + 4 + 8 + 16 + … x - 2x = 1 -x = 1 x = -1 ---
**Author:** ![Latino\_Yoshi](https://avatars.discourse-cdn.com/v4/letter/l/5e9695/32.png) [@Latino\_Yoshi](https://boards.straightdope.com/u/Latino_Yoshi)\ **Post date:** [October 17, 2003, 6:45am UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/26 "2003-10-17T06:45:44Z")
okay, when i first looked at the OP, I thought it equaled zero. My thinking was that you could reduce the equation to 1\*(n/2)-1\*(n/2). Of course, that would have assumed that the terms in the problem came in pairs like (1-1) + (1-1) + (1-1) + … = (0) + (0) + (0) + … = 0. But it doesn’t work that way. Anyways, I have nothing new to add to this thread, except to say that common sense should be allowed to say that it can’t equal anything but 0 or 1. ---
**Author:** ![Gary\_the\_Goat](https://avatars.discourse-cdn.com/v4/letter/g/b5ac83/32.png) [@Gary\_the\_Goat](https://boards.straightdope.com/u/Gary_the_Goat)\ **Post date:** [October 17, 2003, 9:32am UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/27 "2003-10-17T09:32:50Z")
So if you had: 2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2+2-2… ( i think you get it…) it is? ---
**Author:** ![Shade](https://avatars.discourse-cdn.com/v4/letter/s/2bfe46/32.png) [@Shade](https://boards.straightdope.com/u/Shade)\ **Post date:** [October 17, 2003, 9:52am UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/28 "2003-10-17T09:52:09Z")
> [@](#): > > \*Originally posted by Moe \* > \*\*Boy, am I stupid. > > I really don’t get well over half of what’s been said here, but I still don’t see how it can be anything but 0, +1, or -1. The consensus here is that there is no answer. But can’t “0, +1, OR -1” be an answer? I mean, it’s not a single answer, but doesn’t it simplify things more than saying there’s no answer? \*\* I suppose you could say ‘0 or 1’ on the grounds that subsequences of the partial sums converge to those. Firstly, in basic standard mathematics, there is quite definately no answer. ‘…’ is defined as ‘the limit of the partial sums’, which there isn’t. There are extensions which do define this (cf. Hari Seldon’s post.) I think the question you’d like addressed is _why_ mathematics doesn’t define a sum. I’d say that it’s because it’s useful to have the notion of a convergent sum. For instance, if (a[sub]1[/sub]+a[sub]2[/sub]+…)=1, you can say 2.(a[sub]1[/sub]+a[sub]2[/sub]+…)=(2.a[sub]1[/sub]+2.a[sub]2[/sub]+…)=2, and perform other operations. Most other suggestions would throw this structure out of the window, making the sum almost useless. Is that a fair summary? ---
**Author:** ![Phage](https://avatars.discourse-cdn.com/v4/letter/p/a4c791/32.png) [@Phage](https://boards.straightdope.com/u/Phage)\ **Post date:** [October 17, 2003, 12:12pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/29 "2003-10-17T12:12:05Z")
**Moe** , _If you take the series to infinity why can’t you conclude that for every +1 there will be a -1 to cancel it out?_ No, because you can assume with equal weight that for every -1 there will be a +1 to cancel it as well. The sequence alternates, but will never settle on one side or the other. ---
**Author:** ![Jay\_Belanger](https://avatars.discourse-cdn.com/v4/letter/j/e0b2c6/32.png) [@Jay\_Belanger](https://boards.straightdope.com/u/Jay_Belanger)\ **Post date:** [October 17, 2003, 1:17pm UTC](https://boards.straightdope.com/t/1-1-1-1-1-1/207719/30 "2003-10-17T13:17:03Z")
> [@](#): > > \*Originally posted by Hari Seldon \* > **Yes, it is Cesaro summable to 1/2 and that is probably the most reasonable answer, but it is still a convention, not a fact.** It is a fact that it is Cesaro summable to 1/2. I assume that “by a fact” you mean whether or not you want to consider the Cesaro sum to be the sum of the series. > [@](#): > > \*\* > As for 1 + 2 + 4 + 8 + … = -1, well that is nonesense > \*\* Well, 1 + 2 + 4 + 8 + … isn’t equal to -1 using the modern notion of the sum of a series, but that doesn’t make it nonsense. [Previous page](https://boards.straightdope.com/t/1-1-1-1-1-1/207719.md?page=1)