# 1 + 2 + 4 + 8 + 16 + 32... = -1 ?!
**URL:** https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341
**Category:** Factual Questions
**Created:** [March 6, 2008, 4:54pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341 "2008-03-06T16:54:28Z")
**Posts on this page:** 20
**Page:** 1
### Author:  [@Frylock](https://boards.straightdope.com/u/Frylock)
#### Post date: [March 6, 2008, 4:54pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/1 "2008-03-06T16:54:28Z")
In [this thread](http://boards.straightdope.com/sdmb/showthread.php?p=9549171#post9549171) someone points out you seemingly can derive the equation 1+2+4+8+16+… = -1 as follows:
x = 1+2+4+8+16+…
x = 1+2(1+2+4+8+…)
x = 1+2x
x = -1
My question is, what do mathematicians take to be the right way to respond to this.
Is there a consistent way to build a mathematics of such “sums”? Or do there turn out to be more than one way to “sum” some infinite series? If the latter, then do we rule out such sums on this basis, or come up with a math which takes into account not only the members of the list of addends but also their order and groupings?
Or do we just call these kinds of sums “undefined” and move on?
It appears to me that there are some infinite sums which do not come out to a nice neat answer like this one does (I think 1+2+3+4+5… is an example of this), while some (like this one) do. Am I right to think this is the case? Or is there some way to find such a “sum” for _any_ arbitrary infinite list of addends?
And so on.
-FrL-
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### Author:  [@Giles](https://boards.straightdope.com/u/Giles)
#### Post date: [March 6, 2008, 4:59pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/2 "2008-03-06T16:59:16Z")
The series 1+2+4+8+16+… does not have a limit, so it’s meaningless to do arithmetic with it.
[/QUOTE]
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### Author:  [@Frylock](https://boards.straightdope.com/u/Frylock)
#### Post date: [March 6, 2008, 5:01pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/3 "2008-03-06T17:01:53Z")
[QUOTE=Giles]
The series 1+2+4+8+16+… does not have a limit, so it’s meaningless to do arithmetic with it.
[/QUOTE]
[/QUOTE]
So mathematicians say that a “sum” like this is undefined?
It seems so weird that it come out so nicely and neatly to a simple answer of -1. I’m used to seeing “undefined” appear wherever an answer to a problem ends u being something that intuitvely looks to me like infinity. In my head I think I have almost unconsciously conflated the two.
So am I right to suspect that if you changed the order of the numbers in the list, or changed the way you group them in your derivation, you can get nice neat little answers other than -1?
-FrL-
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### Author:  [@Frylock](https://boards.straightdope.com/u/Frylock)
#### Post date: [March 6, 2008, 5:04pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/4 "2008-03-06T17:04:15Z")
Hmm… can anyone explain anything from the following: [Theorems on methods for summing divergent series](http://en.wikipedia.org/wiki/Divergent_series#Theorems_on_methods_for_summing_divergent_series) to a(n intelligent) layman?
-FrL-
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### Author:  [@Frylock](https://boards.straightdope.com/u/Frylock)
#### Post date: [March 6, 2008, 5:07pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/5 "2008-03-06T17:07:17Z")
Ah… I bet if someone could explain this comment from [here](http://cornellmath.wordpress.com/2007/07/28/sum-divergent-series-i/):
> [@](#):
>
> In fact, if instead of using the absolute value for our metric we used the 2-adic valuation, the sum honestly converges to -1.
then I’d start to understand. Whats a metric? Whats a 2-adic valuation?
-FrL-
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### Author:  [@Sunspace](https://boards.straightdope.com/u/Sunspace)
#### Post date: [March 6, 2008, 5:09pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/6 "2008-03-06T17:09:24Z")
But you can also do this:
x = 1+2+4+8+16+…
x = 1+2+4(1+2+4+8+…)
x = 1+2+4x
x-4x = 3
-3x = 3
x = -1
Oh, wait. It comes out the same. So… um… I dunno. I thought it was going to come out with a different value of x, and then it would be like dividing by zero, where if you allow it, you can prove that 1 = 2.
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### Author:  [@Napier](https://boards.straightdope.com/u/Napier)
#### Post date: [March 6, 2008, 5:10pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/7 "2008-03-06T17:10:01Z")
I think that the problem is that you are dealing with infinities. It’s obvious that x is infinite in the first line. What is less obvious is that infinity would be a solution to the equation x = 1+2x. And I’m not sure how correct I am in saying that. But, I don’t see how you could say that twice infinity plus one would not be infinity, so infinity would have to be a solution, wouldn’t it? Though, this doesn’t explain why -1 wouldn’t also be a solution.
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### Author:  [@Giles](https://boards.straightdope.com/u/Giles)
#### Post date: [March 6, 2008, 5:17pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/8 "2008-03-06T17:17:21Z")
There’s some weird voodoo going on there. If you assume that the limit is a number s, you can prove that it’s -1, and that’s consistent with some other facts about this kind of series. However, the partial sums are clearly always positive, so if it had a limit, the limit would have to be positive (or possibly zero) – the limit can’t be -1. So what G.H. Hardy seems to have done might work in some axiom system, but it doesn’t work in ordinary real or complex analysis.
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### Author:  [@Colophon](https://boards.straightdope.com/u/Colophon)
#### Post date: [March 6, 2008, 5:26pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/9 "2008-03-06T17:26:53Z")
Well, it’s obvious. You keep adding in an infinite series, you’ll go right round the back of the number line and come up the other side at -1. 🙂
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### Author:  [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)
#### Post date: [March 6, 2008, 6:15pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/10 "2008-03-06T18:15:06Z")
Isn’t the “2-adic valuation” bit related to 2s complement? For a 4 **bit** signed number, you have the following binary representations
0 = 0000
1 = 0001
…
7 = 0111
-8 = 1000
-7 = 1001
…
-1 = 1111
so -8 + -4 + -2 + -1 becomes(\*) in binary
1000 + 1100 + 1110 + 1111 = 0001
or -8 + -4 + -2 + -1 = 1
This generalizes for N-bit 2s complement numbers.
I think this is why **Sunspace** is getting the same result, instead of being able to get an arbitrary result, like you’d expect to be able to get from infinity - infinity.
(\*)I can’t sum 8 + 4 + 2 + 1 because there is no +8. If you interpret 1000 as +8 instead of -8, you could.
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### Author:  [@Giles](https://boards.straightdope.com/u/Giles)
#### Post date: [March 6, 2008, 6:28pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/11 "2008-03-06T18:28:34Z")
[QUOTE=ZenBeam]
Isn’t the “2-adic valuation” bit related to 2s complement?
[/QUOTE]
I think that works (in this case), except that you need to represent integers with an unlimited number of bits. For positive integers, you have a infinite number of 0’s, followed by the usual binary representation; for negative integers, you have an infinite number of 1’s, followed by the usual 2’s-complement representation, e.g.:
+8 = …0000000001000
-8 = …1111111110111
So the sum looks like this:
…0000000000001 (+1)
…0000000000010 (+2)
…0000000000100 (+4)
…0000000001000 (+8)
…
* * *
…1111111111111 (-1)
Of course, that’s not normal arithmetic, or even normal computer arithmetic.
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### Author:  [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)
#### Post date: [March 6, 2008, 6:31pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/12 "2008-03-06T18:31:04Z")
The problem is that, as Napier points out, the sum on the right diverges, so you
ve got infinity on the right. Then you equate that infinity to 1 plus two times itself, whivh is correct – it’s just as infinite, and you can make a one-to-one correspondence of the terms (except the one). What you can’t do, themn, is subtract the one from the other as you could in “normal” math. Infinities don’t behanve nicely when you do that, no matter how they’re written.
Notyice that, if you replace this infinite sum with something similar that converges, like x = 1 + (1/2) + (1/4) + (1/8)+… and do the same steps, everything works out fine, and you get x = 2. But that’s because you;re not trying to subtract one infinity from another, but one finite sum from another (even though both have an infinite number of sums).
As it stands, yor “proof” is like those proofs that rely on a “hidden” division by zero – you get in trouble because you’re trying to use infinity as if it’s just another number. It ain’t.
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### Author:  [@gazpacho](https://boards.straightdope.com/u/gazpacho)
#### Post date: [March 6, 2008, 6:44pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/13 "2008-03-06T18:44:12Z")
[QUOTE=ZenBeam]
Isn’t the “2-adic valuation” bit related to 2s complement? For a 4 **bit** signed number, you have the following binary representations
0 = 0000
1 = 0001
…
7 = 0111
-8 = 1000
-7 = 1001
…
-1 = 1111
so -8 + -4 + -2 + -1 becomes(\*) in binary
1000 + 1100 + 1110 + 1111 = 0001
or -8 + -4 + -2 + -1 = 1
This generalizes for N-bit 2s complement numbers.
I think this is why **Sunspace** is getting the same result, instead of being able to get an arbitrary result, like you’d expect to be able to get from infinity - infinity.
(\*)I can’t sum 8 + 4 + 2 + 1 because there is no +8. If you interpret 1000 as +8 instead of -8, you could.
[/QUOTE]
The series does not converge with twos complement number either. It just cycles through 1,3,7 -1 etc.
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### Author:  [@Hypnagogic\_Jerk](https://boards.straightdope.com/u/Hypnagogic_Jerk)
#### Post date: [March 6, 2008, 6:51pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/14 "2008-03-06T18:51:37Z")
[QUOTE=Frylock]
then I’d start to understand. Whats a metric? Whats a 2-adic valuation?
[/QUOTE]
A metric is a function that gives you the distance between two points (in a _metric space_). Now, given that every rational x can be written as x = 2[sup]n[/sup] u/v, with u and v indivisible by 2, with n being a unique choice, we can then define the _2-adic valuation_ in this way: |x|[sub]2[/sub] = (1/2)[sup]n[/sup]. If x = 0, then |x|[sub]2[/sub] = 0. In the same way that the absolute value defines a metric on the rationals (d(x,y) = |x-y|), the 2-adic valuation does the same.
It’s been some time since I last worked with the [p-adic numbers](http://en.wikipedia.org/wiki/P-adic_number), but I’ve posted in the other thread to invite other mathematicians whose knowledge of the subject is better than mine to come explain why this series does converge (and to -1) in the 2-adic numbers.
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### Author:  [@Hypnagogic\_Jerk](https://boards.straightdope.com/u/Hypnagogic_Jerk)
#### Post date: [March 6, 2008, 6:57pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/15 "2008-03-06T18:57:48Z")
Missed the edit window.
Of course, the convergence of sequences and series is defined in a metric space. A sequence x[sub]n[/sub] converges to x if, for every e \> 0, then there exists a point N such that if n \> N, then |x[sub]n[/sub] - x| \< e. From this, we can see how different metrics lead to different definitions of convergence, and we can probably explain how this series converges in the 2-adic numbers.
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### Author:  [@Hypnagogic\_Jerk](https://boards.straightdope.com/u/Hypnagogic_Jerk)
#### Post date: [March 6, 2008, 7:03pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/16 "2008-03-06T19:03:04Z")
Let x[sub]n[/sub] = 1 + 2 + 4 + … + 2[sup]n[/sup] = 2[sup]n+1[/sup] - 1. Then x[sub]n[/sub] - (-1) = 2[sup]n+1[/sup], and therefore |x[sub]n[/sub] - (-1)|[sub]2[/sub] = |2[sup]n+1[/sup]|[sub]2[/sub] = (1/2)[sup]n+1[/sup], which converges to 0 as n goes to infinity. This isn’t a very formal proof (I haven’t even proved that the 2-adic valuation really defines a metric on the rationals), but it can help you see how this series converges to -1 in the 2-adic numbers.
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### Author:  [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)
#### Post date: [March 6, 2008, 7:21pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/17 "2008-03-06T19:21:42Z")
> [@](#):
>
> The series does not converge with twos complement number either.
With N-bit numbers, there are always N terms to sum, so it converges. For all finite N, it converges to
-2^(N-1) + … + -8 + -4 + -2 + -1 = 1.
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### Author:  [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)
#### Post date: [March 6, 2008, 7:32pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/18 "2008-03-06T19:32:20Z")
[QUOTE=Frylock]
So mathematicians say that a “sum” like this is undefined?
It seems so weird that it come out so nicely and neatly to a simple answer of -1. I’m used to seeing “undefined” appear wherever an answer to a problem ends u being something that intuitvely looks to me like infinity. In my head I think I have almost unconsciously conflated the two.
[/QUOTE]
The standard Calculus II definition for the sum of an infinite series is that it’s limit of the sequence of partial sums, **if** such limit exists. So the sum of 1 + 2 + 4 + 8 + … would be the limit, as n approaches infinity, of 1 + 2 + 4 + … + 2[sup]n[/sup]. Since this limit doesn’t exist, the series diverges: it doesn’t have a sum.
In the earlier days of infinite series, mathematicians played around with them without being all that careful about whether or not they converged, leading to some weird, contradictory or controversial results.
> [@](#):
>
> So am I right to suspect that if you changed the order of the numbers in the list, or changed the way you group them in your derivation, you can get nice neat little answers other than -1?
Well, there’s a theorem that says that if you have a **conditionally convergent** series (i.e. a convergent series in which some of the terms are positive and some negative, which would diverge if you made all the terms positive), you can, by a suitable rearrangement of the terms, get it to add up to any sum you want.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [March 6, 2008, 8:31pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/19 "2008-03-06T20:31:39Z")
I’m playing fast and loose with p-adic integers here, but this should express the basic idea:
Fix a prime p, and consider sequences of integers _a_ = (_a_[sub]0[/sub], _a_[sub]1[/sub], _a_[sub]2[/sub], …). We introduce an equivalence relation such that _a_ and _b_ are equivalent iff _a_[sub]i[/sub] = _b_[sub]i[/sub] mod p except at finitely many points. The p-adic integers are the equivalence classes of this relation.
In the 2-adic integers, (1, 1, 1, 1, 1, …) = (0, 1, 1, 1, 1, …), (2, 2, 2, 2, 2, …) = (0, 0, 2, 2, 2, …), (4, 4, 4, 4, 4, …) = (0, 0, 0, 4, 4, …), (8, 8, 8, 8, …) = (0, 0, 0, 8, …), and so on. If you add together the right-hand side of every power of two in this form, you get (0, 1, 3, 7, 15, …), which is the same as (0, -1, -1, -1, -1, …), and that’s the same as (-1, -1, -1, -1, -1, …). Again, this is just a sketch, but the details work out well.
In the case of complex numbers, the power series 1 + r + r[sup]2[/sup] + r[sup]3[/sup] + … converges to 1/(1 - r) whenever |r| \< 1. This is an analytic function on the unit disk, meaning that it’s defined there and its derivatives behave nicely, so there’s a unique analytic function defined on the complex plane that agrees with it on the unit disk. That function happens to be 1/(1 - r), which is defined everywhere except r = 1. That allows us to say in some sense that the sum of any geometric series is 1/(1 - r) as long as the common ratio r is not 1.
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### Author:  [@ErmesMarana](https://boards.straightdope.com/u/ErmesMarana)
#### Post date: [March 6, 2008, 8:49pm UTC](https://boards.straightdope.com/t/1-2-4-8-16-32-1/440341/20 "2008-03-06T20:49:28Z")
It might seem as though we can arbitrarily change the limits of sequences by changing the metric, but this is not the case. If we want to change the limit of even a single sequence, then the new metric will not be able to provide a limit for the exact same sequences the old one could.
If we had another metric that gave us limits for exactly the same sequences as 2-adic valuation, then the limit here would still be -1.
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