I saw the puzzle on Hoffstader’s “Godel Escher Bach” (I could not solution in book). I think the series diverges since the partial sum does not approach a limit as the number of terms increases. That’s just my guess though. What is the answer?
You pegged it right on.
1/2, of course.
We know
1 + x + x^2 + x^3 + … = 1/(1-x)
So
1 - 1 + 1 - 1 + 1 - 1 … = 1 + (-1) + (-1)^2 + (-1)^3 + (-1)^4 + (-1)^5 + … = 1/(1-(-1)) = 1/2
Don’t believe me? Ask Euler (or was it Leibniz?) They didn’t worry so much about having partial sums converge. So they’d also have
1 + 2 + 4 + 8 + 16 + 32 + … = -1
How 'bout that?
But now the series is defined to be the limit of the partial sums, so when you say that the partial sums don’t approach a limit, you are saying exactly that the series doesn’t converge. Which it doesn’t. (And so the first series I wrote only converges for |x| < 1.)
This was a very contentious problem before the introduction of limits. There were people who felt very strongly that it summed to 1, -1, 0, or 1/2.
It only proves that the Universe is running on a two’s-compliment machine.
Now the only question is… big-endian or little-endian?
It is Cesàro summable, with Cesàro sum 1/2.
So, exactly how stupid am I if I think the answer is clearly zero?
Pretty damn stupid. 
Seriously, this one’s a little tough to handle without the appropriate background. It’s a trivial exercise for an undergraduate math major, but if you’re not, I wouldn’t expect you to see it right away.
Will people throw things at me if I say 1 - 1 + 1 - 1 + 1 … == 0.999…?
Is there any simple way to explain this? Because I was thinking the same exact thing as Iteki. Why don’t all the +1’s and -1’s simply cancel each other out?
(1 - 1) + (1 - 1) + (1 - 1) + … seems to be equal to 0. But 1 + (-1 + 1) + (-1 + 1) + … seems to be equal to 1. Which should we choose as correct?
Well, that simple type of reasoning leads to rapid problems. If you say
1 - 1 + 1 - … + ( 1 - 1) + ( 1 - 1) + … = 0,
then why not instead
1 - 1 + 1 - … = 1 - ( 1 - 1) - ( 1 - 1) - … = 1
So we need a more rigorous definition of the sum of an infinite series, one which will avoid such problems. One of the consequences of this definition is that any convergent series has terms which converge to 0. Since the terms of our series don’t do this, the series does not converge.
There are some complications over series which converge conditionally, but these are by way of counter-intuitive results. They are still perfectly rigorous
Oooo, I am so tempted. People have been thrown off boats for lesser remarks, ya know.
For no good reason, I’ll note in passing that 1 - 1 + 1 - 1 + … can be regarded as the number 1.111… in base -1. Or it could be, if it were actually a number, which we know it isn’t, because the series has no limit, so never mind.
And thus I hereby nominate base -1, with digits 0 and 1, as the most useless basis for a positional number system. Not only is it unable to express non-integers — a drawback it shares with base +1 — but in addition it’s about twice as inefficient at expressing the integers it can express. To wit:
Base 10 Base 1 Base -1
---------------------------
0 0 0
1 1 1
2 11 101
3 111 10101
4 1111 1010101
...
Well, that’s enough geeky silliness (and silly geekiness) for one day.
So, how does positional notation work in base 1? And how do you take the base-1 log of a number? (Remember that log-n x is equal to ln x/ln n, and that ln 1 == 0.)
I don’t regard base 1 as being truly comparable with base 2 or 10 or any other nonzero, nonone base. It is a rather degenerate special case, in that it breaks positional notation (1000 == 1 == 10 == …) and the logarithm function (division by 0 is undefined).
Unary isn’t a positional number system, and Bytegeist’s version of base -1 seems to be based off the assumption that it is.
Another way to think of it. What is:
1 x -1 x 1 x -1 x 1 x -1 x 1 x -1 x 1 x -1 x 1 x -1 …
If you can’t find an answer to that, then why should you be able to calculate:
1 + -1 + 1 + -1 + 1 + -1 + 1 + -1 + 1 + -1 + 1 + -1 + …
Just because you can write something out doesn’t mean you can calculate it meaningfully (unless you have an infinite amount of time of course, in which case the answer is…? No, there still isn’t one: if you spend an infinite amount of time on it you’ll obviously never come to an answer.)
Most of the posts assume that the question has an answer. Let me say that it doesn’t as posed.
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. Yes, it can be thought of as a continuation of the series for 1/1-x to the domain where it doesn’t converge and if you substitute x = 1, you get 1/2. That does not make 1/2 the value of the series, though. As for 1 + 2 + 4 + 8 + … = -1, well that is nonesense and illustrates the folly of using the series argument. There is, incidentally, a different notion of convergence (called 2-adic) in which that series does converge and to -1, but that is quite a different matter.
Anyway, the bottom line is that you have to say what notion of convergence you are interested in and then you may get an answer. But there is no god-given answer.
I’ll take a shot at explaining how it is usually done with partial sums. It’s important to remember what Hari Seldon said and understand that what I say isn’t the only way to do it, it’s just that this is the most common approach to convergence. It’s useful to point out the distinction between sequences (the succession of terms, e.g. 1, -1, -1) and series (the sum of terms, 1 - 1 + 1 -1).
Suppose we have an infinite series (summation) and some formula that describes each term (e.g. the ‘nth’ term = n[sup]2[/sup] / ln (n+1) . In the OP’s case, we could use (-1)[sup]n+1[/sup]. ). Rather than tackle the whole infinite thing at once, we take some of it, and figure out where it’s going.
The sum of the first 2 terms will be some value, and the sum of the first 3, first 20, first 490 terms will each be some value, etc. It’s these subsequent sums that are called the ‘partial sums’, since they sum part of the series. Now, to get the infinite series we’d have to add terms forever, but we expect that, if the series converges to some value, our sums are going to get inexorably closer to that value the more and more partial sums we do.
This is more or less what it means to say ‘the sequence of partial sums converges’. If we kept doing partial sums and the sequence never gets close to any one number, then we can’t say that the series is going to converge to that (or any) particular number. A series can only be said to converge to a single number.
It’s like looking at a car that drives off on an infinite road. Unfortunately it’s being driven by a penguin, and swerves around a lot (but there’s a curb keeping it on the road). Also imagine that the steering of the car doesn’t work properly, and once the steering wheel is turned, from then on it can only be turned less than that. We can say that the penguin will eventually be traveling in a straight line, since while it keeps swerving, it swerves less and less each time. But if the steering wheel were normal, all we would know is that the penguin’s going to keep wandering, while hitting the curbs a lot.
Okay, maybe it’s not like that.
Derleth & ultrafilter:
I wasn’t being very serious in that post, or even slightly serious, as I hope you could tell. Still, I shouldn’t have misapplied the term “positional number system” to unary notation, which doesn’t really fit the description.
Likewise to “base -1”, or “nega-unary” notation, if that’s what we can call it. Not that there’s much call to call it anything.
Interesting though that MathWorld fails to exclude 1 or -1 as candidates for bases, saying that “any integer number” can be used.
Wow, really? Even zero?
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?