@ Indistinguishable & @ Asympotically fat I think I get it … so would it be fair to say that -1/12 is an answer to the summation, not necessarily the only answer?
Thanks - sorry for dropping in not having read the thread, but I saw the Chu-Carroll post and realized I’d seen the same thing on the GQ 1st page for several days now. 
Yes, but it’s a reasonable answer in the sense that the methods that we use to assign that sum have properties that we consider desirable. If you want to say that it’s something else, you need to explain how you’re making that assignment, and you need to show that the methods that you’re using have equally nice properties.
I’m not aware of any divergent series that is assigned multiple values by reasonable notions of summation. Anyone?
Thanks, ultrafilter - I wasn’t really wondering is another value specifically would make a good answer, but whether ‘undefined’ is just as much an answer as ‘-1/12’ is… (I’m guessing yes?)
That’s probably my favourite post in this thread.
Damn, leafing through those linked blogs and their comment threads makes me somewhat despair of the quality of discourse on the net, especially given that they are all part of the so-called ‘sceptical’ community that prides itself in rationality…
But maybe Asympotically fat and his friends were all wankers. Then his post doesn’t prove anything. 
You could just insist that divergent series don’t have sums, but in doing so you’d be throwing out a tool that is useful in making predictions about actual physical systems (albeit quantum ones).
I’ve been looking for some information on the specifics of how divergent series are used in quantum mechanics, and I found a very nice explanation of a modern mathematician’s view of divergent series. In particular, I want to quote this bit:
Ah! That helps explain a bit. Thanks, ultrafilter!
BTW in case this hasn’t been mentioned before…
In some kind of sum, 1 + 2 + 4 + 8 + 16 +… = -1
Ack, I’ve been failing to find time to keep up with this thread, but …
Yeah, exactly. One thing I wanted to note, way back when, was that I was quite heartened by this thread; the SDMB seemed to be the one place on the Internet where people by and large really got it, so to speak (the vast majority of the discussion elsewhere, even from people one might expect better from, being terribly, terribly disappointing, as you note).
Back in this post in an earlier thread, Indistinguishable wrote
Using exp(-nh) seems arbitrary. We could use a different regularization function in place of exp(nh), like 1/(1+n*h). Then we get
f(h) = a1/(1+1h) + a2/(1+2h) + a3/(1+3h) + a4/(1+4h) + …
Do we still get 1 - 1 + 1 - 1 + … = 1/2? Playing around with it numerically, that looks to be the case. If I were willing to get a pen and piece of paper, I could probably show that it does.
Since 1/(1+nh) decays less rapidly than exp(-nh), there are going to be series which converge using the later function, but not using the former, so you could perhaps argue that exp(-nh) is “better”. But if our series we want to sum diverges too rapidly, exp(nh) won’t decay rapidly enough either, and we’d have to pick a different regularization function to get a value. i don’t think you can find one function that will always work.
There are many different definitions of integration. I read somewhere over a hundred. They don’t all work all the time. Some functions aren’t integrable using one definition, but are integrable using a different one. But (as far as I know (and correct me if I’m wrong!)) all the definitions that can integrate some function agree on the value for that integral.
Similarly, there are an unlimited number of different regularization functions one could use for a given series. Can it be shown that the ones which converge to a finite value all give the same answer for that series? Or can someone give a counter-example? (No pathological ones, like 0*h, and all integrals equal zero.)
I guess I’m thinking here mostly in terms of series whose partial sums alternate in sign, diverging in magnitude, but not just towards +infinity or just towards -infinity.
Yes, you will get the same results for series like 1 - 1 + 1 - 1 + … and 1 - 2 + 3 - 4 + … using other decay functions (including the one you chose).
As noted above, it’s convenient to define Abel summation using exponential decay, but
Using exponential decay amounts to thinking in terms of Fourier transforms; we are then essentially taking the sum of a function to be the limiting value of its Fourier transform at 0. Actually, you can use the perspective of Fourier theory more generally, to understand the “any sufficiently smooth decay function” result; handwavily, as the decay coefficient sequence d[sub]0[/sub], d[sub]1[/sub], d[sub]2[/sub], … moves towards constantly 1 (no decay), its Fourier transform “moves towards the Fourier transform of the constantly 1 function (i.e., the Dirac delta)”, so that the sum using such a decay is equal to the convolution of the Fourier transform of the series being summed (i.e., the associated function as we’ve defined it above using exponential decay (slightly rotated)) with a function “moving towards the Dirac delta”, which amounts to the limiting value of this associated function at 0. (To make this argument analytically rigorous, we need to clarify our understanding of “moves towards the Dirac delta”; the smoothness requirements will come in here.)
Missing words reinstated in bold.
(Perhaps this post by Terence Tao will be of use in clarifying the universality properties of smoothed sums)
Yeah, they are. You’re using the one for real numbers, which gives the same answers as for the ones for integers on the domain of integers. I’d say the addition of real numbers is what most assume when they see the plus sign.
I’d include complex numbers, but I’m not entirely sure you can say that ordered pairs that cannot be reduced really count as addition. If so, then throw those in. The point is that people do not in any way think it involves extending some function. Addition is assumed to be one of the simplest operations, and thus can’t consist of all the operations needed to do all these other things.
People can’t envision getting from finite real summation to Abel summation as easily as they can get from finite real summation to infinite limit-based summation.
No, the plus sign for real numbers can’t be applied to the domain of integers. It’s true that 2.0+3.0=5.0, and that 2+3=5, but those are two different statements, because 2 is not the same object as 2.0, nor is 3 the same object as 3.0, nor 5 the same as 5.0 .
Now, in actual practice, this is often glossed over, because there is a very strong sort of correspondence between the objects 2 and 2.0 , 3 and 3.0 , and so on (Indistinguishable can tell us the name of that correspondence; I always get the something-morphisms mixed up). And so, given that the real-domain plus works the same way on those corresponding objects that the integer-domain plus works on integers, we can usually get away with pretending that the integer-domain plus is a special case of the real-domain plus, even though it isn’t. But by the same token, we can similarly get away with pretending that the sum of a convergent series is a special case of the sum of various sorts of more general series.
Something interesting as well.
Take any number, any number at all and half that number, then half the answer ( 10,000,5,000,2500,1250,625,312.50…………) all the way to infinity (I used excel). If you add all those numbers up, it comes out to exactly double the original starting number. So if I star with 10 million and continue dividing by 2 until infinity, and add them all up, I will get 20 million
The series you describe will approach twice the original value, but never quite reach it.
He specified “all the way to infinity.” The sum of the infinite series is exactly 2n, in exactly the same way that 0.999999… is exactly 1.
Not that I know what I’m talking about, but although he said “all the way to infinity” he said he demonstrated it with Excel, which has got to be a bit short of infinity, no?
At any rate, I’ll concede your point, the Excel spreadsheet notwithstanding.