Why does the sum of all natural numbers equal -1/12?

Factual Questions
61

Sure. Again, there are multiple stories we can tell; however, the one which accords with the account we’ve been using is like this:

1 + 3 + 6 + 10 + … = the average of (1 + 2 + 3 + 4 + …) and (1 + 4 + 9 + 16 + …).

1 + 2 + 3 + 4 + … = -1/12, as noted above.

As for 1 + 4 + 9 + 16 + …, we’ll calculate this analogously to how we calculated 1 + 2 + 3 + 4 + … .

Let’s start with the alternating series X = 1 - 4 + 9 - 16 + … . Let’s call this X. Note that if we shift X over and add it to itself, we get 2X = 1 - 3 + 5 - 7 + … . Let’s call this Y. Note that if we shift Y over and add it to itself, we get 2Y = 1 - 2 + 2 - 2 + … . Let’s call this Z. Note that if we shift Z over and add it to itself, we get 2Z = 1 - 1 + 0 + 0 + 0 + 0 + … = 0. Thus, in some sense, Z = 0, Y = Z/2 = 0, and X = Y/2 = 0.

Now take W = 1 + 4 + 9 + 16 + … . If we subtract X from W, we get 0 + 8 + 0 + 32 + 0 + 72 + … = zeros interleaved with 8 * (1 + 4 + 9 + … ). Thus, in some sense, W - X = 8W, which tells us W = -X/7 = 0.

So 1 + 3 + 6 + 10 + … is the average of -1/12 and 0, which is -1/24.

If one wanted to use the characteristic function technique noted above to calculate this instead, we could note that our characteristic function would be f(h) = 1e[sup]-1h[/sup] + 3e[sup]-2h[/sup] + 6e[sup]-3h[/sup] + 10e[sup]-4h[/sup] + … = e[sup]-h[/sup]/(1 - e[sup]-h[/sup])[sup]3[/sup] = h[sup]-3[/sup] + h[sup]-2[/sup]/2 - 1/24 - h/240 + h[sup]2[/sup]/480 + …, whose degree 0 term is -1/24.

[Note, as a word of warning, that this characteristic function account of summation is sensitive to index-shifting on series which are not ordinary Abel summable, as explained in the old thread. Thus, if, for example, we had chosen to consider 1, 3, 6, …, to be the 0th, 1st, 2nd, …, terms of the series, rather than the 1st, 2nd, 3rd, …, terms of the series, respectively, we would get a characteristic function of f(h) = 1/(1 - e[sup]-h[/sup])[sup]3[/sup] and a degree 0 term of 3/8 instead.]

Not dumb at all! Just the sorts of questions an inquisitive mind would ask.

62

Are there are other harder ways to extract this value, or is this the only approach?

63

Indistinguishable, is there any “natural” notion of addition which encompasses both the example I gave and the ones you’ve been giving? Is there any “natural” notion under which any of these sums exists and is finite, but has a different value?

64

Sure. It’s not so terribly unnatural to smash together the ideas which would separately encompass these series (I did this at the end of the old thread), taking the summation of a[sub]n[/sub] over all n to be given by first constructing the function f(h) = sum of a[sub]n[/sub] e[sup]-nh[/sup] (which if convergent anywhere is absolutely convergent and analytic for sufficiently large h), meromorphically continuing it along the positive real axis to 0, and then extracting the degree 0 term of its Laurent series at 0. We can phrase this differently to make it look more or less natural, but at least to me, that’s natural enough.

Well, given linearity, the normalization 1 + 0 + 0 + 0 + … = 1, and shift-invariance for the series 1 + 2 + 4 + 8 + …, we must have that 1 + 2 + 4 + 8 + … = -1, in the ordinary way. I can’t think of any particularly compelling stories where we give any of these up right now.

The story for 1 + 2 + 3 + 4 + … is different, and against the backdrop of linearity and normalization, giving that series a finite value already necessitates giving up shift invariance for it (since subtracting that series from the average of its forwards and backwards shifts would give us 1/2 + 0 + 0 + 0 + … = 0); what we actually use is shift-invariance for 1 - 2 + 3 - 4 + … (which is very nearly absolutely convergent to 1/4, as this sum follows using any twice continuously differentiable decay rate), along with the principle that doubling all indices changes nothing.

As noted before, with a different starting index, the method I’ve been discussing in this thread/post would give 1 + 2 + 3 + 4 + … different results (e.g., if we take 1, 2, 3, … to be the 0th, 1st, 2nd, …, terms respectively, the sum by this method would be 5/12). But apart from such re-indexing, I’m not aware of any particularly compelling story for any other value, other than 0.

The story for 0 can be put like so: instead of taking the degree 0 term of a power series expansion around h = 0 of a[sub]n[/sub]e[sup]-hn[/sup], take the degree 0 term of a power series expansion around x = 1 of a[sub]n[/sub]x[sup]n[/sup] instead. (This methodology is equivalent to Abel summation + “The sum of nth powers of the naturals is 0 for each natural n”, in the same way that the characteristic function methodology I’ve discussed throughout the thread is equivalent to Abel summation + “The sum of nth powers of the positive integers accords with the Riemann zeta function for each n”)

Of course, 0 is not the most interesting result, and one can do myriad other ad hoc manipulations, or even variants on the above systematic ideas, to get whatever values one likes. I’m just not yet aware of any such stories that are particularly compelling (though I have no reason to rule them out… In fact, I’d be shocked if we couldn’t come up with some, for some purposes.).

65

I realized something that might be a little more easy for the skeptical to swallow than the recent discussion.

One of the objections that often comes up in discussions like these is that you can’t add up positive integers and get a negative fraction. That’s true for finite sums, but when you’re adding up infinitely many terms, things can get a bit squirrelly.

In general, the sum of finitely many rational numbers will be rational. However, if you look at the sum 1 + 1/2 + 1/6 + 1/24 + 1/20 + …, the limit is e, which is not a rational number. That’s an absolutely convergent series, which is the nicest class of infinite sums that anyone’s ever considered. And moving to even that forces you to give up something that you thought was obvious.

So if it can happen for nice series, what’s the issue with having it happen for thornier ones?

66

Thanks ultrafilter. I like the idea. However that wouldn’t be enough to resolve my objections (if I had any.)
In my mind, adding up an infinite series of rationals and getting an irrational is quite a different beast from adding up an infinite series of positives and getting a negative.
At the core of the issue is that old chestnut of thinking of infinity as a number (and a very big one). Replace that with the paradigm of “infinite is without limit and behaves weird”, and these results are less surprising.
If it helps anyone, you could consider a function like f(x)=tanx or f(x)=1/x. Both of these have discontinuities – if you approach from one side the limit is positive infinite, and from the other it is negative infinite.
(I like tan because you can have a geometrical construct where tanx represents the length of the tangent of a circle where it intersects a ray with angle x, originating at the circle centre. With this in mind, it is easy to see tangent flip from positive infinity to negative infinity as the ray passes π/2. So what exactly is tan(π/2)? is it positive infinite or negative infinite? Neither. It is just infinite.)
Given that “infinity” can behave as a positive or a negative on a whim, it shouldn’t be that surprising that an infinite series can sum to a negative number.

67

I first came across this concept in the book Prime Obsession by John Derbyshire (which is well worth a read for anyone interested.)

In introducing the Riemann zeta function with the standard definition zeta(s)=sum(1/n[sup]s[/sup]) he shows its validity for real values greater than 1. A bit of time is spent establishing that it is undefined for s=1. A bit of discussion is used to establish that there are values for real arguments between zero and one. Then there is a bit of a hand-wave under the title of “domain stretching” to say that the function has values for negative reals and complex arguments.
It is nice to have a thread like this that fills in a few of those details (but not so many as to spoil the surprise should I decide to devote a proper amount of time to the matter.)

68

There’s no “last” in an infinite series.

69

I started that earlier thread and I still don’t have even negative one-twelfth of a clue.

70

Fellow 12 year old with brain damage here! I didn’t read all the posts in detail, but I didn’t see anyone making this basic point that may help. The sequence actually includes all negative natural numbers. So the question really is:

. . . - 2 + -1 + 1 + 2 … = -1/12

So prior to seeing the video, I would have guessed the answer would be 0 because for every positive natural number, there is a negatively equal natural number. Lo and behold it doesn’t! It equals a smidge less than 0. Pretty damn cool from my naive experience!

71

No, it doesn’t, and it isn’t.

72

Not in a position to argue, so genuine question here: his proofs depend on the inclusion of negative numbers. So how are those justified?

73

The proof is based on manipulating several different series, these other series include negative numbers, but the series that equals -1/12 is 1 + 2 + 3 + 4 + 5 + 6 … and that is a series with no negative numbers.

74

Alrighty. Thank you. And just as I thought I sort’ve understood what was going on! :wink:

75

With the nitpicks that you forgot the 0! term, and that you typoed 120 as 20.

76

Not the way I was conceptualizing it. The whole point is that I’m using finite numbers and sets (specifically because I can’t grasp infinite ones.) What I do to conceptualize convergence is to take an finite subset and generalize it. I find the pattern and continue it.

But I think a better strategy is to use the 0.99999… paradigm, and see the number as …1111 repeating. That does exactly what you say, meaning there’s always another 1 on the left. I can visualize that, and work accordingly. Bravo!

Unfortunately, I don’t see how to take that conceptualization and generalize it to 1 + 2 + 3 + 4 … = -1/12. We’re using what I would call double addition here, where we’re adding one to every number, and then adding that number back to the previous answer. I can’t figure out how to convert that to the 0.9999… conceptualization I mentioned earlier.

Heck, I’d love to see a derivation from your -1 series to the -1/12 series. Why does adding all those extra numbers reduce the sum by a factor of 12?

77

BTW, is there an easier, less technical way to show when a series is or is not shiftable? It wouldn’t have to be particularly complete or rigorous, just a general idea. The full version is too dense for me.

Because I’m pretty sure that shifting is what was messing with a lot of people in the YouTube comments. They keep on resorting to counterproofs that apparently utilize non-Abel summable series.

78

You don’t, at least not directly. As Indistinguishable said, that series and the one I posted use two different generalizations of summation. I posted my example not as an example of the same method that Indistinguishable was using, but as an example that there are other definitions of addition that can lead to similarly-counterintuitive results.

79

The short, unrigorous version is this: A series is shiftable if it’s “very nearly” convergent in the standard sense, where “very nearly” means the only problem with convergence in the standard sense is the discreteness of the cutoffs in the partial sums; smoother decay would lead to convergence. So, for example, the series 1 - 1 + 1 - 1 + 1 - 1 + … is shiftable, because even though the ordinary partial sums “Add up the first n terms at 100% strength, the next term at 0% strength” oscillate between 0 and 1, if instead we decide to look at the smoother partial sums “Add up the first n terms at 100% strength, then the next term at 50% strength, then the remaining terms at 0% strength”, we will always get 1/2.

As it happens, any series whose nth term is given by a sum of polynomial functions of n times exponential functions of n whose bases aren’t 1 (that is, a sum of functions of the form cn[sup]p[/sup]b[sup]n[/sup] for constants c, p, and b with b not equal to 1) will be shiftable. This covers every example in this thread other than 1 + 1 + 1 + 1 + … and 1 + 2 + 3 + 4 + …, both of which are not shiftable (in these cases, c = 1, p = 0 or 1, and b = 1).

Note that no series of positive terms summing to a negative value will ever be shiftable. For a series to be shiftably summable but not absolutely convergent is for it to display a balanced cancellation of its positive and negative terms.

(If you just want jargon, the super-short version is this: a series is shiftable if it’s Abel summable)

80

Er, what I mean, clunkily worded here, is that every series examined in this thread is of the form “My nth term is a sum of functions of the form cn[sup]p[/sup]b[sup]n[/sup]”, though some involve taking b = 1. Those which don’t involve taking b = 1 (including 1 - 1 + 1 - 1 + …, 1 - 2 + 3 - 4 + …, and 1 + 2 + 4 + 8 + …) are all shiftable; those which do involve taking b = 1 (including 1 + 1 + 1 + 1 + … , 1 + 2 + 3 + 4 + …, and 1 + 3 + 6 + 10 + …) are all non-shiftable.