## Question about 1+2+3+4...=-1/12 (please hear me out)

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DR6
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### Question about 1+2+3+4...=-1/12 (please hear me out)

(Yeah, I've read the sticky about these kinds of topics. This is not going to be the usual flame war, I swear).

So I know that 1+2+3+4... has no actual sum, because their partial sums have no limit. However, saying that the series equals -1/12 is getting a lot of traction, and because of a good reason: you can use a generalization of summation using analytic continuations and the Riemann zeta function which does give that result. A lot of proponents argue that -1/12 is the only value we can assign to the divergent series. My question is: is this true? That is, are there not other reasonable summation methods which give different results? I have tried some googling and not much has come out of it.

Note that I don't consider rearranging the series a "reasonable" summation method, since it's inconsistent for divergent sums. Analytic continuations do not suffer from that problem.

jaap
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

Apart from the sticky threads, there is also this recent thread, with a fair amount of discussion:
viewtopic.php?f=17&t=108909

DR6
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

jaap wrote:Apart from the sticky threads, there is also this recent thread, with a fair amount of discussion:
viewtopic.php?f=17&t=108909

That thread does not seem to answer my question, however. They only talk about the validity of 1+2+3+4..=-1/12, and the things said are things I was already aware to.

Farabor
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

To turn the question around on you....what value would making this definition have? Anytime we add a definition to something, we generally do it because it is useful/leads to greater insights/etc. Regardless on the methodology, this 'result' seems to be valueless.

mike-l
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

You need to define precisely what you mean by reasonable summation methods. Namely, what properties do you want to have.

In any case, I asked a similar question in the linked thread and I still don't know the answer. Under what conditions does the analytic continuation method give a unique answer. It's possible to have two analytic continuations that give different answers even when starting from the same function (eg the power series for ln z, which has a branch cut, has infinitely many analytic continuations, so eg, the sum of 2n/n has infinitely many answers: all odd multiples of i*pi). This doesn't apply to the zeta function (which is analytic away from z = 1), so it's possible -1/12 is the unique answer from that method, but I don't know. And I don't know the answer to the broader question allowing any 'reasonable' method of summation.

However, -1/12 is a very 'natural' answer, and one that has proved to be incredibly useful in a number of fields.
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mike-l
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

Actually, I can give conditions that make -1/12 unique. Regularity, linearity and invariance under inserting (infinitely many) 0s into the sequence. (The latter is a stronger condition than stability (assuming regularity and linearity) so in particular a number of "interesting" methods fail this)

The proof in the numberphile video becomes rigorous with those assumptions.
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DR6
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

Farabor wrote:To turn the question around on you....what value would making this definition have? Anytime we add a definition to something, we generally do it because it is useful/leads to greater insights/etc. Regardless on the methodology, this 'result' seems to be valueless.

Yeah, and the method that result in -1/12 is very useful and gives a lot of insight. What I'm asking is if there are other methods that are similarly useful but give different results.

mike-l wrote:You need to define precisely what you mean by reasonable summation methods. Namely, what properties do you want to have.

Honestly, I don't even know. Giving a unique answer and generalizing normal summation are the most obvious ones. Linearity and similar properties would also be interesting. My question is intentionally open-ended there.

mike-l wrote:Actually, I can give conditions that make -1/12 unique. Regularity, linearity and invariance under inserting (infinitely many) 0s into the sequence. (The latter is a stronger condition than stability (assuming regularity and linearity) so in particular a number of "interesting" methods fail this)

The proof in the numberphile video becomes rigorous with those assumptions.

Yeah, this is the kind of stuff I'm talking about. In fact, my curiosity sparked from an article a guy from numberphile wrote: in the second sentence he claims that "[-1/12] is the only sensible value one can attach to this divergent sum": the rest of the text supports that -1/12 is a sensible value, but it's not obvious to me that this should be the only sensible one.

So with regularity, linearity and invariance under insertion of zeros, we can prove that -1/12 is the only sensible result? How can we prove this? What "interesting" methods fail invariance under insertion of zeros?

mike-l
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

Insertion of 0s (even finitely many) plus linearity and regularity imply stability

Bessel summation, for example, is not stable.

As for showing -1/12 follows from those three conditions,

Let S be 1+2+3+... And suppose our method sums this.
Let S1 be 0+1+0+2+0+3...
Then by invariance these sum to the same

S2 = S-4S1 sums by linearity, but this is 1-2+3-4...

Let S3 = 0 + 1 -2 +3 ... which sums the same as S2 by invariance

By linearity, S4 = S2 + S3 sums, and this is 1-1+1-...

By invariance, linearity, and regularity, 1-sum(S4) = sum(S4), whence sum(S4) = 1/2, sum(S2) = 1/4 and sum(S) = -1/12
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DR6
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

mike-l wrote:Insertion of 0s (even finitely many) plus linearity and regularity imply stability

Bessel summation, for example, is not stable.

As for showing -1/12 follows from those three conditions,

Let S be 1+2+3+... And suppose our method sums this.
Let S1 be 0+1+0+2+0+3...
Then by invariance these sum to the same

S2 = S-4S1 sums by linearity, but this is 1-2+3-4...

Let S3 = 0 + 1 -2 +3 ... which sums the same as S2 by invariance

By linearity, S4 = S2 + S3 sums, and this is 1-1+1-...

By invariance, linearity, and regularity, 1-sum(S4) = sum(S4), whence sum(S4) = 1/2, sum(S2) = 1/4 and sum(S) = -1/12

But linearity and stability are too strict to be able to sum 1+2+3+4+5...

Assuming stability and linearity:

S = 1+2+3+4+5... | definition
S1 = 0+1+2+3+4+5... | stability, we have sum(S) = sum(S1)
S2 = 1+0+0+0... | S - S1, aka linearity

S2 should be 1(because it's convergent to 1), but with linearity and stability we can get a proof for sum(S2) = sum(S) - sum(S1) = 0, so we can't use a summation method which uses both to sum this series. Proof from Wikipedia(yeah, I should have found this before asking here, bla bla, sorry).
I tried to make the proof more concise and fucked it up, go to wikipedia or scroll a bit further down to find the corrected version

Yet there is something I don't get about all summation methods for this series give the same result, -1/12. I take it no deeper reason behind this is known? Wikipedia doesn't seem to draw attention to this fact.
Last edited by DR6 on Thu Oct 16, 2014 5:44 pm UTC, edited 2 times in total.

mike-l
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

Fair enough. I was assuming that our method summed the series, so it turns out that that assumption is vacuous. (It is, though, of course true that all such methods give -1/12, just that there is no such method)
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Nicias
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

DR6 wrote:But linearity and stability are too strict to be able to sum 1+2+3+4+5...

Assuming stability and linearity:

S = 1+2+3+4+5... | definition
S1 = 0+1+2+3+4+5... | stability, we have sum(S) = sum(S1)
S2 = 1+0+0+0... | S - S1, aka linearity
....

Shouldn't S-S1 = 1+1+1+1+1.....?

mike-l
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

Yeah but that still isn't summable if you have stability, as it would sum to 1 + itself
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DR6
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

Nicias wrote:Shouldn't S-S1 = 1+1+1+1+1.....?

Yeah, actually I missed some steps. The actual proof from wikipedia is:
S = 1+2+3+4+5... | definition
S1 = 0+1+2+3+4+5... | stability, we have sum(S) = sum(S1)
S2 = S-S1 = 1+1+1+1+1+1 ... | by linearity, S2 = 0
S3 = 0+1+1+1+1+1... | stability, still 0
S4 = S2-S3 = 1+0+0+0... = 1 | this should actually be 0, so QED

mike-l wrote:Yeah but that still isn't summable if you have stability, as it would sum to 1 + itself

No, actually the proof was supposed to work with a weaker form of stability that only needs to be able to append 0s to the start of the sequence.

mike-l
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

But then you need regularity to say 1+0+0... = 1, and combined with linearity that gives full stability

Edit: If you only require stability under adding 0s, and linearity, then there is a trivial method to sum everything: always assign a sum of 0
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DR6
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

mike-l wrote:But then you need regularity to say 1+0+0... = 1, and combined with linearity that gives full stability

Edit: If you only require stability under adding 0s, and linearity, then there is a trivial method to sum everything: always assign a sum of 0

No, because one of the requirements for something to be a summation method is that, for all convergent series, it has to output the actual result. Otherwise it's just an arbitrary function on sequences. This is why 1+0+0+0... = 1, and why assigning 0 to everything doesn't work.

mike-l
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

That's called regularity

The point though is that I'd you have that and linearity, full stability and stability under prepending 0s are equivalent

Edit: Just to make this explicit, you said:

DR6 wrote:
mike-l wrote:Yeah but that still isn't summable if you have stability, as it would sum to 1 + itself

No, actually the proof was supposed to work with a weaker form of stability that only needs to be able to append 0s to the start of the sequence.

The proof requires linearity, and the proof requires that 1+0+0+... sums to 1.

If we assume those two facts, then, calling your 'weaker' property 0-stability, we have 0-stability if and only if stability.

Pf:

0-stability is a special case of stability, so one direction is obvious. Conversely, suppose a summation method is linear, sums 1+0+0+... to 1, and is 0-stable.

Suppose a_n is a sequence, and b_n is the same sequence starting one term later (ie it omits a_0). Let c_n be the sequence obtained by replacing a_0 with 0, or equivalently obtained by inserting a 0 at the start of b_n. Let d_n = a_0 + 0 + 0 + ....

Then by 0 stability, Sum(c_n) = Sum(b_n)
Since d_n = a_0 * (1+ 0 + 0 + ...), we have by linearity that d_n sums to a_0.
Since c_n + d_n = a_n, we have that a_n sums to a_0 + sum(b_n), which is the statement of stability.

Note that this doesn't even require full regularity, just finite regularity (ie if a sequence is 0 almost everywhere, it sums to the sum of its nonzero terms), which itself is a consequence of stability and linearity. This shows that even if you lift the regularity restriction and only require linearity and stability that you cannot sum 1+2+3+... (or 1+1+1+...)
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Boyd1
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

What is the integral of xe^x from -infinite to 0?

DR6
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

Boyd1 wrote:What is the integral of xe^x from -infinite to 0?

It's -1. What does that have to do with the thread?

Moole
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

mike-l wrote:Insertion of 0s (even finitely many) plus linearity and regularity imply stability

Bessel summation, for example, is not stable.

As for showing -1/12 follows from those three conditions,

Let S be 1+2+3+... And suppose our method sums this.
Let S1 be 0+1+0+2+0+3...
Then by invariance these sum to the same

S2 = S-4S1 sums by linearity, but this is 1-2+3-4...

Let S3 = 0 + 1 -2 +3 ... which sums the same as S2 by invariance

By linearity, S4 = S2 + S3 sums, and this is 1-1+1-...

By invariance, linearity, and regularity, 1-sum(S4) = sum(S4), whence sum(S4) = 1/2, sum(S2) = 1/4 and sum(S) = -1/12

That still doesn't prove that -1/12 is the only reasonable value; it could simply be that no such function sum(S) exists - that the requirements on it are contradictory and could yield different answers, and it's certainly far from obvious that this is not the case (especially were we allowed to add infinitely many zeros to a sequence. That just strikes me as asking for trouble)
Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.

mike-l
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

I never said it proved that it's the only reasonable answer. I said that these are conditions under which it's the only answer. And indeed, it turns out those conditions can't be satisfied. All of this is stated above.
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keed
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### about sum of natural numbers (lamer)

Sorry if this is not for here. Delete it.

I saw a video in youtube , prooving that the sum of all natural numbers is -1/12. (youtube:numberphile)

It was quite neat.
It is based on the proof that series:

S = 1 -1 + 1 -1 +1 -1 ... so on, is = 1/2

The proof was simple (youtube:numberphile)

But I have a trouble with it.

The proof was:

Saying S = 1 -1 +1 -1 +1 -1 ... (infinity)

So if we group the members in that way:

(1 -1) + (1 -1) + (1 -!) ...
We expect zero (it's not correct but anyway)

If we group members like this:

1 + (-1 +1) + (-1 +1) + (-1 +1) ...

the result would be 1 (incorrect again)

So we do this:

What if we try:
1- S

1-S = 1 - (1 -1 +1 -1 +1 -1 ...)

Which is:
1 -1 +1 -1 +1 -1 +1 ... (etc.)

The fact is that it is exactly S.
So:

1 -S = S

Therefore:
2S = 1

And:

S = 1/2

Everything is OK, but in fact it says that S = -S

- (1 -1 + 1 -1 +1 -1 +1 -1 ...) = -1 +1 -1 +1 -1 ...

It is the same because addition does not have order.

May be one cannot reorder infinite series (regarding the sum ) nevertheless addition properties.

So in 1 -S, it is not exactly addition of 1 and -S.
It is just 1- S, which is S.

I need help here.

I am sorry, it is lamer at all.

DR6
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### Re: about sum of natural numbers (lamer)

keed wrote:Sorry if this is not for here. Delete it.

I saw a video in youtube , prooving that the sum of all natural numbers is -1/12. (youtube:numberphile)

It was quite neat.
It is based on the proof that series:

S = 1 -1 + 1 -1 +1 -1 ... so on, is = 1/2

The proof was simple (youtube:numberphile)

But I have a trouble with it.

The proof was:

I need help here.

I am sorry, it is lamer at all.

Yeah, this is the point from which I was starting the thread. See, actually, that numberphile video was "lying". 1 + 2 + 3 + 4... does not actually sum to -1/12, and 1-1+1-1+1... does not actually sum to 1/2.

The sum of an infinite series is defined mathematically as the limit of the partial sums. That is, if S = 1 - 1 + 1 - 1 + 1, we can make partial sums like s1 = 1, s2 = 1+1, s3 = 1+1-1, s4 = 1+1-1+1... and if those converge to a value, that is the sum of S. Under this definition, neither of the series for which the numberphile video give a "sum" are actually summable: 1+2+3+4+5... grows unboundedly and 1-1+1-1+1... oscillates between 0 and 1 . Your manipulations don't work because not all series can be manipulated that way without altering the result. The series where there is a sum are called "convergent", and the others are "divergent".

The thing is that there are a lot of situations where "summing" one of those series does make sense. In that case, we talk about "summation methods": new ways of defining the sum of an infinite series which extend the normal sums, without any contradictions. For example, one such summation method would be calculating the limit of the average of those partial sums: this would give 1-1+1-1+1...=1/2. There is nothing guaranteeing that those summation methods should agree with each other.

It turns out that all the arithmetic manipulations that are done in the Numberphile video(which would otherwise be worthless) map directly to steps on a more complicated summation methods, which makes them mathematically justified.

Elmach
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### Re: Question about 1+2+3+4...=-1/12 (please hear me out)

Silly question:

Obviously, sum of r^n for nonnegative n is 1/1-r. (For abs r less than 1) (analytic continuation around one gives an answer though)
Less obviously, sum of r^n for negative n is 1/r-1. (For abs r greater than 1) (but analytic continuation again?)

So the sum of r^n for all integers n is 0 (not convergent anywhere). (But we subtract poles at one... Is this valid?)

Does this have any meaning?

edit by epiphany:

Take r one
We have doubly infinite sum of one is zero
So singly infinite sum of one is negative one half
Which is one possible answer (Ramanujan?)
Assuming you are rigorous enough

keed
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### Re: about sum of natural numbers (lamer)

DR6 wrote:
keed wrote:Sorry if this is not for here. Delete it.

I saw a video in youtube , prooving that the sum of all natural numbers is -1/12. (youtube:numberphile)

(...)
I am sorry, it is lamer at all.

Yeah, this is the point from which I was starting the thread. See, actually, that numberphile video was "lying". 1 + 2 + 3 + 4... does not actually sum to -1/12, and 1-1+1-1+1... does not actually sum to 1/2.

The sum of an infinite series is defined mathematically as the limit of the partial sums. That is, if S = 1 - 1 + 1 - 1 + 1, we can make partial sums like s1 = 1, s2 = 1+1, s3 = 1+1-1, s4 = 1+1-1+1... and if those converge to a value, that is the sum of S. Under this definition, neither of the series for which the numberphile video give a "sum" are actually summable: 1+2+3+4+5... grows unboundedly and 1-1+1-1+1... oscillates between 0 and 1 . Your manipulations don't work because not all series can be manipulated that way without altering the result. The series where there is a sum are called "convergent", and the others are "divergent".

The thing is that there are a lot of situations where "summing" one of those series does make sense. In that case, we talk about "summation methods": new ways of defining the sum of an infinite series which extend the normal sums, without any contradictions. For example, one such summation method would be calculating the limit of the average of those partial sums: this would give 1-1+1-1+1...=1/2. There is nothing guaranteeing that those summation methods should agree with each other.

It turns out that all the arithmetic manipulations that are done in the Numberphile video(which would otherwise be worthless) map directly to steps on a more complicated summation methods, which makes them mathematically justified.

Thank you very much for your answer. My post is full of things that are not extremely clever, but nevermind. This field is much broader and it is not just about numberphile’s videos. So I will say my lamer opinion as short as I can.
Of course it is about methods of summation and in this particular case – the sum of all natural numbers. It is known that proofs for it do not sustain linearity and stability at the same time. None the less I “believe” that they are true and give sufficiently correct results. (It sounds stupid, I know.)
You say that “The sum of an infinite series is defined mathematically as the limit of the partial sums” and that “ 1 + 2 + 3 + 4... does not actually sum to -1/12”.
When thinking about a limit in mathematical sense I see something like “for f(x), when x approaches something (it could be plus or minus infinity too) f(x) approaches L, and L is the limit”. But this doesn’t mean that the limit is always something very abstract. If f(x) = x then lim f(x) when x -> 5 is 5. But for this function f(5) is exactly 5. Let’s imagine another function: f(x) = 5 (a constant). Then
Lim f(x) when x-> [+infinity] is 5. I am sure (well, almost sure) that f(x) here is exactly 5.
So, of course 1 + 2 + 3 … is a divergent series. And the partial sums grow infinitely. I mean when we here grow limitless we get limitless big result. But when we have all the numbers of the series (i. e. all natural numbers) their real sum is -1/12. There are different divergent series and one could think that their sums all tend to infinity. But when you sum really all of their members you get different finite values. (This is not a generalization, just something like an example).
The series 1 + 1 + 1 + 1 + … = -1/2
(Ok, it’s Wikipedia.)
I personally am not going to write in this forum too often, I feel so.
Thanks for everything!

brenok
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### Re: about sum of natural numbers (lamer)

keed wrote:(...)
When thinking about a limit in mathematical sense I see something like “for f(x), when x approaches something (it could be plus or minus infinity too) f(x) approaches L, and L is the limit”. But this doesn’t mean that the limit is always something very abstract. If f(x) = x then lim f(x) when x -> 5 is 5. But for this function f(5) is exactly 5. Let’s imagine another function: f(x) = 5 (a constant). Then
Lim f(x) when x-> [+infinity] is 5. I am sure (well, almost sure) that f(x) here is exactly 5.
I don't see how that's relevant to the rest of the post.

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