Is my "reasoning" for Sum of Natural Numbers = -1/12 valid?

Hello everyone,

I appologise in advance if I am making a headache inducing error in my reasoning here. Feel free to ask for clarification if needed. Posting from phone makes it difficult to format and edit things nicely.

I was watching a Numberphile video on the proof that the sum of all Natural numbers equals -1/12 (Link: https://youtu.be/w-I6XTVZXww). I am fine with the proof itself. However reading the comments on everyone not fine with it, I had this little thought experiment pop into my head.

The main problem people had with the idea is that the sum is divergent with no limit, and that shifting the sequence was not allowed by the rule of maths. To this I say that addition doesn't care about the order things are added together, so shifting things around won't change the result.

However, my thought experiment, for the lack of a better term, aims to show that the value of the sum of all natural numbers is finite.

Let:

S_I = sum of all Integers
S_N = sum of all Natural numbers.

S_I can be written as

S_I = 0 + S_N +(-S_N ) or S_N - S_N

If S_N is undefined, that means that there is no solution to this problem. But that doesn't make intuitive sense as this can be written as S_I = (n-n) for all values of n.

Now that last step proves that S_I = 0, but doesn't it also show that S_N must be defined? Or can infinity - infinity be valid if the infinities itself are identical? That doesn't feel right with me.

If the infinite sum of the first series (1 - 1 + 1 - 1 + 1 - 1 + 1...) is in fact 1/2 as he claims, then the rest of the proof seems valid to me. The problem is that no part of mathematics that I am familiar with supports that first infinite sum. There's a link in the video that supposedly explains a proof of it, but I'm not currently interested enough to spend the time on it.

Quote:

---

Originally Posted by Douglas

If the infinite sum of the first series (1 - 1 + 1 - 1 + 1 - 1 + 1...) is in fact 1/2 as he claims, then the rest of the proof seems valid to me. The problem is that no part of mathematics that I am familiar with supports that first infinite sum. There's a link in the video that supposedly explains a proof of it, but I'm not currently interested enough to spend the time on it.

---

Short version, there are mathematics that let you "sum" non-converging series, and that particular series does work out to 1/2. There are some algebra tricks that get you to somewhere similar, but there are proper techniques that apply and they agree.

What breaks the symmetry between positive and negative numbers?

Quote:

---

Originally Posted by NichG

What breaks the symmetry between positive and negative numbers?

---

Natural numbers don't even include negative numbers.

Quote:

---

Originally Posted by Bavarian itP

Natural numbers don't even include negative numbers.

---

I meant in the 1-1+1-1+1-1... sum.

I looked into it more, and its basically a function of how the sum is written out. In divergent sums you lose things like associativity and commutativity. Writing it -1+1-1+1-1+1... would get you -1/2 by the same derivation, even though the full expression contains exactly the same terms.

Quote:

---

Originally Posted by georgie_leech

Short version, there are mathematics that let you "sum" non-converging series, and that particular series does work out to 1/2. There are some algebra tricks that get you to somewhere similar, but there are proper techniques that apply and they agree.

---

If you approach it like a series of numbers to be used one by one. When you're doing the sum 1-1+1-1+1-1+etc half the time your sum will be one, the other half of the time it will be zero, and thus the expected result when going to infinity is a half.

This seems to me like the wrong approach for determining the the average of all natural numbers. Granted, you can't just take all natural numbers, add them up and take the end result, because there are an infinite number of them, but you could at least take the same amount of positive as negative numbers, making the average zero.

In other words, I haven't seen the video yet, but it seems to be like you'd get the opposite result if you start your line of thinking from the sum -1+1-1+1-etc.

Then again, it's math, so it's probably perfectly true yet in this particular instance quite useless.

Quote:

---

Originally Posted by Lvl 2 Expert

If you approach it like a series of numbers to be used one by one. When you're doing the sum 1-1+1-1+1-1+etc half the time your sum will be one, the other half of the time it will be zero, and thus the expected result when going to infinity is a half.

This seems to me like the wrong approach for determining the the average of all natural numbers. Granted, you can't just take all natural numbers, add them up and take the end result, because there are an infinite number of them, but you could at least take the same amount of positive as negative numbers, making the average zero.

In other words, I haven't seen the video yet, but it seems to be like you'd get the opposite result if you start your line of thinking from the sum -1+1-1+1-etc.

Then again, it's math, so it's probably perfectly true yet in this particular instance quite useless.

---

The proper method involves finding the limit of the averages of the partial sums. But the method that demonstrates the 1/2 solution via algebra tricks is actually pretty straightforward. Let's call the sum of 1-1+1-1+1... S. Then let's ask what 1-S is. Well, that's 1-(1-1+1-1+1...), which is the same as 1-1+1-1+1... hey, that looks a lot like the initial series doesn't it? That means 1-S = S, which means 1 = 2S, which means S = 1/2.

Quote:

---

Originally Posted by Mith

The main problem people had with the idea is that the sum is divergent with no limit, and that shifting the sequence was not allowed by the rule of maths. To this I say that addition doesn't care about the order things are added together, so shifting things around won't change the result.

---

You're not just changing the order in which things are added - you're not adding the entire set to the entire set, a term is getting dropped from the second one. You can see this if you include the nth term* or just consider the Riemann series theorem at several points in the process.

*Video link.

Quote:

---

Originally Posted by georgie_leech

The proper method involves finding the limit of the averages of the partial sums.

---

I've read of their existence. But they go well beyond what I know.

Quote:

---

Originally Posted by georgie_leech

But the method that demonstrates the 1/2 solution via algebra tricks is actually pretty straightforward. Let's call the sum of 1-1+1-1+1... S. Then let's ask what 1-S is. Well, that's 1-(1-1+1-1+1...), which is the same as 1-1+1-1+1... hey, that looks a lot like the initial series doesn't it? That means 1-S = S, which means 1 = 2S, which means S = 1/2.

---

Yeah, with algebra tricks, there is the most obvious:

0=Sum(0) for n times=Sum(1-1) for n times=(1-1)+(1-1)+(1-1)...=1-1+1-1+1..=S
Therefore S=0

The problem is that algebra tricks are ill-defined in the "usual" math, if the series is not convergent and has no limit.

And in "usual" math this has no limit, according to the definition (which should be more or less as follows)

(lim S(n)=X for n-> ∞) ⇔ (∀ d > 0 Ǝ M > 0 / ∀ N > M ⇒ |S(N)-X| < d)

(I spent a good deal of time to find and copy all the unicode chars for the mathematical definition of limit, I hope they are visualized correctly, at least).

So, if I choose d=0.1, you must give me a point (M) from where whatever number N that is chosen (with N>M) the differences between the Nth-element of the series and the limit X is less than 0.1.

And this point doesn't exist for any value of X, for this series.

Quote:

---

Originally Posted by Dr.Zero

I've read of their existence. But they go well beyond what I know.

---

Quick version, you take the first partial sum, divide it by 1 (not really, but patterns.) You add up the first 2 partial sums, and divide that by 2. You add the first 3, and divide by 3, and so on. You can put those averages into a Series and and find the limit for that one. This method agrees with the usual methods for finding limits on converging series, but this works on some that don't converge. Like this one. The partial sum average things for this series work out to 1, 1/2, 2/3, 1/2, 3/5, 1/2.... and that series does converge, in this case on 1/2.

Quote:

---

Originally Posted by georgie_leech

Quick version, you take the first partial sum, divide it by 1 (not really, but patterns.) You add up the first 2 partial sums, and divide that by 2. You add the first 3, and divide by 3, and so on. You can put those averages into a Series and and find the limit for that one. This method agrees with the usual methods for finding limits on converging series, but this works on some that don't converge. Like this one. The partial sum average things for this series work out to 1, 1/2, 2/3, 1/2, 3/5, 1/2.... and that series does converge, in this case on 1/2.

---

I'm still on the side that something that eventually sends math in overflow (or, to say it in belkarish: "I'll sum you so much positives that you'll end up in negatives!") has some hidden problem, but this is a great explanation, thanks!

Oh man oh man. I need to bust out my old Functional Analysis book. In that class we generalized limits (and hence series). I forget the terminology.

The series consisting of the sum of natural number is divergent. Period. End of story. (Well, maybe not end of story, this is assuming you're using the accepted definitions of series and divergent... if you aren't, you really should clarify that.)

However, you can *think* of it as -1/12. The easiest way to do this is to use the Euler-Riemann zeta function. zeta(s)=sum of 1/n^s for all natural numbers n, for s>1. Thus if you were to define 1+2+3+4+... as something, a natural candidate would be zeta(-1), which is indeed -1/12.

Quote:

---

Originally Posted by Mith

The main problem people had with the idea is that the sum is divergent with no limit, and that shifting the sequence was not allowed by the rule of maths. To this I say that addition doesn't care about the order things are added together, so shifting things around won't change the result.

---

This is true for finite sums. It's even true for unconditionally convergent infinite sums. It is decidedly untrue for a lot of other infinite sums. In particular, if you have a convergent sum where one subsequence converges to infinity, and another converges to negative infinity, then it is possible to rearrange the terms such that they converge to literally any real number. Some infinite series that don't unconditionally converge have no issue with reordering. The sum of all natural numbers, for example, will always diverge to infinity, regardless of how you order the terms.

1-1+1..., however, is pretty easy to reorder such that you get a different result. Let's take every pair of terms, the first and second, the third and fourth, and so on, and swap them. Now, the sequence is -1+1-1... If you take the partial sums of this sequence, you get -1, then 0, then -1, then 0, and so on. At the very least, this is clearly diverging to two different places than the original sequence, and, if you apply the averaging method from the video (which isn't precisely a standard sum, but can be formalized in a way that is useful), then you get -1/2 where once you got 1/2. We're still hitting every term of the original sequence exactly once, but the result has changed completely.

Infinity is weird.

Quote:

---

However, my thought experiment, for the lack of a better term, aims to show that the value of the sum of all natural numbers is finite.

Let:

S_I = sum of all Integers
S_N = sum of all Natural numbers.

S_I can be written as

S_I = 0 + S_N +(-S_N ) or S_N - S_N

If S_N is undefined, that means that there is no solution to this problem. But that doesn't make intuitive sense as this can be written as S_I = (n-n) for all values of n.

Now that last step proves that S_I = 0, but doesn't it also show that S_N must be defined? Or can infinity - infinity be valid if the infinities itself are identical? That doesn't feel right with me.

---

This is circular logic, plain and simple. You assume that you can subtract these two sets from each other, and then use that to prove that these are the sort of sets you can subtract from each other. If S_N needed to be defined in order for you to do that subtraction, then you needed to prove it was defined before you subtracted, not after.

Gotta say, this is the video that made me really dislike Numberphile, though I do still enjoy some of their videos on occasion. It's a video that tries to make math mystical, instead of telling you why it all makes sense, which skips over the context in which these things work in favor of telling the lie that they just do, and promises you that there's no possible intuition for how this sum operates instead of putting the work in and providing the intuition.

Because, guess what, it is absolutely possible to provide that intuition. Here's a 22 minute long video by the incredible 3Blue1Brown about exactly what that result is derived from, how it functions in mathematics as a whole, and how this differs from standard summation. He tells you what mathematicians should be telling you. That this stuff, for all its complexities, makes sense.

Quote:

---

Originally Posted by Mith

Hello everyone,

However, my thought experiment, for the lack of a better term, aims to show that the value of the sum of all natural numbers is finite.

Let:

S_I = sum of all Integers
S_N = sum of all Natural numbers.

S_I can be written as

S_I = 0 + S_N +(-S_N ) or S_N - S_N

If S_N is undefined, that means that there is no solution to this problem. But that doesn't make intuitive sense as this can be written as S_I = (n-n) for all values of n.

Now that last step proves that S_I = 0, but doesn't it also show that S_N must be defined? Or can infinity - infinity be valid if the infinities itself are identical? That doesn't feel right with me.

---

You have to be very careful with infinities. Here the obvious approach would be to pair each number and show it's true for each range. S(-n,n)= S(0,n)-S(0,n) = 0

But for instance the same logic as above, introducing the sum of odd and even numbers (but only using them to partition the positive numbers) gives a different answer.
S_I = 0 + S_E + S_O - S_N = 2 (S_N) + 1+2 (S_N) + |N| - S_N > 3 S_N

Quote:

---

Originally Posted by eggynack

This is circular logic, plain and simple. You assume that you can subtract these two sets from each other, and then use that to prove that these are the sort of sets you can subtract from each other. If S_N needed to be defined in order for you to do that subtraction, then you needed to prove it was defined before you subtracted, not after.

---

Fair point. The reason why I was asking was because it seemed to easy, but at the time of asking, I wasn't seeing the flaw. I don't have as much experience with set series as I would like.

If I recall the video correctly, it isn't considered a valid proof, just a rough idea on on how to get to -1/12

Quote:

---

Originally Posted by Balain

If I recall the video correctly, it isn't considered a valid proof, just a rough idea on on how to get to -1/12

---

Do you mean it isn't considered a valid proof within the video, or outside of it? The latter is true, but I'm not so sure on the former. Especially because, y'know, the sum of natural numbers is not -1/12, meaning no proof could ever be valid.

I took some free time, so here we go:

Proof that S=1+2+...N is not -1/12


Disclaimer: Please notice that there are plenty of easy ways and considerations to prove that S is not -1/12.
This doesn't want to be easy, it wants to play the same game of the video: using algebra tricks and mixing series to prove that there is a very good reason if you cannot do that, if not under some specific circumstances.
Also, keep in mind that I'm very prone to mistakes due to lack of attention, so, meh, if you find something wrong which invalidates all of this, tell me without any worry for my feelings.
(I mean, something wrong aside the obvious fact that I'm working under wrong premises)


Let's say that we want to verify if there exist some possible definition of limit which can be used to compute the sum of all natural numbers and giving a finite value (namely, -1/12).

Let's define S=1+2+3+4+5...+N+...

Let's assume there is a definition fo limit which gives S=V (with V=-1/12)

Let's at this point define Se=Sx2
Thus S = 1 + 2 + 3 + 4 +5...
x 2
gives Se= 2 + 4 + 6 + 8 +10

Se is, indeed, the sum of all the even natural numbers (the 'e' stands for even)

Let's define
U= 1+ 1 + 1 +1 + 1

Now let's define So=Se-U
Thus
Se = 2 + 4 + 6 + 8 +10...
-U = -1 - 1 - 1 - 1 - 1
=
So= 1 + 3 + 5 + 7 +9 ('o' stands for odd)

Now we can notice that
Se = 2 + 4 + 6 + 8
+
So = 1 + 3 + 5 + 7

Which gives exactly S
Thus S=Se+So (the sum of all even and odds natural numbers, gives the sum of all natural numbers)


S= Se+So = Se+Se-U = 2xSe-U
But
Se = Sx2 -> S= 4xS-U -> 3S = U

But now, let's try to subtract U from S.

S = 1 + 2 + 3 + 4 +5...
-U=-1 - 1 - 1 - 1 - 1
S-U=0 + 1 + 2 + 3 + 4 ... = S
So:
S-U=S

But if S has a finite value then U=0
But 3S=U thus even S=0 which proves that S is not -1/12
(Morever Se=2xS thus even Se=0; and So=Se-U thus So=0 too)

Sums that do not converge on a single result have no meaning.

It doesn't matter how clever you are, sums that do not converge on a single result have no meaning.

Quote:

---

Originally Posted by Jay R

Sums that do not converge on a single result have no meaning.

It doesn't matter how clever you are, sums that do not converge on a single result have no meaning.

---

I wouldn't go that far. Sums that don't converge on a single result don't have this meaning, a sum as defined in the traditional sense. But they can have a bunch of other meanings. Senses in which they do converge, alternate definitions of sum, analytic continuations, all kindsa meanings that, despite not relying on the generic sense of sum, have meaning within mathematics. The -1/12 thing is about as far as you can get from a result derived from standard summing, but it's a real result that does things.

Quote:

---

Originally Posted by eggynack

but it's a real result that does things.

---

What things?
So far we've seen that it and it's negative add up to zero, which is also (possibly) true for the series.

Quote:

---

Originally Posted by jayem

What things?

---

Well, the specifics are outside my pay grade, but my understanding is that the Riemann-Zeta function, which is fundamentally based on analytic continuation and the source for this particular result, has absolute tons of utility. There's a few uses listed in the wikipedia article, for whatever that's worth. And, beyond that, while complex analysis is not something I've studied (which is why the specifics were outside my pay grade in the first place), I'm taken to understand that analytic continuation pops up a lot in the field and is super important and useful. Analytic continuation is all about getting results where the function itself would be undefined, so any function defined by infinite series like this, undefined over some part of the range, and allowing for an analytic continuation over that range, would be another example.

Quote:

---

So far we've seen that it and it's negative add up to zero, which is also (possibly) true for the series.

---

Not all that sure that the sum of all naturals adds to the sum of all negative naturals to give you zero. While I disagree with Jay R's contention that infinite series that diverge are totally useless, that doesn't mean you should go about subtracting them willy nilly, even if the result seems intuitively obvious. We could maybe create a really well specified sense in which this subtraction of divergent series works, but even then we would only have the result for this strange pseudo-subtraction. Infinite series are not unworkable, but you gotta be super careful. Lack of care is how you get stuff like this video, where what are essentially lies about math get spread out to a surprisingly large audience.

Oh I fully agree with a-c in general. It was the -1/12 case that I was wondering about.

And yes I wasn't certain about opposite series canceling, and fairly sure you could have similar situations when they apparently don't.
Though I'm tempted to say, if there was a system that worked that way, all else being equal that would be a better system, and even if things went a bit messy elsewhere... (however that's just a feeling)

Mathologer put up a new video on this topic today: https://www.youtube.com/watch?v=YuIIjLr6vUA

Quote:

---

Originally Posted by jayem

Oh I fully agree with a-c in general. It was the -1/12 case that I was wondering about.

---

Far as I can tell, there's nothing all that special about the -1/12 case that makes it more or less accurate. Raise every single term in the series to 2n, where n is some integer, and the result, going by this method, becomes zero.

Quote:

---

And yes I wasn't certain about opposite series canceling, and fairly sure you could have similar situations when they apparently don't.
Though I'm tempted to say, if there was a system that worked that way, all else being equal that would be a better system, and even if things went a bit messy elsewhere... (however that's just a feeling)

---

It's not so much about having a single system, and then in that system this is the result or not the result. Standard subtraction is a well defined thing, and I'm pretty sure it's not defined for the subtraction of one divergent series from another. Instead, as with most things in math, it's about coming up with axioms that make sense and seeing what happens. So, if you want to add the negative integers to the positive integers and have it be zero, then we should create some new form of subtraction, clearly define how it operates in all circumstances, and then be clear when we're using this form of subtraction as opposed to the generic form. Maybe the results from this form of subtraction will be useful, or maybe not, but the first step is trying it out and seeing what happens.

Quote:

---

Originally Posted by eggynack

Far as I can tell, there's nothing all that special about the -1/12 case that makes it more or less accurate. Raise every single term in the series to 2n, where n is some integer, and the result, going by this method, becomes zero.

---

There is. Essentially you can define a function F(n) where n is the number of terms summed, then fit a continuous curve through all points (F(n),n), then use symmetry to extend the curve into the negative region even though the series itself makes no sense whatsoever there. -1/12 is a significant term on that extended curve.

This is a totally different idea than "the sum of all natural numbers" though.

Quote:

---

Originally Posted by Knaight

There is. Essentially you can define a function F(n) where n is the number of terms summed, then fit a continuous curve through all points (F(n),n), then use symmetry to extend the curve into the negative region even though the series itself makes no sense whatsoever there. -1/12 is a significant term on that extended curve.

This is a totally different idea than "the sum of all natural numbers" though.

---

I'm aware of the meaning of that result. I'm saying that that particular result doesn't seem more significant than other results gained through the exact same method. For example, that 1+4+9+16...=0.