The Sum of All the Natural Numbers

[Office_Shredder]

It just seems silly to me to say "We don't know what the last digit of 1-1+1-1+1... is so we'll just go in the middle and say the sum is a half..." Surely that sum should be undefined? If we don't know what the last digit is then we can't just say "oh what the hell we'll take the average and call that the answer..."

The standard definition of an infinite sum gives that this sum is divergent, so you cannot attach a number to it. There are other ways to define the value of an infinite series that gives 1/2 but when you're in this region you have to be very careful that the manipulations that you do (like lining up two series and canceling terms) are actually valid.

I don't know how often this sort of stuff is accepted in math but it just seems like extremely bad mathematics. It might work in equations ect but that doesn't mean to say that it actually does equal a half. Just like 0.999 recurring doesn't actually equal 1 because somewhere from thin air you're getting a extra 1...

If 0.999 recurring equals 1 then 0.34999 reucurring equals 0.35 I think it's just for sake of keeping things neat? I don't know I'm not nearly experienced enough to comment lol but I just thought I'd share why I think it's incorrect to say such things.

Actually, .9999... repeating is equal to exactly 1, there is no extra 1 coming from anywhere. In the exact same way .3499999 is exactly equal to .35. There is an excellent post on the forum in the Math FAQ in which all the mathematical rigor involved in the statement .999999... = 1 is presented

https://www.physicsforums.com/showthread.php?t=507002

 

[jbunniii]

It just seems silly to me to say "We don't know what the last digit of 1-1+1-1+1... is so we'll just go in the middle and say the sum is a half..." Surely that sum should be undefined? If we don't know what the last digit is then we can't just say "oh what the hell we'll take the average and call that the answer..."

I don't know how often this sort of stuff is accepted in math but it just seems like extremely bad mathematics.

I don't see anyone advocating this. Of course the sum as written is undefined, and it is not "accepted in math." However, as LcKurtz noted in post #23, there are more generalized notions of summation in which we use averaging. These generalized sums give the same answer as an ordinary sum when the ordinary sum converges, but they can also assign values in other divergent cases such as the example you listed.

In examples like $1 -1 + 1 - 1 \ldots$, the idea may seem hokey and rather pointless, but it's actually a very valuable technique in analysis: for example, it can allow us to recover a function from its Fourier series in some instances where that series does not converge.

It might work in equations ect but that doesn't mean to say that it actually does equal a half.

Correct. The series does not converge. But the series of averages (more formally known as the Cesaro summation) DOES converge to 1/2, and that can be a useful concept as well.

Just like 0.999 recurring doesn't actually equal 1 because somewhere from thin air you're getting a extra 1...

Unfortunately, here you are simply wrong. See the FAQ:

https://www.physicsforums.com/showthread.php?t=507002

 

[zoobyshoe]

I don't know if this will shed light on this or obscure it further, but I just finished the novel The Indian Clerk which is about Ramanujan. That the sum of all natural numbers can be expressed as -1/12 is one of the first things he asserts in a letter to British mathematician G.H. Hardy. This confuses Hardy and his collaborator, Littlewood.

The next letter from Ramanujan clarifies it (to their satisfaction). Hardy explains to Littlewood:

"Essentially, it's a matter of notation. His is very peculiar. Let's say you decide you want to write 1/2 as 2^-1. Perfectly valid, if a little obscurist. Well, what he's doing here is writing 1/2^-1 as 1/1/2, or 2. And then, along the same lines, he writes the sequence 1 + 1/2^-1 + 1/3^-1 + 1/4^-1 +...as 1/1/1 + 1/1/2 + 1/1/3 + 1/1/4+..., which is of course, 1 + 2 + 3 + 4 +...So what he's really saying is 1 + 1/2^-1 + 1/3^-1 + 1/4^-1 +...= -1/12."

"Which is the Riemannian calculation for the zeta function fed with -1."

Hardy nods. "Only I don't think he even knows it's the zeta function. I think he came up with it on his own."

I, myself, don't know what any of that means, but it might make sense to some of you. (What I understood was, "Essentially, it's a matter of notation.")

 

[1MileCrash]

It just seems silly to me to say "We don't know what the last digit of 1-1+1-1+1... is so we'll just go in the middle and say the sum is a half..." Surely that sum should be undefined? If we don't know what the last digit is then we can't just say "oh what the hell we'll take the average and call that the answer..."

I don't know how often this sort of stuff is accepted in math but it just seems like extremely bad mathematics. It might work in equations ect but that doesn't mean to say that it actually does equal a half. Just like 0.999 recurring doesn't actually equal 1 because somewhere from thin air you're getting a extra 1...

If 0.999 recurring equals 1 then 0.34999 reucurring equals 0.35 I think it's just for sake of keeping things neat? I don't know I'm not nearly experienced enough to comment lol but I just thought I'd share why I think it's incorrect to say such things.

Oh my goodness!

Statements like "we don't know the last digit of an infinite series" and denying a very simple and obvious equality demonstrate a profound misunderstanding of infinity.

1- 1 + 1 -... does not converge to a value, but the assignment of one half is not meaningless, it has nothing to do with "not knowing the last digit" (because there is no such thing as the last digit, it's like saying we don't know what color the series is )

0.99.. = 1 is trivially and obviously true, it is not even worth talking about among mathematicians.

 

[FlexGunship]

Flex, consider the following. In the second video you posted in General Discussion they showed
$$S_1 = 1 - 1 + 1 - 1 + 1 -1... = 1/2$$
$$S_2 = 1-2+3-4+5-6... = 1/4$$From here, consider
$$S_2 + S_2$$
in the following way: Take the second S₂ and bump it over by three spaces when lining it up under the first, so you get after canceling vertically
1 -2 + 3 - 4 + 5 -6...
_________1 - 2 + 3...
$$1 -2 +3 -3 -3 -3 -3... = 1/2$$
$$2 -3(1+1+1+1+...) = 1/2$$
solving gives
$$1+1+1+1+... = 1/2$$
Now consider
1+2+3+4+... - (1+1+1+1+...) by canceling the first, second, third etc. terms to get
0+1+2+3+4+...
From which we have proven that
S - 1/2 = S
So -1/2 = 0. Doing only the same thing they were doing in the video, only very slightly differently (aligning the $S_2+S_2$ step two steps further to the right).

Again, I'm not a mathematician, but I did the same calculation and it resolved consistently. I believe your error was incorrectly signing the series; I added placeholder zeros to show that you missed the correct offset initially and thereby changed the series:

You wrote this:
$+(1 - 2 + 3 - 4 + 5 - ...)=\dfrac{1}{4}$
$+(0 - 0 + 0 + 1 - 2 + 3 - 4 + ...)=\dfrac{1}{4}$
Which has two consecutive terms of addition and changes the series. (I think it's the same as multiplying by -1? Hadn't gotten that far.)

What you wanted to do was this:
$+(1 - 2 + 3 - 4 + 5 - 6 + ...)=\dfrac{1}{4}$
$+(0 - 0 + 1 - 2 + 3 - 4 + ...)=\dfrac{1}{4}$
______________________________
$+(1 - 2 + (4 - 6 + 8 -10 +...))=\dfrac{1}{2}$
Which simplifies to (notice the important sign change):
$1-2-2(-2+3-4+5-...)=\dfrac{1}{2}$

Move the "-2" inside the series by factoring a "-2" to create the leading "+1" in our series:
$1-2(1-2+3-4+5-...)=\dfrac{1}{2}$

You end up with:
$1 - 2(S_2)=\dfrac{1}{2}$

Which gets your back to the same identity as before:

$\dfrac{1}{2}=2(S_2)$ → $S_2 = \dfrac{1}{4}$

I mean this sincerely, I could be in the wrong on all of this... but this just seems like a case of being diligent and meticulous.

If I re-run your calculation to try to get your result (single value offset emulating $S_1$):
$+ (1-2+3-4+5-...)=\dfrac{1}{4}$
$- (0-1+2-3+4-5+...)=-(0-\dfrac{1}{4})=-(-\dfrac{1}{4})$
______________________________
$(1 - 1 + 1 - 1 + 1 - 1+...) = \dfrac{1}{4}--\dfrac{1}{4} = \dfrac{1}{2}$Writing it all on one line steps for simplicity:
$S_2+S_2 = (1-2+3-4+5-...) + (1-2+3-4+5-...)$
$S_2+S_2 = (1-2+3-4+5-...) - (0-1+2-3+4-5+...)$
$S_2+S_2 = (1-1+1-1+1-1+...)$
$S_2+S_2 = \dfrac{1}{2}=S_1$

 

[FlexGunship]

Writing it all on one line steps for simplicity:
$S_2+S_2 = (1-2+3-4+5-...) + (1-2+3-4+5-...)$
$S_2+S_2 = (1-2+3-4+5-...) - (0-1+2-3+4-5+...)$
$S_2+S_2 = (1-1+1-1+1-1+...)$
$S_2+S_2 = \dfrac{1}{2}=S_1$

You can also bump it over farther (as you did):
$S_2+S_2 = (1-2+3-4+5-...) + (1-2+3-4+5-...)$
$S_2+S_2 = (1-2+3-4+5-...) - (0-0+0-1+2-3+4-5+...)$
$S_2+S_2 = (1-2+3-3+3-3+...)$
$S_2+S_2 = 1-2+3(1-1+1-1+...)$
$S_2+S_2 = 1-2+3(S_1)$
$S_2+S_2 = 1-2+3(\dfrac{1}{2})$
$S_2+S_2 = -1+\dfrac{3}{2}$
$S_2+S_2 = \dfrac{1}{2}=S_1$
If you recall:
$S_1=1-1+1-1+1-...$
Same rule applies if you want to get your original identity:
$S_1+S_1=(1-1+1-1+1-...)+(1-1+1-1+1-...)$
$S_1+S_1=(1-1+1-1+1-...)-(0-1+1-1+1-...)$
$S_1+S_1=1-0+0-0+0-...$
$S_1+S_1=1$
$2S_1=1$
$S_1=\dfrac{1}{2}$****EDITED****

From here, consider
$$S_2 + S_2$$
in the following way: Take the second S₂ and bump it over by three spaces when lining it up under the first, so you get after canceling vertically
1 -2 + 3 - 4 + 5 -6...
_________1 - 2 + 3...
$$1 -2 +3 -3 -3 -3 -3... = 1/2$$

Actually, I take it back... this original post is just wrong. (5-2) = +3. The fifth term in your series has the wrong sign, and from there you've incorrectly generalized the pattern.

$1-2+3-4+5-6+7-8+...$
$0+0+0+1-2+3-4+5-...$
-------------------------------------
$1-2+3-3+3-3+3-3+...$

$1-2+3(1-1+1-1+1-1+...)=\dfrac{1}{2}$ <-------- This is correct
$-1+\dfrac{3}{2}=\dfrac{1}{2}$
$\dfrac{1}{2}=\dfrac{1}{2}$

 

[Curious3141]

1- 1 + 1 -... does not converge to a value, but the assignment of one half is not meaningless, it has nothing to do with "not knowing the last digit" (because there is no such thing as the last digit, it's like saying we don't know what color the series is )

A bit OT, but I like the image from a Tool video (is it Sober?) in your avatar. Tool has a song (well, more of an interlude, really) called "Cesaro Summability" which is apropos here.

 

[1MileCrash]

A bit OT, but I like the image from a Tool video (is it Sober?) in your avatar. Tool has a song (well, more of an interlude, really) called "Cesaro Summability" which is apropos here.

Wow! I never knew that. Very appropriate.

And yes, it's from Sober. I just can't find a song to knock that out of my #1 slot.

 

[Office_Shredder]

Flex, you're I screwed that up let's try this again. If
$$S = 1+2+3+4...$$
is a number that we can do operations on, then S-S = 0
1 + 2 + 3 + 4 +5...
...- 1 - 2 - 3 - 4...
= 1+1+1+1+... = 0
OK, great. Let 1+1+1+... = T. So we have T = 0, and T-1 is
1+1+1 + 1 +...
-1
=
1+1+1+1+... = T

So T-1=T which gives -1 = 0

 

[consciousness]

I don't know if this will shed light on this or obscure it further, but I just finished the novel The Indian Clerk which is about Ramanujan. That the sum of all natural numbers can be expressed as -1/12 is one of the first things he asserts in a letter to British mathematician G.H. Hardy. This confuses Hardy and his collaborator, Littlewood.

I, myself, don't know what any of that means, but it might make sense to some of you. (What I understood was, "Essentially, it's a matter of notation.")

A Google search led me to this-

Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a sum to infinite divergent series. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties which make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined.

See http://en.wikipedia.org/wiki/Ramanujan_summation

I can't understand any of this stuff.

 

[Curious3141]

A Google search led me to this-

Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a sum to infinite divergent series. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties which make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined.

See http://en.wikipedia.org/wiki/Ramanujan_summation

I can't understand any of this stuff.

There are basically quite a few summation techniques that allow a divergent series to be "assigned" to a single finite "sum". Examples include the Cesaro, Abel, Borel, Ramanujan summations and Zeta function regularisation. They're not sums in the "usual" fashion (a divergent series does not have a defined sum), but the "sums" they arrive at are useful in certain areas of advanced math and physics (including String Theory).

 

[consciousness]

There are basically quite a few summation techniques that allow a divergent series to be "assigned" to a single finite "sum". Examples include the Cesaro, Abel, Borel, Ramanujan summations and Zeta function regularisation. They're not sums in the "usual" fashion (a divergent series does not have a defined sum), but the "sums" they arrive at are useful in certain areas of advanced math and physics (including String Theory).

Yes, but isn't the assigned sum incorrect? I will perhaps see the significance of these sums when I study more advanced topics.

 

[Curious3141]

Yes, but isn't the assigned sum incorrect? I will perhaps see the significance of these sums when I study more advanced topics.

They're meaningless in a "usual" sense, but they're apparently quite useful when certain problems involve divergent infinite sums. I have no personal experience with such problems, unfortunately, so this is the limit of my knowledge.

 

[dipole]

First of all, why are two physics professors doing a video about pure mathematics? One of which is clearly unqualified to do so...

When you analytically continue a function, there's no gaurentee that it will have the same form outside the region of regular convergence, When the Reiman-Zeta function is continued beyond the positive real axis, it no longer can be represented by the sum $\sum_{n=1}^{\infty} \frac{1}{n^s}$. So yea, $\zeta(-1) = -1/12$ says nothing about the sum of that series.

I agree with Office and mfb, these guys are manipulating infinite sums in very obviosiously incorrect ways to arrive at false conclusions. You could probably make those sums converge to almost anything using their rules of manipulation. I would like to see a convincing physical example, besides the Casimir effect - which already involved the dubious idea of subtracting infinities from each other, so who knows exactly in what sense this sum appears, and string theory which should never be used a physical justificiation IMO.

If someone can provide one physical example where the physics is clear and experimentally demonstrable, I'll be satisfied, but they won't be able to, because any child could tell you that to claim $1 + 2 + 3 + 4 + \cdots = -1/12$ in the normal sense of summation, is absolute nonsense.

 

[Office_Shredder]

Yes, but isn't the assigned sum incorrect? I will perhaps see the significance of these sums when I study more advanced topics.

As an example of why assigning numbers to sums without caring too much about what the number is (just that it's consistent), there is an object called a multiple zeta value
$$\zeta(s_1,...,s_k) = \sum_{n_1 > ... > n_k \geq 1} \frac{1}{n_1^{s_1}...n_k^{s_k}}$$

These satisfy a bunch of relationships called the shuffe and quasi-shuffle relations for multiplying them together. The relations can be described entirely combinatorially in the input (take the numbers $s_1,...,s_k$ and shuffle them around and add them together in very well defined ways). Once you have these relations, the following things appear to be true:
$$\zeta(2) \zeta(1) = \zeta(2,1) + \zeta(1,2) + \zeta(3)$$
and
$$\zeta(2) \zeta(1) = \zeta(1,2) + 2 \zeta(2,1).$$
When you combine these equalities together you get the relationship
$$\zeta(2,1) = \zeta(3).$$

This is a mathematically true statement and a very surprising result at first glance. The only problem with what I wrote above is that a multiple zeta value requires that s₁ be at least 2, and each other s_i be at least 1 to converge. So $\zeta(1)$ and $\zeta(1,2)$ don't converge in the first place, and what I wrote above to derive the true statement is just gobbledygook. So the question is how do I invent some way of attaching numbers to these series so that the above math works - the final calculation is completely independent of the numbers I attach to the divergent series, I just need to know that I can attach a number to them and do manipulations in a way that is mathematically consistent.

There are other examples I am sure (this is just the one I know of off the top of my head) where you have divergent series that you want to do calculations with, not because you care about what number they are equal to but because they just show up as an intermediate step in something bigger you are interested in. Often you won't be able to do anything and just have to find another way around, but if you have a good way of consistently attaching a number to what appears to be a divergent series then you can save yourself some hassle.

Note that when I say "consistently attaching a number" I mean you can do things like add series and multiply series and you are guaranteed that the numbers you get will add and multiply together correctly, and other things like that.

 

[mfb]

@Office_Shredder and FlexGunship: See post 14, where I got a contradiction with the same methods used in the video.

 

[FlexGunship]

FYI, I've been away for the last couple of days. I'm still looking at this. I'min way over my head, but it still makes for a fun intellectual challenge.

My latest notebook scribbling seem to imply fundamentally different behaviors for infinite sums that have alternating signs and sums that don't. This leads me to think that sums with non-alternating signs have some complex component to them that is not being written explicitly. This complex component becomes obvious when you "shift by 1 term". Certainly not a valid argument (you can always say, "fine, that works with complex terms, but now do the sum without them"), but it sure makes for some good brain exercise.

However, if you accept that a "shift by 1" doesn't work, and try a "shift by 2"...

$S = +1+2+3+4+5+...$
$S = -0-0-1-2-3-4-5-...$
--------------------------------------
$0 = +1+2+2+2+2+2+...$

$0 = 1 + 2(T)$
$-\dfrac{1}{2}=T$

This result might also be arbitrary (or arbitrarily wrong), but at least it matches:

$S_1 = 1-1+1-1+1-1+... = \dfrac{1}{2}$ via the following:

$+(T = 1+1+1+1+1+1+... = -1/2)$
$-(S_1 = 1-1+1-1+1-1+... = 1/2)$
---------------------------------------
$T-S_1 = 0+2+0+2+0+2+0+2-... = -1$
$T-S_1 = 2(1+1+1+1+1+1+1+...) = -1$
$T-S_1 = 2T=-1$
So, again: $T=-\dfrac{1}{2}$

The fundamental problem is that, in both cases, you have to "pretend" there are intervening complex terms in non-alternating infinite sums and just arbitrarily avoid mixing them.

And even more problematically... I have no idea what the "generally accepted" sum of 1+1+1+1+1+... is because I don't know how to enter it into Wolfram-Alpha and there's no Wikipedia page on it... and -1/2 doesn't make an iota of sense.

 

[atyy]

For those interested in the abuse of mathematics, here's David Tong's string theory notes. The relevant pages to read are his p39-40 and p85. http://www.damtp.cam.ac.uk/user/tong/string.html

 

[marcusl]

E. T. Jaynes addresses exactly this sort of mis-summation of infinite series in his book Probability Theory: The Logic of Science. Section 15.2 opens this way:

As a kind of introduction to fallacious reasoning with infinite sets, we recall an old parlor game by which you can prove that any given infinite series $S=\sum_i a_i$ converges to any number x that your victim chooses.

So we aren't limited to -1/12; he goes on to show how to sum it to anything you want. He identifies the crux of the fallacy thus:

Apply the ordinary processes of arithmetic and analysis only to expressions with a finite number n of terms. Then after the calculation is done, observe how the resulting finite expressions behave as n increases indefinitely. Put more succinctly, passage to a limit should always be the last operation, not the first. In case of doubt, this is the only safe way to proceed.

The jokers in the video violate this by manipulating and "evaluating" infinite sums without considering the proper limiting process.

 

[FlexGunship]

For those interested in the abuse of mathematics, here's David Tong's string theory notes. The relevant pages to read are his p39-40 and p85. http://www.damtp.cam.ac.uk/user/tong/string.html

So, this strikes me as representative of the discussion thus far.

I read p39-40, and I'm okay with it. I don't know zeta function regularization, but I accept that everyone is getting -1/12 out of this particular operation, and that everyone acknowledges that this is famously unconvincing. Fine.

However, the implication is that anyone working in the field of string theory must first dispose of common mathematical practice prior to "getting anything done" yet some of the giants of mathematics (like Euler, Riemann, and Ramanujan) seem to concur on this particular point.

But there's a genuine irony here. Physics is applied mathematics. Specifically, it is the application of mathematics to our observation of reality. Often, physics calls upon mathematics to formalize and/or generalize an observation. But in this ONE area, the discussion proceeds like this:

Physics: "Hey, math!"
Math: "What?"
Physics: "I've got this thing. It looks like infinity but, uh, it needs to be -1/12. I know that's crazy and random... but..."
Math: "Yeah, I have one of those."
Physics: "REALLY?!"
Math: "Yeah."
Physics: "Well, let me use it!"
Math: "Naw, it's just a trick I know."
Physics: "Oh..."
Math: "Well, actually, it's like... three tricks I know... super important tricks!"
Physics: "Oh?!"
Math: "But you still can't use it."

 

[D H]

I don't see anyone advocating this.

Euler, Hardy, Ramanujan advocated exactly this. More recently, here's Terrance Tao advocating this: http://terrytao.wordpress.com/2010/...tion-and-real-variable-analytic-continuation/

That said, the youtube video cited in the original post is an example of "physicists doing math (badly)". To see "mathematicians doing math (formally)", see the link that I posted. Or see the other links where $1+2+3+4+\ldots = -1/12$ is formally defined via zeta function regularization. That's an example of "physicists doing math (correctly, surprisingly)".

What you cannot do is manipulate those formal sums on a term-by-term basis. Do that and it's easy to arrive at contradictions. Unfortunately, that's exactly what was done in video cited in the OP, hence my label "physicists doing math (badly)." Just because their manipulations happened to arrive at the formal result does not mean that what they did was valid. It isn't.

 

[atyy]

So we have $1 + 2 + 3 + ... = -1/12$ being made sensible by the Riemann zeta function $\zeta(-1)=-1/12$.

How about the starting point in the video $1 - 1 + 1 -1 + ... = 1/2$ ? After some googling, I found the the Dirichlet eta function $\eta(0) = 1/2$. Wikipedia calls this the Abel sum of Grandi's series. Is that the right notion here?

Edit: Looking at D H's link in his post #56 to Terry Tao's article, it looks like it is.

 

[atyy]

That said, the youtube video cited in the original post is an example of "physicists doing math (badly)".

I just came across the Cesaro sum http://en.wikipedia.org/wiki/Summation_of_Grandi's_series which Wikipedia says is "rigourous", and seems very close to what was being presented in the OP's video.

 

[D H]

Nope. Everything after the 2:50 mark in that video is invalid. You can't do that with conditionally convergent series, let alone divergent ones.

 

[Office_Shredder]

atyy, Cesaro summation is a rigorous method of assigning values to series. The series 1+2+3+... is not Cesaro summable though, so it doesn't help with the original problem. It is used in some areas to great effect; for example partial sums of Fourier series can be quite bad on continuous functions even (diverging everywhere), but the Cesaro sum ends up converging for functions that are way worse than continuous. It ends up that when you want to prove results about Fourier transforms it's a lot better to consider the Cesaro sum of the Fourier series rather than the regular sum.

Notice how the wikipedia article then goes on in a later section to describe a whole host of issues in which inserting zeroes into sums can change the Cesaro sum, so when doing Cesaro summation you have to be extremely careful that you are being actually rigorous, and not just taking the word rigorous and slapping it onto a bad argument.

 

[Dadface]

Look at how he adds S2 to S2. He writes one series of numbers (from 1 to 7 in the example given)and writes an identical second series of numbers which he places below the first series.He then pushes one of the two series along by one and concludes that the sum is given by 1-1+1-1 etc .The sum, however, is not given by that. In the example given where the last digit equals 7 the sum is equal to 8. He ignored the seven at the end of the pushed along series. This number 7 was not overlapping with any other number.
When the non overlapping number at the end of the series is taken into account one can see that 2s2 approaches infinity as the number of numbers in each series approaches infinity (plus or minus infinity depending on the number of digits). 2s2 has numerical values which increase with the length of the series as follows:
+2,-2,+4,-4,+6,-6, etc

 

[atyy]

atyy, Cesaro summation is a rigorous method of assigning values to series. The series 1+2+3+... is not Cesaro summable though, so it doesn't help with the original problem. It is used in some areas to great effect; for example partial sums of Fourier series can be quite bad on continuous functions even (diverging everywhere), but the Cesaro sum ends up converging for functions that are way worse than continuous. It ends up that when you want to prove results about Fourier transforms it's a lot better to consider the Cesaro sum of the Fourier series rather than the regular sum.

Notice how the wikipedia article then goes on in a later section to describe a whole host of issues in which inserting zeroes into sums can change the Cesaro sum, so when doing Cesaro summation you have to be extremely careful that you are being actually rigorous, and not just taking the word rigorous and slapping it onto a bad argument.

Is it right to say that 1+2+3+... is also not Abel summable?

So one really has to use zeta regularization, or the smooth cut-off mentioned in the link that D H gave in #56?

 

[pwsnafu]

Is it right to say that 1+2+3+... is also not Abel summable?

Correct. It is mentioned in Wikipedia page as well.

 

[1MileCrash]

'After taking a course in mathematical physics, I wanted to know the real difference between Mathematicians and Physicists. A professor friend told me "A Physicist is someone who averages the first 3 terms of a divergent series"'

I thought you'd all enjoy that one.

 

[AlfieD]

'After taking a course in mathematical physics, I wanted to know the real difference between Mathematicians and Physicists. A professor friend told me "A Physicist is someone who averages the first 3 terms of a divergent series"'

Hmmm, interesting!

 

[pjantoniopj]

Here's where i saw the first mistake with Numberphile;

Given:
S1 = 1 - 1 + 1 - 1 + 1 - 1 + ...

Numberphile folks concluded that they could not just do the regular math (addition, subtraction) to obtain S1 = 0.
So they introduced their next step; add S1 to S1(shifted to the right)

S1 = 1 - 1 + 1 - 1 + 1 - 1 + ...
S1 = 1 - 1 + 1 - 1 + 1 - 1 + ...

And then, magically, they concluded that Yes, now we can to regular math here. They proceeded to add and subtract in a vertical direction.
Wait, what? You couldn't add and subtract horizontally but you can add and subtract vertically?

If you can add 1+0, then -1+1, then 1-1, then -1+1, etc., FOREVER, then why couldn't you do that in the first place with just S1?

 

[cryptist]

Algebraic manipulations on the video may be wrong, however, it is TRUE that finite and physically useful results can be obtained from divergent sums. In fact, mathematicians themselves are using it. There is a technique called analytic continuation that provides an extention of the domains of analytic functions. Note that, these odd results also consistent within the complex analysis. These divergent sums do not have contradictory finite results, they are actually valid!

Secondly, Casimir effect is a physical phenomenon that experimentally verified and well understood. In these effect, infinite summation of frequencies gives Zeta[-3]=1/120 where Zeta[-3] is a divergent sum and this result is verified experimentally. So in nature infinite summation of some "things" can REALLY give a finite result, which is consistent with mathematics.