Talk:1 + 2 + 3 + 4 + ⋯/Archive 2

This is an archive of past discussions about 1 + 2 + 3 + 4 + ⋯. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Archive 1

Archive 2

The harmonic series, for example

The harmonic series

$\sum _{n=1}^{\infty }{\frac {1}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots \!=\infty$

Using the comparison test it is easy to see that each term in the series below is greater than or equal to corresponding term of the harmonic series

$1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots \!$

1 + 2 + 3 + 4 + 5 + $\cdots \!$

It follows that the sum of the second series must be infinite as well

$1+2+3+4+5\cdots =\infty$

The above proof is precisely how the harmonic series is itself proven to be infinite in the first place. From the WP article Harmonic series#Comparison test:

One way to prove divergence is to compare the harmonic series with another divergent series, where each denominator is replaced with the next-largest power of two:

{\begin{aligned}&{}1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+{\frac {1}{9}}+\cdots \\[12pt]>{}&1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots \end{aligned}}

Each term of the harmonic series is greater than or equal to the corresponding term of the second series, and therefore the sum of the harmonic series must be greater than the sum of the second series. However, the sum of the second series is infinite:

{\begin{aligned}&{}1+\left({\frac {1}{2}}\right)+\left({\frac {1}{4}}+{\frac {1}{4}}\right)+\left({\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}\right)+\left({\frac {1}{16}}+\cdots +{\frac {1}{16}}\right)+\cdots \\[12pt]={}&1+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+\cdots =\infty \end{aligned}}

It follows (by the comparison test) that the sum of the harmonic series must be infinite as well. More precisely, the comparison above proves that

\sum _{n=1}^{2^{k}}{\frac {1}{n}}\geq 1+{\frac {k}{2}}

for every positive integer $k$ .

This proof, proposed by Nicole Oresme in around 1350, is considered by many in the mathematical community to be a high point of medieval mathematics. :It is still a standard proof taught in mathematics classes today. Cauchy's condensation test is a generalization of this argument.

So what am I missing here?? --71.81.74.166 (talk) 23:21, 25 May 2017 (UTC)

You're not missing anything. The first sentence of the article says the series is divergent, i.e., is not equal to a finite number. Gap9551 (talk) 23:26, 25 May 2017 (UTC)

What is -1/12? --71.81.74.166 (talk) 23:31, 25 May 2017 (UTC)

Had you read as far as the second paragraph of the article, you would know the answer! --JBL (talk) 23:45, 25 May 2017 (UTC)

I would have assumed a result so ridiculous would say more about the way it was derived (like there might be something wrong somewhere). I find it interesting that there are methods which not only give absurd results, but also can be disproven so easily; and yet these methods are useful somehow? (though I'm certain there is nothing fundamentally wrong with these methods, as it would have been pointed out many years ago) But not being a mathematician, I guess I still don't really get what's going on here. (and I have read the second paragraph) Thanks and sorry I've wasted anyone's time--71.81.74.166 (talk) 00:03, 26 May 2017 (UTC)

Further down an informal derivation is provided, and some applications to other fields. See also the links to the summation articles. Gap9551 (talk) 00:29, 26 May 2017 (UTC)

Proof of Consecutive Integers Starting at Any Positive Integer

I'm not sure if this would be the appropriate page, but it is related. May I add a proof of the following?

\sum _{i=m}^{n}i

Jeanlovecomputers (talk) 07:43, 29 March 2018 (UTC)

This is a particular case of a sum of an arithmetic progression and that article is the more likely place for such information. --JBL (talk) 12:34, 29 March 2018 (UTC)

That makes sense. Would it be ok if I place the proof on that page? Or do I have to ask in the "Talk" tab on that page? Jeanlovecomputers (talk) 03:02, 30 March 2018 (UTC)

Probably it depends on the content of your proof, how it relates to the other content on the page, and its sourcing. (I cannot promise what other editors will and will not do.) Per WP:BOLD, there is no harm in trying as long as you are prepared for the possibility that it might be reverted (and at that point the correct thing to do would be to discuss on the talk page). --JBL (talk) 12:41, 30 March 2018 (UTC)

I have no source, per se. I came up with the proof myself. It originated with me thinking about partial sum of integers from 1 to n. Then, I got to thinking what if it didn't necessarily start at 1 and I came up with a valid proof for a partial sum of positive integers from m to n. I thought it would be nice to add to Wikipedia since I can't find the proof on here anywhere. Jeanlovecomputers (talk) 10:51, 31 March 2018 (UTC)

253=0???

1+2+3+4+5+6+...=-1/12

12+24+36+48+60+72+...=-1

1+1+1+1+1+1+...=-1/2

11+23+35+47+59+71+...=-1/2

10+22+34+46+58+70+...=0

9+21+33+45+57+69+...=1/2

8+20+32+44+56+68+...=1

...

1+13+25+37+49+61+...=9/2

1+2+3+4+5+6+...=(-1)+(-1/2)+(0)+(1/2)+(1)+...+(9/2)=21

Thus 21=-1/12, 252=-1, 253=0??? — Preceding unsigned comment added by 101.10.35.19 (talk) 16:30, 7 April 2018 (UTC)

Yes, you've rediscovered the fact that this series can't be summed by linear, stable summation methods. This is discussed already in the article. --JBL (talk) 16:40, 7 April 2018 (UTC)

Excuse me. I don't understand this.

I think 1 + 2 + 3 + 4 + ⋯ should equal to infinity because when it is plused again and again, it should be more and more.

Why do it equal to -1/12?

Thank you.Manzzzz^(talk) 05:50, 28 October 2014 (UTC)

A more formal way to state your intuition is to say that the partial sums increase without bound and the series diverges to infinity. That's the topic of the first paragraph of the lead section and the first section of the body, titled "Partial sums".

The connections between the series and the number -1/12 are the topic of... well, the rest of the article! Melchoir (talk) 07:35, 28 October 2014 (UTC)

You can't understand why 1+2+3+...=-1/12 because it simply isn't true. But there is an eerie connection between the divergent series and the negative number. Read the article carefully. See also these Wikiversity lessons:

Divergent series
Casimir effect in one dimension|.
MATLAB/Divergent series investigations.

If I do remember correctly from elementary school, I believe the = operator means equal to, not connected to... — Preceding unsigned comment added by 69.43.65.114 (talk) 20:17, 1 December 2014 (UTC)

I agree 100%. Mathematicians shouldn't use the "=" sign, but instead something like "→". But keep in mind two points:

They probably adopted the equal sign because its more fun.
Wikipedia is not supposed to tell the mathematicians what symbols they should use.--guyvan52 (talk) 02:12, 2 December 2014 (UTC)

I think it's not so much the "=" symbol that's misleading, but the "+" symbol. Summation methods like Ramanujan summation are not the same as conventional addition and are only used by mathematicians. --Florian Blaschke (talk) 02:48, 21 April 2018 (UTC)

Vote for featured article

The infinite sum stepping stair postulate with solution of -1/12 is so amazing after performing a smoothed curve regularization. I would like to suggest this article be featured on front page Wikipedia.

I think it's a good idea! 1Mmarek (talk) 15:01, 16 August 2018 (UTC)

URL

Note that the last character of the URL of this page requires Unicode, which is not nice — Preceding unsigned comment added by Microbizz (talk • contribs) 16:41, 4 February 2019 (UTC)

1 + 2 + 3 + 4 + ⋯ = -1/9.

https://www.youtube.com/watch?v=Piamci9xnEM John W. Nicholson (talk) 23:48, 30 March 2019 (UTC)

This method uses linearity and something similar to stability (actually, more general). This is known to fail, as explained in the article. Vincent Lefèvre (talk) 00:32, 5 July 2019 (UTC)

1 + 2 + 3 + 4 + ⋯ = -1/8.

[[1]] Not a trained mathematician, it seems wise to include these similarly valid results in the article. 96.240.226.129 (talk) 03:13, 26 May 2020 (UTC)

I am a trained mathematician, and the gist here is that for finite sums it doesn't matter how you re-arrange the ordering you always get the same answer. For infinite series, you can make finitely many reorderings without changing the result, but the series must be absolutely convergent (i.e. the sum of absolute values must converge) to be able to make an infinite number of changes to the ordering. There's a theorem that states if a series does not converge absolutely, you can make it converge to any number you want by rearranging the order of summation.

This video showing -1/8 and the one showing -1/12 use these illegal "parlor tricks" (i.e. an infinite number reordering on a non-convergent series) to get the number they're after. You can use the same techniques to get any number you'd like. Mr. Swordfish (talk) 15:25, 26 May 2020 (UTC)

Would you agree that Srinivasa Ramanujan was an authority on mathematics?

Here is a scan of Ramanujan's NoteBook 1, chapter VIII, Page 3:

[ https://www.imsc.res.in/~rao/ramanujan/NoteBooks/NoteBook1/chapterVIII/page3.htm ]

Look at the 6th line. What did Ramanujan write after the "∴" (Therefore sign)?

The article says

"Because the sequence of partial sums fails to converge to a finite limit, the series does not have a sum.

"Although the series seems at first sight not to have any meaningful value at all, it can be manipulated to yield a number of mathematically interesting results.

"For example, many summation methods are used in mathematics to assign numerical values even to a divergent series. In particular, the methods of zeta function regularization and Ramanujan summation assign the series a value of ⁠−112⁠, which is expressed by a famous formula,

1+2+3+4+\cdots =-{\frac {1}{12}},

where the left-hand side has to be interpreted as being the value obtained by using one of the aforementioned summation methods and not as the sum of an infinite series in its usual meaning."

It hardly seems fair to characterize summation methods such as zeta function regularization, Ramanujan summation, and cutoff regularization as "parlor tricks", nor is it accurate to say "You can use the same techniques to get any number you'd like." --Guy Macon (talk) 19:56, 26 May 2020 (UTC)

Excuse me, but could you point out the place where I describe zeta function regularization, Ramanujan summation, or cutoff regularization as "parlor tricks"? Because I said no such thing.

What I did say is that performing invalid (but reasonable looking to non- mathematicians) symbol manipulation is a parlor trick.

As for the assertion ""You can use the same techniques to get any number you'd like." Please see Riemann_series_theorem. Application to the current discussion is left as an exercise for the reader. Mr. Swordfish (talk) 22:14, 26 May 2020 (UTC)

Probably you shouldn't use such a condescending tone: the series that is the title of this article does not satisfy the hypotheses of the Riemann series theorem. --JBL (talk) 22:50, 26 May 2020 (UTC)

Upon re-reading my response, I realize that I didn't clearly state the main thrust of it: There is nothing special about the -1/8 figure, at least not that I am aware of. It is possible to manipulate the symbols to get any results you want if you're willing to do some invalid symbol manipulation. That's why it would not be appropriate to include that in the article; the -1/12 result is the number arrived at by several mathematically valid summation methods. The video from Numberphile also arrives at that result but their "proof" is not mathematically valid. It's the logical equivalent of calculating 16/64 by cancelling the sixes in the numerator and the denominator - sure, it gives the right result, but for the wrong reason.

Clearly, if there are mathematically valid summation methods that give answers other than -1/12 (or that the series diverges) then we should include those results in the article. I am unaware of any, clever YouTube videos notwithstanding. Mr. Swordfish (talk) 22:28, 26 May 2020 (UTC)

Note: Something that is missing is that these summation methods have been applied only to one particular series: 1 + 2 + 3 + 4 + ⋯. This alone does not build a theory. One legitimate question is: If you do the same kind of things on some other series, assuming that you also get values for each method, will you necessarily get the same value like here? And if you find other consistent summation methods (e.g. something based on analytic continuation, but with another function rather than zeta), will you necessarily obtain −1/12 again for this series 1 + 2 + 3 + 4 + ⋯? Before asserting 1 + 2 + 3 + 4 + ⋯ = −1/12, you need to ensure some universality. Vincent Lefèvre (talk) 22:25, 26 May 2020 (UTC)

I don't think it's fair to say that "these summation methods have been applied only to one particular series". Granted, this article only concerns itself with one series, but the summation methods discussed are more generally applicable than one might infer from this article alone. It's probably worth re-reading with that possible misconception in mind to ensue we're not indaverdently conveying the wrong idea. I agree that your question is legitimate, but I'm pretty sure the answer is "yes" and think the overall treatment in various Wikipedia articles reflects that. Mr. Swordfish (talk) 22:42, 26 May 2020 (UTC)

I've eventually found the answer. If one is free to choose the regularization method by analytic continuation, 1 + 2 + 3 + 4 + ⋯ can at least give any value ≥ −1/12. So there isn't much magic about −1/12. It is just related to some summation methods, but other similar ones can give a different value. Something on which this WP article is quiet. Vincent Lefèvre (talk) 23:17, 26 May 2020 (UTC)

Sorry, does any of this have anything to do with proposed edits to the article? If not, this is not the right venue. --JBL (talk) 22:50, 26 May 2020 (UTC)

The question at hand is whether we include answers other than -1/12, e.g. -1/8. My response was that there's nothing special about -1/8 while there is something special about -1/12 (i.e. zeta(-1) = -1/12). So we include -1/12 but not all the other numbers that can be obtained by invalid symbol manipulation. At least that's my understanding. I think Vincent Lefèvre may be on to something when he says that the result depends on the method of summation. If so, we should address that in the article. Mr. Swordfish (talk) 13:35, 27 May 2020 (UTC)

Thanks, Mr. Swordfish, that's exactly the point. If all these are spurious answers to one degree or another, all these should be addressed in the article. As no one is correct, all should get appropriate and considered weight. 96.240.227.171 (talk) 03:15, 28 June 2020 (UTC)

I have added a section to the purpose of couching these alternate values without assigning too much value to them, while acknowledging the points expressed in this talk section. 96.240.227.171 (talk) 03:39, 28 June 2020 (UTC)

And I have reverted, because nothing you added is compatible with WP:OR. --JBL (talk) 11:27, 28 June 2020 (UTC)

For those who have watched the wrong YouTube video, here is the right one

Two of them, actually:

1 + 2 + 3 + 4 + 5 + ... = -1/12 (7.65 million views)
Sum of Natural Numbers (second proof and extra footage) (1.48 million views)

The above videos are referenced in this Smithsonian Magazine article:

The Great Debate Over Whether 1+2+3+4..+ ∞ = -1/12: Can the sum of all positive integers = -1/12? It can, sort of...

...and The New York Times referenced them:

In the End, It All Adds Up to - 1/12

I hope this helps (but of course we know that it won't) :( -Guy Macon (talk) 04:41, 26 May 2020 (UTC)

I haven't followed this talk page for a while so don't know what the "wrong YouTube video" refers to, but the above appear to be the Numberphile videos which are the wrong ones to watch. Instead, see Mathologer who really understands the math and is able to present it extraordinarily well:

https://www.youtube.com/watch?v=YuIIjLr6vUA • points out what's wrong (and by "wrong" I mean WRONG) with Numberphile's treatment
https://www.youtube.com/watch?v=jcKRGpMiVTw • Mathologer's first on the topic

Johnuniq (talk) 22:53, 26 May 2020 (UTC)

I found the second Mathologer video to be more enlightening than the first. Interesting quote from Mathologer's 2016 video:

"[Srinivasa Ramanujan] is a genius and [G. H. Hardy], well he's no dummy either, So somehow they can take these things seriously, so we better have a closer look."

Also, Numberphile made a third video ( https://www.youtube.com/watch?v=0Oazb7IWzbA ) about the first two ( https://www.youtube.com/watch?v=w-I6XTVZXww https://www.youtube.com/watch?v=E-d9mgo8FGk ) featuring Edward Frenkel, who also is no dummy. --Guy Macon (talk) 20:09, 28 May 2020 (UTC)

This is a big debate! My solution is that it can be any value, by the Riemann series theorem Nononsense101 (talk) 01:25, 23 December 2020 (UTC)

Nononsense101: The Riemann series theorem cannot be used here since the series is not conditionally convergent. Anyway, https://math.stackexchange.com/a/2190184/459608 shows that there are regularization methods that can give other results. — Vincent Lefèvre (talk) 01:46, 23 December 2020 (UTC)

Whoopsie! You actually can't use the Riemann series theorem. I mean CAN'T. Sorry for the error and the tone of this post;assume good faith.Nononsense101 (talk) 02:02, 23 December 2020 (UTC)

There is no reason for this response to be in a separate section, so I have removed the section header and changed the indentation; I hope no one objects. --JBL (talk) 03:50, 1 January 2021 (UTC)

Add this integration and a graph showing it

\int _{-1}^{0}{\frac {x(x+1)}{2}}dx=-{\frac {1}{12}}

A graph for this is available here: https://commons.wikimedia.org/wiki/File:Oe2Pma5.png — Preceding unsigned comment added by 2601:18D:680:A64B:0:0:0:B6E8 (talk) 13:36, 16 April 2019 (UTC)

What is the f(x) ( as in y=f(x) ) of the smoothed line in https://en.wikipedia.org/wiki/File:Sum1234Summary.svg ? 37.219.189.203 (talk) 17:11, 28 July 2019 (UTC)

I think it's probably what you'd get if you smooth using the cutoff function used in Cesaro summation, which is given in the Terence Tao reference [1]. So it would be f(x)= -1/12+x^2/6. So actually the graph is pretty misleading because f(x) would soon dip well below the discrete values. Maybe this should be mentioned somewhere. I guess the cutoff function has a stronger and stronger effect as you go on, which makes the smoothed function rise less steeply. 88.109.94.143 (talk) 22:46, 24 September 2020 (UTC)

Punctuation

(I think, this is a trivial issue, but JayBeeEll insisted on discussing it here...)

I would like to insert commas (shown here in red) in these sentences:

If
1 + 2 + 3 + ⋯ = x,

then adding 0 to both sides gives

0 + 1 + 2 + ⋯ = 0 + x = x
by stability.

and

For a function f, the classical Ramanujan sum of the series $\textstyle \sum _{k=1}^{\infty }f(k)$ is defined as
$c=-{\frac {1}{2}}f(0)-\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(0),$
where f^(2k−1) is the (2k − 1)-th derivative of f, and B_2k is the 2k-th Bernoulli number: B₂ = ⁠1/6⁠, B₄ = ⁠−+1/30⁠, and so on.

Please see this discussion at the language reference desk for explanations.

Any serious arguments against (or in support of) these commas? — Mikhail Ryazanov (talk) 15:52, 1 June 2021 (UTC)

It is very strange that you insist on personalizing what is a basic matter of WP policy, viz., that if you make an edit and someone objects to it, the correct next step is to discuss the edit on the article talk page (as you have now, finally, done). I will happily remove these first two sentences from my response if you remove the pointless first parenthetical to which they refer.

Substantively, repeating my comments from elsewhere: In the first example, the comma is optional, and in my opinion it needlessly breaks the flow in very short conditional sentence. I will be very happy to restore it if the consensus other editors of this article is opposed to my view. I find the second comma completely unforgiveable: it misleads a reader into pausing and trying to interpret the equation too early, when in fact it is absolutely necessary to continue reading in order to understand the displayed equation. However, this also suggests that the entire sentence should be rewritten to avoid the problem of introducing multiple pieces of notation after they are used. Concretely, one option is to move either the definition of the notation B_2k or the definition of the notation f^(2k−1) into a separate sentence preceding "For a function ...": "Let B_2k denote the 2k-th Bernoulli number: B₂ = ⁠16⁠, B₄ = ⁠−130⁠, and so on. For a function f, the classical Ramanujan sum of the series

\textstyle \sum _{k=1}^{\infty }f(k)

is defined as ...." --JBL (talk) 16:28, 1 June 2021 (UTC)

Apparently, Viennese Waltz was absolutely correct – nobody here cares... So JayBeeEll, would you now admit that the arguments from the language reference desk and two reputable style guides outweigh your unique opinion? Or we shall escalate this dispute to a higher level, per "WP policy"? — Mikhail Ryazanov (talk) 19:01, 18 June 2021 (UTC)

AFAIK, according to the punctuation rules, these commas are needed (but I agree that the second one can be misleading). Even though they don't make much difference when one reads the text, I'm wondering whether they are more important with non-visual web browsers. — Vincent Lefèvre (talk) 19:37, 18 June 2021 (UTC)

@Vincent Lefèvre: About the second case, how do you feel about rewriting in one of the ways I suggested? --JBL (talk) 21:37, 18 June 2021 (UTC)

@JayBeeEll: Yes, it may be better to define the notation for the Bernoulli numbers first (this is a standard notation, but I'm not sure that most readers know about them). — Vincent Lefèvre (talk) 22:39, 18 June 2021 (UTC)

Other finite results?

Are there manipulations that result in 1 + 2 + 3 + 4 + ⋯ having a finite value different from -1/12 ? Isambard Kingdom (talk) 15:45, 23 December 2016 (UTC)

Yes. If you allow invalid manipulations like the ones shown in the video you can make it add up to whatever you want it to. Simply add in a conditionally convergent series and apply the Riemann series theorem. Mr. Swordfish (talk) 16:39, 25 January 2017 (UTC)

But what if one only allows methods similar to those regarded as valid? For instance, the article mentions zeta function regularization. But could one find a different analytic function that would yield a different result? More precisely, the function f would be defined by a series such that, e.g. for f(1), the terms would be 1, 2, 3, 4, etc., but by analytic continuation, f(1) would be defined with a value different from −1/12. Is that possible? Vincent Lefèvre (talk) 00:41, 5 July 2019 (UTC)

This has been answered at https://math.stackexchange.com/a/2190184/459608 and yes, with other regularization methods, one can obtain other values. — Vincent Lefèvre (talk) 01:59, 23 December 2020 (UTC)

Surely it is important for the article to reflect the above? Only having -1/12 looks to be bias and WP should not display bias Dan88888 (talk) 09:16, 1 July 2021 (UTC)

I'm not Ramanujan but... It seems to be a strange result

In my opinion the series of 1 + 2 + 3 + 4 + 5 + ... is obviously the sum of Natural numbers, obviously it can be written as a result of the series defined by Ramanujan as 1+ (1 + 1) + (1 + 1 + 1) + (1 + 1 + 1 + 1) + (1 + 1 + 1 + 1 + 1) + ... Being sum of Natural numbers, a negative number can never come as a result of the sum. On the other hand, it would be enough to take a look at the well-known formula found by Gauss regarding the sum: (n (n + 1)) / 2 Trying to do the study of function as a function of n (easy, because it is a parabola) anyone can note that the function has negative values only for Real numbers between -1 and 0. Since the sum instead requires Natural numbers (rational numbers from 0 to infinity with denominator equal to 1), we can only take the part of this parabola relative to the first quadrant. Not only that: from the parabola only discrete values can be taken because it is a valid formula for Natural numbers and not for Real numbers. Basically I believe (perhaps wrongly) that probably the series 4s does not make 1 and that therefore s is not worth 1/4. How can there be two different results for the sum of the natural numbers? How can a divergent series have a convergent result at -1/12, among other things impossible because the sum concerns natural and non-rational numbers, negative or real integers? Perhaps the theory behind infinite series should be revised.

Yes, maybe I'm wrong, but I ask myself at least two questions ...

ONLY THE INTEGRAL (IN REAL) FROM -1 to 0 COMES -1/12

Enorab Otrebor — Preceding unsigned comment added by 217.172.199.105 (talk) 14:54, 5 September 2021 (UTC)

Wikipedia is an encyclopedia. Encyclopedia authors are supposed, at least on paper, to take already published and verified information and compile it in a "short" descriptive and neutral article. In order words, wikipedia is not a place for original research. If you have an original idea and fix the current theory, you should first publish it on a scientific paper or book. If it's a valid research, or at the very least noticeable research, it'll eventually have it's own section or even article. — Preceding unsigned comment added by 240d:1a:98a:3600:6de1:dca2:56ac:ec42 (talk) 04:55, 6 September 2021 (UTC)