Talk:1 − 2 + 3 − 4 + ⋯

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1 − 2 + 3 − 4 + ⋯ is a featured article; it (or a previous version of it) has been identified as one of the best articles produced by the Wikipedia community. Even so, if you can update or improve it, please do so.
This article appeared on Wikipedia's Main Page as Today's featured article on April 15, 2007.
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March 11, 2007 Peer review Reviewed
March 16, 2007 Featured article candidate Promoted
A fact from this article appeared on Wikipedia's Main Page in the "Did you know?" column on March 2, 2007.
Current status: Featured article
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This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
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Field:  Analysis

Cauchy product

Hi. Great page. What does "(C,3) summable" in the Cauchy product section mean? Robinh 08:43, 2 March 2007 (UTC)

Whoops, that should have been explained in Cesàro summation and briefly here. I'll work on it… Melchoir 16:59, 2 March 2007 (UTC)

I just have to say: excellent page! These sums are one of my favourite parts of mathematics (1 + 2 + 3 + 4 + ... = -1/12 is another wonderful example), and I think that this certainly deserves to be on DYK. Well done, all who've contributed! --3M163//Complete Geek 13:09, 2 March 2007 (UTC)

• obviously i am too dumb to know how to post my own comment, but can someone sum it all up then explain to someone like me?

thnks -mls** —Preceding unsigned comment added by Mylifesucks (talkcontribs) 20:41, 5 February 2008 (UTC)

Error?

In the generalisations section it states

For odd n,

${\displaystyle 1-2^{2k+1}+3^{2k+1}-4^{2k+1}+\cdots =(-1)^{k}{\frac {2^{2k+2}-1}{2k+2}}B_{k+1}}$

where Bn are the Bernoulli numbers.

But if we substitute k=0 on both sides we get

${\displaystyle 1-2+3-4+\cdots =(-1)^{0}{\frac {2^{2}-1}{2}}B_{1}}$
${\displaystyle ={\frac {3}{2}}B_{1}}$
${\displaystyle =-{\frac {3}{4}}}$

as ${\displaystyle B_{1}=-{\frac {1}{2}}}$ according to these values for Bernoulli numbers which I am assuming to be correct.

Now it works if we use ${\displaystyle B_{2}={\frac {1}{6}}}$, so should the expression be?

For odd n,

${\displaystyle 1-2^{2k+1}+3^{2k+1}-4^{2k+1}+\cdots =(-1)^{k}{\frac {2^{2k+2}-1}{2k+2}}B_{k+2}}$

Athosfolk 14:02, 2 March 2007 (UTC)

Actually having read through the page on the Dirichlet eta function which this was linked to, I now think this should read.

For odd n,

${\displaystyle 1-2^{2k+1}+3^{2k+1}-4^{2k+1}+\cdots =(-1)^{k}{\frac {2^{2k+2}-1}{2k+2}}B_{2k+2}}$

So that it matches the expression there.

Athosfolk 14:16, 2 March 2007 (UTC)

Ah, yes. That was just a stupid typo, and I'll fix it now. Thanks for the catch! Melchoir 16:56, 2 March 2007 (UTC)
…on second thought, it wasn't just a typo; I think Hardy's exposition is wrong. I've switched to a source that agrees with everything else. Melchoir 17:22, 2 March 2007 (UTC)

1 + 2 - 3 + 4 - · · ·

Is the Abel sum of the above -1/4? Nil Einne 14:27, 2 March 2007 (UTC)

If you mean -1 + 2 - 3 + 4 - · · ·, then yes. The Abel sum of 1 + 2 - 3 + 4 - · · · is 7/4, being 2 greater than the previous. Melchoir 16:58, 2 March 2007 (UTC)
Ooops yes -1... Nil Einne 14:03, 19 March 2007 (UTC)

OR valve

I'd like to write the following things in the article, but I can't, so I will write them here instead. Hopefully someone will find them amusing.

• 1 − 2 + 3 − 4 + · · · = 1/4 is probably the Euler characteristic of the Klein four-group and its classifying space.
• ${\displaystyle 1-2+3-4+\cdots ={\frac {\hat {1}}{(1+S)^{2}}}}$
• In the language of the literature on sequence transformations, one would probably call 1/4 the "antilimit" of the partial sums of 1 − 2 + 3 − 4 + · · ·, since it is not the unique number they approach but the unique number they run away from.

That's all for now. Melchoir 19:20, 5 March 2007 (UTC)

Image format

Five of the PNGs on this article have SVG equivalents, which look better and should be used in the long run. However, current server-side software issues make the SVGs unreliable. Melchoir 17:44, 13 March 2007 (UTC)

So what, are we avoiding SVGs now? How long is this supposed to last?--Pharos 23:20, 14 April 2007 (UTC)
I wish I had an answer! Melchoir 18:55, 15 April 2007 (UTC)

Article title

Is it standard/appropriate to use formulae and mathematical expressions as the titles of articles on an English Wikipedia? It's my feeling that the title of an article should express a concept in some way that has intrinsic meaning in English wherever a proper noun name is unavailable or ambiguous. For example, please see the renaming discussion at E=mc². If the series 1 - 2 + 3 -4 + ... has some name, it should be given that name. Further, is the use of three spaced bullet characters an acceptable substitute for an ellipsis? Robert K S 09:45, 14 March 2007 (UTC)

I doubt there is any guidance on this at Wikipedia. The usual advice in style guides is to avoid formulas in titles of articles and sections, but that's not universally adhered to (for instance, there is a book called "SL2(R)"). Part of the reason is that it's very hard to avoid them in some cases. In this case, there does not seem to be any name for the series.
The ellipsis should be at the same height as the minus signs, as in
${\displaystyle 1-2+3-4+\cdots }$
The proper character to use for the ellipsis is perhaps ⋯ (Unicode U+22EF), so perhaps the page should be moved to 1 − 2 + 3 − 4 + ⋯ . However, it may well be the case that this character does not display properly for many people. On my font, it looks worse than the three bullet characters (the spacing between the dots is too small). So, yes, the bullets are perfectly acceptable to me. -- Jitse Niesen (talk) 12:03, 14 March 2007 (UTC)
There's also relevant discussion, however brief, at Talk:1 + 2 + 4 + 8 + · · · and [1]. I see that the E=mc^2 issue is more of a merger than a move, and in any case the phrase "mass-energy equivalence" is a recognized concept name on its own. For this article, I don't think there's an alternative nearly that good. Melchoir 18:48, 14 March 2007 (UTC)
Well, whatever the name of the article is, this one needs several redirect pages. No one who doesn't already know about the article is going to type in 1 − 2 + 3 − 4 + · · ·. --Cryptic C62 · Talk 15:49, 15 April 2007 (UTC)
It has a lot of redirects already. Did I miss one? Melchoir 18:56, 15 April 2007 (UTC)
The ones that I've noticed fall under two categories:
--Cryptic C62 · Talk 02:59, 16 April 2007 (UTC)

FA status

How come this article appears in the Wikipedia:Featured Articles list (under the mathematics section), but doesn't have a star on it? (The FA list seems to be the only source that says this article is an FA) Leon math 23:25, 16 March 2007 (UTC)

I added it; it has to be done manually. Melchoir 23:33, 16 March 2007 (UTC)

How encyclopedic

[[Image:P3001234.jpg|thumb|King Leonidas promptly devouring the first few thousand terms and partial sums of 1 − 2 + 3 − 4 + · · ·]] What a wonderful encyclopedia. That image all the more so proves that Wikipedia is an encyclopedia.

Why thankyou :] PhoenixJ 00:44, 15 April 2007 (UTC)

Ugh. Boring, pointless, essentially incoherent past the introduction, utterly nerdy. Vranak
I'm a mathematics student. I find it most interesting. —The preceding unsigned comment was added by 147.8.16.89 (talkcontribs).
Thanks for sharing. We'll make sure to have only featured articles on pretty actors and actresses in future. --Saforrest 14:18, 15 April 2007 (UTC)
Why would you think that's what I'm interested in? Those are almost as bad. Vranak

You nerds, he's talking about that deleted image on the side. He must have seen it when he viewed the page and thought it was part of the article.--Stockynever 21:45, 15 April 2007 (UTC)

Talk page has been mirrored

Take a look at Talk:1 − 2 3 − 4 · · ·. It appears to have been created by HagermanBot, as it has the only edit. --LuigiManiac 00:59, 15 April 2007 (UTC)

I've left User:Hagerman a note; I'm not bot expert, but it looks like it's something about the code dealing with '+' signs. -GTBacchus(talk) 01:03, 15 April 2007 (UTC)
Yup, I've taken notice of this before (User talk:Hagerman#Bug report). Unfortunately, the bot creator has left the site, so the only options are to block the bot or to just fix it when it happens. --- RockMFR 03:18, 15 April 2007 (UTC)
Yeah, I saw that. Well, at least there can't be that many pages with "+" in the title, right? --LuigiManiac 03:36, 15 April 2007 (UTC)

Seen this somewhere before...

This section from the front page:

where h is the "sum" of the series

 h = 1 − 1 + 1 − 1 + · · · = 1 − (1 − 1 + 1 − · · · ) = 1 − h.

Is the same tricky sidestep they used in the false 0.999...=1 argument on Wikipedia. They are removing a finite member out of an infinite set of numbers and then using the ". . ." to ignore what was done. Infinity cannot be manipulated in this way. To say 1-1+1-1+1-1... = 1/2 is silly. If their math leads them to this conclusion, they should rethink their methods before trying to ooo and awe people with tricks.

Who is Hardly? The references on this page are atrocious. You can't reference a book by just giving the author's name until you have at least one reference of the book title. I want to find out how this works, but I can't without knowing what the book is called. —The preceding unsigned comment was added by JohnLattier (talkcontribs).

Have you tried reading the References section. You'll find that the details of the books footnoted are all there. Harryboyles 05:08, 15 April 2007 (UTC)
I've seen a good parable explaining why it's kind of intuitive that 1 - 1 + 1 - 1 + . . . = 1/2.
There's a king, you see, who dies, and his will declares that all his wealth is to be divided equally between two sons. They divide up the kingdom, and the castles, and the land, and everything, but there's one gem that's worth more than the rest of the kingdom put together. They can't split it in half, so they're stymied, until a mathematician suggests they take turns: one has the gem for one year, and then the other for the next year, and so on. Thus, each will have the gem for 1 - 1 + 1 - 1 + . . . = 1/2 of the time!
It's certainly not a mathematically rigorous argument, but it's a cute story that might help explain why such expressions can be considered meaningful. -GTBacchus(talk) 06:18, 15 April 2007 (UTC)
For anyone else who's interested in GTBacchus' example, check out History of Grandi's series. Melchoir 18:54, 15 April 2007 (UTC)
Thanks, Melchoir; I'd looked for that, but couldn't find it. -GTBacchus(talk) 03:14, 16 April 2007 (UTC)
IIRC, there are like 10 or so arguments in the 0.999...=1 case. Some more mathematically rigirous then the others. There were like 100 different people trying to claim it was false, most of them who didn't even appear to read the whole article. It appears based on your references complaint that you haven't properly read this article either... Edit: Forgot to mention, wikipedia does not publish original thought. As such, there is no such thing as a wikipedia argument. Rather these are arguments presented by other reliable sources which wikipedia mentions. If you have any arguments against 0.999...=1 or 1-2+3-4+=1/2 you should try to get them published in a reliable source. Good luck with that... Alternatively, if you just want to publish your ideas without having to ensure they're sound, you might want to try Conservapedia. They don't appear to care much about whether what they publish is sound, and they don't appear to like maths they don't understand Nil Einne 07:25, 15 April 2007 (UTC)
Hardy is G. H. Hardy, and the work in question is Divergent Series, as mentioned in the References section (the page reference you cited was from 'Notes').
There is no problem with the idea pulling one term out of an infinite summation in itself: surely you would agree that since 1+(1/2)+(1/4)+...=2, then (1/2)+(1/4)+(1/8)+...=1. The problem is when the infinite summation in question is not defined because it does not converge (for whatever definition of 'converge' you want to use). --Saforrest 14:30, 15 April 2007 (UTC)

I don't get it

Wikipedia should feature articles the average user can understand... or maybe i'm a dumbass 71.135.79.58 09:11, 15 April 2007 (UTC)

Wikipedia is not exclusively for the ignorant or stupid.—The preceding unsigned comment was added by 81.155.140.254 (talkcontribs).
Well, neither is it exclusively for the math geeks. Rosa 21:14, 15 April 2007 (UTC)
And the very occasional featuring of a mathematics article makes it exclusively for people interested in maths? Now that’s an “intelligent” deduction!—The preceding unsigned comment was added by 81.155.140.254 (talkcontribs).
...and the fact that some of us aren't familiar with advanced mathematical terms makes us stupid or ignorant? I was just replying to your previous argument by the same logic dude...will you mind signing your comments btw? Rosa 22:03, 15 April 2007 (UTC)
Maybe you may find reading this helpful too. Rosa 22:17, 15 April 2007 (UTC)

Nor I. How can you add and subtract whole numbers and end up with a fraction? Or is that not what the article is trying to say? —The preceding unsigned comment was added by 60.224.136.155 (talkcontribs).

The answer to what you're asking: When you have more and more terms, the series should diverge. So one way of looking at it would be the series have no sum.
On the other hand -- and this series ties in with the 1-1+1-1+... one... you can draw a conclusion that the sum is 1/4 provided you can do so. It's one field where the actual results that people understand will differ from theory. - Penwhale | Blast him / Follow his steps 10:19, 15 April 2007 (UTC)
Thanks for trying, but quite frankly I still don't fully understand this article. Well, maybe a lot of you folk in here are familiar with advanced mathematical terms but for those of us who aren't it's just beyond comprehension.Rosa 19:48, 15 April 2007 (UTC)
"a man's reach should exceed his grasp, Or what's a heaven for?" 64.231.51.237 16:49, 15 April 2007 (UTC)
Even the most simple things may be explained in an obscure manner, and this is easy to do. Contrarywise, even the most complex things may be explained in a clear manner, but this is hard to do. Rosa 19:48, 15 April 2007 (UTC)
Rosa, I think this might help. In mathematics, if you can define a set of rules, that's consistent with itself, then you can work with those rules. According to the definition of "sum" and "equals" that most of us are used to, it makes no sense to say that 1 - 2 + 3 - 4 . . . equals anything.
However, there are other, somewhat more generous, ways of defining "sum" and "equals", that don't contradict our ordinary, finite, arithmetic, but which produce some additional surprising results with infinite sums. There are definitions and sets of rules under which expressions like the one in this article can make sense, and add up to a particular number. Within those rulesets, the sum is, oddly enough, 1/4.
I think one good way to see how that could possibly work is to consider this: if you take the series (1 - 2 + 3 - 4 . . .), and write it down 4 times in just the right way, you can see how all the positive terms and all the negative terms cancel out, except for one of the initial 1's. Thus, four copies of the series add up to 1, so each one must equal 1/4. -GTBacchus(talk) 01:30, 19 April 2007 (UTC)
  1 - 2 + 3 - 4 + 5 - 6 + . . . . .
+ 1 - 2 + 3 - 4 + 5 - . . . . .
+ 1 - 2 + 3 - 4 + 5 - . . . . . .
+ 1 - 2 + 3 - 4 + . . . . . . .
--------------------------------------------
= 1 + 0 + 0 + 0 + 0 + 0 + . . .

Ah, finalmente... that was a clear explanation indeed, thanks. Rosa 02:16, 19 April 2007 (UTC)
Just one note: there is no need to redefine "equals". It means what it always means: two things are exactly, identically, logically the same in every possible way. They are, in fact, the very same thing. People try to get around statements like 0.999... = 1 by hoping that = has a different meaning than usual, but it doesn't. Melchoir 03:37, 19 April 2007 (UTC)
Wow, I didn't think I was trying to "get around" anything. Can you help me out with this point, though?
The objects on either side of the equal sign are real numbers, right? And real numbers are equivalence classes of, what, Cauchy sequences of rational numbers? So, when we apply something like Abel summation and find a way to identify a real number with (1 - 2 + 3 - 4 + . . .), aren't we redoing our partition, to include more sequences in each equivalence class? And if you've defined a new equivalence relation, by drawing a new partition, then haven't you just redefined "equals"? Is there a difference between redefining what's on either side of the equal sign, and redefining the equal sign itself? -GTBacchus(talk) 04:12, 19 April 2007 (UTC)
Drawing a new partition would redefine "real number" at the very least, and that's not what we're doing here. Part of the confusion arises from the fact that just about any mathematical expression has multiple meanings, so one has to help the author out by selecting the meaningful one:
• "1 − 2 + 3 − 4 + · · ·" might mean the formal series, which might be defined as the sequence 1, -1, 2, -2, .... It wouldn't make any sense to assert that a series is equal to a number, so we discount that possibility.
• "1 − 2 + 3 − 4 + · · ·" might mean the ordinary sum of the above formal series. This would be a number, but it does not exist, so we discount this one too.
• "1 − 2 + 3 − 4 + · · ·" might mean the Abel sum of the above formal series. This is a number, namely 1. Since 1 = 1, then 1 − 2 + 3 − 4 + · · · = 1.
I wouldn't say that the last option is really a redefinition, since there's no relevant previous definition for a divergent series.
I guess you could mentally assign the responsibility of all interpretations to the equal sign, but then you lose the ability to interpret each symbol separately. Admittedly, this happens in education all the time. I think in the New Math, the sentence "2 - 1 = 1" is interpreted not as saying that 1 subtracted from 2 is 1, but that 1 added to 1 is 2. Seems silly. Also, in calculus, integrals are usually introduced by saying that the S and the dx form a monolithic symbol. This is also unnecessary; each symbol has an independent meaning, and when these are combined in order, you get integration. Melchoir 17:51, 19 April 2007 (UTC)
I like the trick/representation GTBacchus uses above. Was playing a bit, and found that it can be used for the other series mentioned in the article as well provided one uses higher Binomial coefficients. For instance:
  1( 1 - 4 + 9 - 16 + 25 - 36 + 49 - ... )
3(     1 - 4 +  9 - 16 + 25 - 36 + ... )
3(         1 -  4 +  9 - 16 + 25 - ... )
1(              1 -  4 +  9 - 16 + ... }
+ ----------------------------------------
1 - 1 + 0 +  0 +  0 +  0 +  0 + ...   =  0

Congrats with the FA status. It certainly is a wonderful article.JocK 17:33, 22 April 2007 (UTC)

Possible problem?

I could very well be mistaken, but here goes. In the "stability and linearity" section, I do not understand why h = 1 - h follows from stability and linearity. Suppose in general that we have the series a1 + a2 + ... The article would assert (correct?) that given linearity and stability, this is equal to a1 + (a2 + a3 + ...).

The series a1 + a2 + ... corresponds to the sequence of partial sums (a1, a1+a2, ...) Stability (if I understand it correctly) allows us to conclude that the series corresponding to the truncated sequence (a1+a2, a1+a2+a3, ...) has value equal to the original series. Thus a1 + a2 + ... = (a1 + a2) + a3 + ... By linearity, this is equal to (a1 + 0 + 0 + ...) + (a2 + a3 + ...), provided that a1 + 0 + 0 + ... converges. If we invoke REGULARITY, we then have that this is equal to a1 + (a2 + a3 + ...).

Perhaps I misunderstand stability, in which case the entry in http://en.wikipedia.org/wiki/Divergent_series should be improved. Perhaps I'm missing a possible demonstration with just stability and linearity that a1 + a2 + ... = a1 + (a2 + a3 + ...); this demonstration should be made clear in an article of "featured" quality. Or is regularity an implicit assumption? Then it should probably be made more explicit. Kier07 11:01, 15 April 2007 (UTC)

I noticed the same before seeing your comment, so I have now changed it, as well as removing the assumption that 1 - 1 + 1 - ... is summable, since this itself follows from the manipulations. --Ørjan 18:10, 15 April 2007 (UTC)

Not to put it delicately, but you're wrong. Regularity doesn't even enter into the picture, because nowhere is the classical definition of sum used in the manipulation. And the manipulations do prove rather than exemplify; they are not an example of a summation method. All this development, including which axioms are used and how, are found in Hardy as cited. Melchoir 19:01, 15 April 2007 (UTC)
I don't have access to the Hardy book right now. But can you please make it clear in the article (or at least to me, here on the talk page) how the argument leading to h = 1/2 follows from linearity and stability? Which part of the "problem" I identified is, in fact, not true? Do I misunderstand the meaning of stability (in which case that Wikipedia article needs to be changed), or is the argument in the article leaving out steps which would make everything clearer? The trouble is, there are probably a lot of readers like me, who understand enough to question the logic of the proof, but do not have access to the Hardy reference (by the way, Hardy is probably right, but not necessarily). Kier07 19:27, 15 April 2007 (UTC)
OK, I am beginning to think we must be wrong about the definition of stability - it really should remove a constant at the beginning of the series, not the sequence, so that regularity is not necessary just to get things like A(1 + 0 + 0 + ...) = 1. Neither of the versions discussed over at Talk:Divergent series actually get this right, which may be what confused us. I guess we should clarify it over there first. --Ørjan 20:55, 15 April 2007 (UTC)

It has been Reddit-ed, possibly Dugg as well, I'll keep an eye on it to cut vandalism but could someone lock it please?Mutton 12:28, 15 April 2007 (UTC)

The vandalism is probably caused by the increased visibility caused by its mention as "Today's featured article" on the main page. Generally, we do not protect these articles; see Wikipedia:Main Page featured article protection. -- Jitse Niesen (talk) 12:37, 15 April 2007 (UTC)
I haven't been to the homepage yet today... woops. Thanks for the clarification.Mutton 12:41, 15 April 2007 (UTC)

Great article

Nice work, folks! --Doradus 18:24, 15 April 2007 (UTC)

Indeed it is. I don't care to actually do math, but math theory can be awesome. Keep it up, esteemed editors. Paul Haymon 04:04, 18 April 2007 (UTC)

Format of 1/4

How should we format the fraction 1/4 when it occurs within text? In order of my own preference, the choices are

14
1/4
¼

I don't know about other people's fonts, but the last option is less readable. The second option isn't as pretty. Melchoir 19:07, 15 April 2007 (UTC)

14
which might end up being customizable? –Dan Hoey 12:15, 17 April 2007 (UTC)
After looking at Template:frac, I see that (1) They recommend using subst: at this time, because it's still in test, and (b) it expands to your second option except that it wraps the numerator and denominator in <small>, so they are less readable. I now agree that 1/4 is preferable, and I'll take up the limitations of Template:frac on the talk page. –Dan Hoey 13:01, 17 April 2007 (UTC)
Okay, as long as someone's looking into it. Melchoir 19:13, 17 April 2007 (UTC)
I don't know if anyone's really looking seriously; I've put a couple of suggestions at Template_talk:frac that don't seem to be addressed. Meanwhile, I've converted most of the fractions here to m/n form. One of the 1/2's got changed back by User:Tompw, perhaps not noticing that there were five other occurrences this fraction (and many more of 1/4 and other fractions) in the sup/sub format. I reverted that, because I'd like to keep the format consistent as much as possible. I'd be glad to change them all to ½ and ¼ if there's some sort of consensus about it, but when I started the situation was pretty chaotic. –Dan Hoey 13:01, 19 April 2007 (UTC)
I changed the sup/sub version because it wasn't displaying correctly (the 2 wasn't visible). Thinking about it, forlumae this complex should really displayed using [itex] tags with LaTex (see WP:MATH), which would look like:

${\displaystyle {\begin{array}{rcllll}2s&=&&(1-2+3-4+\cdots )&+&(1-2+3-4+\cdots )\\&=&1+&(-2+3-4+\cdots )&+1-2&+(3-4+5\cdots )\\&=&0+&(-2+3)+(3-4)+(-4+5)\\{\frac {1}{2}}&=&&1-1+1-1\cdots \\\end{array}}}$
Does this seem OK? Tompw (talk) 19:09, 19 April 2007 (UTC)

Actually, I have a problem with changes to that section, but I don't really want to get into it without having the time to explain myself clearly. Melchoir 19:19, 19 April 2007 (UTC)
I'm surprised that the 2 did not display correctly. Does this occur in the text as well? What browser and platform?
I can see that the [itex] version might do, as long as we are in a display equation. At first, it seemed like a big tool for a small job, but the job has tended to expand a bit. I'd better look at the WP:MATH guidelines more closely. –Dan Hoey 12:30, 20 April 2007 (UTC)
I run IE and Windows XP (like the majority of WP users). I'll copy the LaTex code in. Tompw (talk) 18:34, 20 April 2007 (UTC)

Mention dump

Melchoir 19:45, 15 April 2007 (UTC)

Why is this important?

I read through the article, but I can't see anywhere that it shows the importance or notability of this series. Maybe I missed it in the formulas, but could someone put some reference up to why this is a significant series in the field of mathematics? Slavlin 04:17, 16 April 2007 (UTC)

Its important in respect to number theory. Like everything, some things are more important than other things in certain medium i.e. if you were a chemist, you would find that chemistry was more important to you than say French. There is no real reason to why this number series is so important but ini that respect mathematics is just about solving logical problems and all the mathematicians that did attempt to must have done it for their own pleasure. --pizza1512 Talk Autograph 06:19, 16 April 2007 (UTC)
If you note, I asked why it was important in regards to the field of mathematics. I want to understand why the material is notable. What does this tell us about or what does this open the door to? Slavlin 13:41, 16 April 2007 (UTC)
Pizza1512 said it was important in number theory. Doesn't the intro explain the significance? The series 1 − 2 + 3 − 4 + · · · is closely related to Grandi's series 1 − 1 + 1 − 1 + · · ·. Euler treated these two as special cases of 1 − 2n + 3n − 4n + · · · for arbitrary n, a line of research extending his work on the Basel problem and leading towards the functional equations of what we now know as the Dirichlet eta function and the Riemann zeta function. Leebo T/C 14:19, 16 April 2007 (UTC)
I am looking for something more accessible. I know that this will be a technical article, but I would like to see something that tells me why it is important without me having to go to these other articles and also figure out why they are important too. I am looking for something like what is described in "Articles that are unavoidably technical".Slavlin 19:40, 16 April 2007 (UTC)
It is important as one of the best-known examples of a divergent series which can be summed in a reasonable way using various summation techniques. It's not just any example, but one which has been discused much in the literature, starting with Euler. So it's got historical significance, too. It's true, that doesn't necessarily become clear in the current intro. --345Kai 21:50, 16 April 2007 (UTC)

Rigor in "Divergence" section

The argument for divergence is given as "It clearly does not settle down and converge to a particular number, so 1 − 2 + 3 − 4 + … diverges." This is very nonrigorous.

Without resorting to even an elementary application of the epsilon/delta definition of convergence, we can improve this to something that makes a real mathematical argument. It should be removed (the link the the term test proves it) or improved.

For an improvement I propose:

If it converged to a particular number n, then the series of partial sums would need to remain close to n as it went to infinity. However, the partial sums get consistently farther from n beyond the nth term of the series.

...though the pronouns in this could use cleanup. —Preceding unsigned comment added by 216.165.132.250 (talk) 18:26, 27 March 2008 (UTC)

The section begins by making the trivially obvious observation that the series diverges by the term test. There's no need for rigor after that. Melchoir (talk) 18:43, 15 October 2008 (UTC)

Abel quote

Is this quote correct? I guess some of the translations atenuated the devilishness of the divergent series:

This last sum became an object of particular ridicule by Niels Henrik Abel in 1826:
"Divergent series are on the whole devil's work, and it is a shame that one dares to found any proof on them. One can get out of them what one wants if one uses them, and it is they which have made so much unhappiness and so many paradoxes. Can one think of anything more appalling than to say that
0 = 1 − 2n + 3n − 4n + etc.
where n is a positive number. Here's something to laugh at, friends."

I don't dispute that 0 = 1 − 2n + 3n − 4n + etc, but that this is not as devilish as saying that 0 = 1 + 2n + 3n + 4n + etc. In fact, the first can be derived from the second, by noticing that 2n + 4n + 6n + 8n + etc = 2n (1 + 2n + 3n + 4n + etc) = 0. Albmont (talk) 17:43, 15 October 2008 (UTC)

I checked the citation to Grattan-Guinness and the quote from there is correct: "devil's work". Markushevich's translation has "a devilish concoction", which is about on par. The original French can be viewed here: "quelque chose de bien fatal".
I for one agree that the positive series are far more devilish than the alternating series, although I'm not sure if Abel would have agreed, or if he simply thought they were all equally meaningless. Melchoir (talk) 18:35, 15 October 2008 (UTC)
Thanks! I copied-and-pasted the original quote (it's obvious public domain) to the french wikiquote in Niels Henrik Abel. Les séries divergentes sont en général quelque chose de bien fatal should be The divergent series are, in general, something quite deadly (if my French is right); there's no religious/satanist implication in Abel's quote (why would he invoke his mother's temptor? better hide before (Mc)Cain kills him). Albmont (talk) 11:24, 16 October 2008 (UTC)
The quote was effaced from the french wikiquote. It seems that there are some crazy rules that prevent quotes from a book 150 years old :-( Well, the quote is there in Google. Albmont (talk) 16:29, 30 October 2008 (UTC)

Possible error?

In the section where assuming we can write s = 1 - 2 + 3 + ..., the computation showing 4s = 1 appears flawed. I don't see how you get from the second to the third line. It looks like the 1 on the left should be a 2. —Preceding unsigned comment added by Particle25 (talkcontribs) 05:36, 28 November 2009 (UTC)

No, it is correct. The 1 on the left comes from taking 1 from the second group, 1 from the third group and 1-2 = -1 from the fourth group, so sum is 1. Gandalf61 (talk) 11:48, 28 November 2009 (UTC)

I added a (what I hope is) a simple explenation for the paradox. The math part may have been correct, but IMHO, its was almost exclusively understandable by people who already understand why "1-2+3-4...=1/4". Hence, as it was, it was a bad wikipedia article. (much like an explenation of basic chemistry is bad if we start mentioning quarks) —Preceding unsigned comment added by Qube0 (talkcontribs) 18:55, 10 March 2010 (UTC)

I agree that it would be helpful to add some explanation of what divergent series are and how they are usually given no numerical meaning. However, there are many points in your edit that aren't right: it isn't meaningful to talk about a formula not having an answer, or "ending up" as "either" one of two values. Also, there are no "infinite numbers" implicit in the series, nor is there a point where we "reach" infinity. Finally, there are no "rules" being broken, and nothing is being hidden: the article says repeatedly that the series diverges.
For those reasons, I'm reverting the edit. This is without prejedice to a description of what convergent versus divergent series are, or how they are typically used. Melchoir (talk) 00:36, 11 March 2010 (UTC)
Then I'm going to add the explenation GTBacchus gave above to Rosa. For what it's worth, you're incorrect in saying it's analog to the 0.999 situation or that they "exactly, identically, logically the same in every possible way". Here, we're assigning numerical values to diverent series - and this can get some pretty interesting results.
If you don't believe me, it's the same logic which proves that
${\displaystyle 1+2+3+4+\ldots =-1/12}$.
(as the infinite sum minus our toggle-serie give 4 times the infinite sum). And then we go on with
${\displaystyle 1+1+1+1+\ldots =1+(2-1)+(3-2)+(4-3)+\ldots =1+2+3+4+\ldots -(1+2+3+\ldots )=-1/12-(-1/12)=0}$
while ' obviously ' it should be ${\displaystyle 1+1+1+1+\ldots =-1/2}$
(what's that? ${\displaystyle +\infty }$? Silly Toto, we're not in Kansas anymore!) — Preceding unsigned comment added by Qube0 (talkcontribs) 15:16, 22 September 2014 (UTC)

Article moved to 1 − 2 + 3 − 4 + · · ·

I've moved this article to 1 − 2 + 3 − 4 + · · ·, for consistency with other similar articles in Category:Divergent series. -- The Anome (talk) 11:19, 22 April 2010 (UTC)

Could some admin please move this to 1 − 2 + 3 − 4 + … The current title looks very ugly on Mac. —Preceding unsigned comment added by A:-)Brunuś (talkcontribs) 12:32, 26 June 2010 (UTC)

What is ? vs. How to ?

1-2+3-4... looks to me to be a Fractal. Then 1/4 would be not a "sum" but some kind of fractal parameter. Just like the other -1/12 or something. Nicolae-boicu (talk) 03:23, 10 August 2012 (UTC)

How is it a fractal? Fractals are (quasi-)self-similar. — Arthur Rubin (talk) 17:56, 10 August 2012 (UTC)

Requested move

The following discussion is an archived discussion of the proposal. Please do not modify it. Subsequent comments should be made in a new section on the talk page. No further edits should be made to this section.

The result of the proposal was no consensus. --BDD (talk) 20:23, 22 April 2013 (UTC) (non-admin closure)

1 − 2 + 3 − 4 + · · ·1 − 2 + 3 − 4 + ⋯ – Refer to the names of other articles in Category:Divergent series. Relisted. Favonian (talk) 17:01, 10 April 2013 (UTC). Yzyzsun (talk) 14:47, 3 April 2013 (UTC)

• oppose all of them should be using "..." so this should be called 1 - 2 + 3 - 4 + ... (and all other articles) per MOS:ELLIPSIS and WP:AT#Special characters ; neither the current nor the proposed titles use common keyboard characters, and can be replaced by such, so use periods instead of the current format, as would typically be typed by most people (or even written by most people) -- 65.92.180.137 (talk) 00:22, 4 April 2013 (UTC)
The above discussion is preserved as an archive of the proposal. Please do not modify it. Subsequent comments should be made in a new section on this talk page. No further edits should be made to this section.

1 - 2 + 3 - 4 ⋅⋅⋅ = 0?

Starting with these two premises:

1 - 2 + 3 - 4 ... = c
1 + 2 + 3 + 4 ... = y


Part 1:

1 + 2 + 3 + 4 ... = y
1 + 2 + 3 ... = y
1 + 3 + 5 + 7 ... = 2y


Part 2:

- 1 - 2 - 3 - 4 ... = -y
- 1 - 2 - 3 - 4 ... = -y
- 2 - 4 - 6 - 8 ... = -2y


Finale:

1     + 3     + 5     + 7     ... =  2y
- 2     - 4     - 6     - 8 ... = -2y
1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 ... =  0
c  =  0


I need an explanation for the calculations and the result above. In fact, I personally don't think I need an explanation, I personally think that this is all wrong from the start: invalid operations, playing around with series that diverge, and so on... But if you do have an explanation, I would be glad to hear it. — manual signature: comment added by ThoAppelsin (talkcontribs) 19:43, 1 February 2014 (UTC)

Part 1 is wrong. If you're going to define y by zeta function regularization, then you can't shift the terms of the series one place to the right without altering the regularized sum. This is shown in explicit detail on Terence Tao's blog.
In Hardy's terminology, zeta function regularization is not a stable summability method.
By contrast, c can be defined by Abel summation, using power series, which are much easier to shift around. Abel summation is a stable summability method. So you can shift the terms of a series to the right and get the same Abel sum. And Abel summation isn't the only relevant technique that obeys this axiom. There's also (C, 2) summation. That's why the article makes this general statement:
What the above manipulations actually prove is the following: Given any summability method that is linear and stable and sums the series 1 − 2 + 3 − 4 + ..., the sum it produces is ​14.
Hardy makes precisely this point, so I'll add a citation to that sentence. Melchoir (talk) 20:38, 1 February 2014 (UTC)

1^-s   +   2^-s   +   3^-s   +   4^-s ... = zeta(s)
- 2*2^-s              - 2*4^-s ... = zeta(s)*2*2^-s
1^-s   -   2^-s   +   3^-s   -   4^-s ... = zeta(s)*(1 - 2*2^-s)

... is absolutely the same, nothing different, if someone is in the end going to replace s with -1. There is nothing different between substituting before the gimmick or after it. In fact, if such an operation is to be performed, the circumstances will be evoking a limiting condition for the further usage of the resultant equation, which would be "true for only s > 1" in this case.
Such a formulization with exponents avails pre-substitution manipulations only through the even/odd number duality, so there is in fact a single way to cook up a random result, without a substitution beforehand. However, as I said, if someone is to substitute a value that should have been banned on the way, the result will be just as wrong it is on my calculation above. — manual signature: comment added by ThoAppelsin (talkcontribs) 23:47, 1 February 2014 (UTC)
Abel summation and zeta function regularization are not the same, and the sets of transformations that they respect are not the same. You cannot apply a transformation which is valid for one method to the other method, observe that you get the wrong answer, and conclude that both methods are invalid. Moreover, neither method merely replaces s with 1 or -1. They transpose the problem in a subtle and powerful way. I strongly recommend that you read Hardy's book, at least up to page 9. Melchoir (talk) 00:03, 2 February 2014 (UTC)

Missing footnote

There is reference 17 . ^Knopp, p.491..., which has no corresponding entry in footnotes section. Paweł Ziemian (talk) 19:51, 13 April 2014 (UTC)

I don't remember if I wrote that or someone else did, but I have a hunch that I know what the reference is. Will check on it later to confirm... Melchoir (talk) 20:57, 14 April 2014 (UTC)
That reference hasn't the entry at footnotes section, somebody knows it? Bye, --Elisardojm (talk) 23:09, 12 November 2017 (UTC)

1 − 2 + 3 − 4 + ⋯ = INDETERMINATE !

1 − 2 + 3 − 4 + ⋯ can'not be 1/4 ! because 1 − 2 + 3 − 4 + ⋯ = (1 + 3 + 5 + ⋯ ) - (2 + 4 + 6 + ⋯) = infinity - infinity = INDETERMINATE ! --ᔕGᕼᗩIEᖇ ᗰOᕼᗩᗰEᗪ (talk) 16:07, 31 August 2014 (UTC)

That reasoning fails, even for the kinds of convergent series one studies in calculus. Consider the alternating harmonic series 1/1 − 1/2 + 1/3 − 1/4 + ⋯ = log 2. The problem is that the various theorems that allow you to decompose a series into two related series require all of the series to be convergent (or summable). When both of the series on the right-hand side of the equation are divergent, we can't conclude anything about the series on the left-hand side. Melchoir (talk) 19:38, 31 August 2014 (UTC)
i did'nt understand what you said. but i just want to show that there are various solution. --ᔕGᕼᗩIEᖇ ᗰOᕼᗩᗰEᗪ (talk) 14:52, 11 September 2014 (UTC)

does it even converge ???

how can 1 − 2 + 3 − 4 + ⋯ be equal to a finite number. even it's finite how can it be a unnatural number (1/4). —_— --ᔕGᕼᗩIEᖇ ᗰOᕼᗩᗰEᗪ (talk) 15:46, 16 October 2014 (UTC)

Nope, the series doesn't converge. The third sentence of the article explains why: "The infinite series diverges, meaning that its sequence of partial sums, (1, −1, 2, −2, ...), does not tend towards any finite limit." Melchoir (talk) 16:54, 16 October 2014 (UTC)

Shifting makes multiplication not reversable by division. Is that going to be a problem?

I have a problem with this 4S = 1 Which basically is:

S = 1 - 2 + 3 - 4 + 5 - 6 + . . . . .
+ 1 - 2 + 3 - 4 + 5 - . . . . .
+ 1 - 2 + 3 - 4 + 5 - . . . . .
+ 1 - 2 + 3 - 4 + . . . . .


If You do that, then yeah, it looks like 4S=1 but I don't think there was 4 times S. Even if they look the same, those wore 3 different sums. So it should look like this: S+T+T+U=1
If You would do this:

4S = 1 - 2 + 3 - 4 + 5 - 6 + . . . . .
+ 1 - 2 + 3 - 4 + 5 - 6 + . . . . .
+ 1 - 2 + 3 - 4 + 5 - 6 + . . . . .
+ 1 - 2 + 3 - 4 + 5 - 6 + . . . . .


Than You would get:

4S = 4 - 8 + 12 -16 +20 -24 +. . . . .


That way if You divide both sides You will get back original S = 1 - 2 + 3 - 4 + 5 - 6 + . . . . .
Let's do another shift:

4S = 1 - 2 + 3 - 4 + 5 - 6 + 7 - . . . . .
+ 1 - 2 + 3 - 4 + 5 - . . . . .
+ 1 - 2 + 3 - 4 + . . . . .
+ 1 - 2 + 3 - 4 + . . . . .


we get:

4S = 1 - 2 + 4 - 4 + 4 - 4 + 4 - . . . . .



— Preceding unsigned comment added by 153.2.247.31 (talk) 09:19, 20 October 2014 (UTC)

${\displaystyle 4S=1-2+4-4+4-4+4-...}$
${\displaystyle 4S=1-2+4(1-1+1-1+...)}$
and now, since ${\displaystyle 1-1+1-1+...=1/2}$, you concluded the same:
${\displaystyle 4S=1-2+4(1/2)}$
${\displaystyle 4S=1}$
Qube0 (talk) 14:22, 27 October 2014 (UTC)

Excuse me. I don't understand this.

Excuse me.

I don't understand 4th line to 5th line.

${\displaystyle {\begin{array}{rclllll}4s&=&&(1-2+3-4+\cdots )&{}+(1-2+3-4+\cdots )&{}+(1-2+3-4+\cdots )&{}+(1-2+3-4+\cdots )\\&=&&(1-2+3-4+\cdots )&{}+1+(-2+3-4+5+\cdots )&{}+1+(-2+3-4+5+\cdots )&{}+(1-2)+(3-4+5-6\cdots )\\&=&&(1-2+3-4+\cdots )&{}+1+(-2+3-4+5+\cdots )&{}+1+(-2+3-4+5+\cdots )&{}-1+(3-4+5-6\cdots )\\&=&1+&(1-2+3-4+\cdots )&{}+(-2+3-4+5+\cdots )&{}+(-2+3-4+5+\cdots )&{}+(3-4+5-6\cdots )\\&=&1+[&(1-2-2+3)&{}+(-2+3+3-4)&{}+(3-4-4+5)&{}+(-4+5+5-6)+\cdots ]\\&=&1+[&0+0+0+0+\cdots ]\\4s&=&1\end{array}}}$

Thank you.Manzzzz(talk) 04:18, 27 October 2014 (UTC)

I was confused at first too, but then I noticed that they were the 'columns': from step 4 to step 5 they combine the columns so you instead of four infinite series, you get one infinite serie (each item of the serie a sum of four parts)
          1    -     2      + . . . . . . .
+ 1 - 2    +     3      - . . . . .
+ 1 - 2    +     3      - . . . . . .
+ 1 - 2 + 3    -     4      + . . . . .
--------------------------------------------
1 +
1    -     2      + . . . . . . .
- 2    +     3      - . . . . .
- 2    +     3      - . . . . . .
+ 3    -     4      + . . . . .
-------------------------------------------- (4th line to 5th line: combine the columns)
+1 + (1-2-2+3) + (-2+3+3-4) + . . . . .

(1-2-2+3) contains each first term of the infinite lists, (-2+3+3-4) contains each second term, etc ...
Qube0 (talk) 14:10, 27 October 2014 (UTC)
I think there is an opportunity for streamlining this equation. I wrote the "Stability and linearity" section a long time ago, and it's been edited by various other authors. The "Explanation of the paradox" section is newer, and it uses a much nicer presentation! Perhaps the equation in "Explanation of the paradox" can be kept the way it is, and the equation in "Stability and linearity" can be replaced by a pair of equations relating this series to Grandi's series. (I apologize if my meaning is unclear; I'll try to show what I mean later.) Melchoir (talk) 18:02, 27 October 2014 (UTC)

As long as we're talking nonstandard summation methods ...

The last sentence of the article reads:

" For example, the counterpart of 1 − 2 + 3 − 4 + ... in the zeta function is the non-alternating series 1 + 2 + 3 + 4 + ..., which has deep applications in modern physics but requires much stronger methods to sum."

But wait! If S = 1 + 2 + 3 + 4 + ... then

2S = 2 + 4 + 6 + 8+... and

-3S = S - 2(2S) = 1 + 2 + 3 + 4 + ... - 2(2 + 4 + 6 + 8+...) = 1 - 2 + 3 - 4 +....

So if the last summation has already been shown to have a nonstandard summation method assigning it a value of 1/4, then the sum S = 1 + 2 + 3 + 4 + ... can be readily assigned the value (-1/3)(1/4) = -1/12 without requiring "much stronger methods to sum".Daqu (talk) 13:24, 22 October 2015 (UTC)