# Buzz Blog

## Correction: Does 1+2+3+4+ . . . =-1/12? Absolutely Not! (I think)

### (If you'd rather just know what

*1+2+3+4+ . . .*actually

*is*equal to, check out our next post in this series)

Brief Summary:

**1+2+3+4+ . . . is not equal to -1/12**, but both the infinite series and the negative number are associated with each other in a way that can be seen in this graph

The area of the little region below the horizontal axis equals -1/12, and the infinite area under the curve on the right gives you 1+2+3+4+. . . , which goes to infinity as you add terms, not to -1/12.

(Update 2-6-14: of course, this graph shows what it looks like if you can only see a finite region of the graph. So 1+2+3+4+...+

**m**is going to infinity, provided

**m**is some finite number. This doesn't say anything about what happens if you could see all the way to infinity - I'd need a much bigger screen for that. For this reason, I am crossing out all the things that I am not so sure about in this post.)

For a longer explanation, read on . . .

Thanks in large part to the patience and persistence of people like Bernd Jantzen, who commented extensively on my previous post on this subject, I have found a way to simply, and graphically explain how the series 1+2+3+4+ . . . is associated with (and not equal to) the number -1/12, (update 2-6-14: provided you are actually taking the limit as you add terms and not looking at all the infinite terms at the same time as the Numberphiles did).

I usually try to avoid writing equations on this blog, so you're going to have to bear with me. But at least there will be pictures, and those are the most important parts.

Here we go . . .

I find it easier to understand things visually, so it occurred to me to plot out the series 1+2+3+4+ . . . at various points as you add the numbers. These points are called partial sums. This is what you get if you stop the sum after each step

for

*n*=1 the partial sum is 1

for

*n*=2 the partial sum is 1+2 = 3

for

*n*=3 the partial sum is 1+2+3=6

for

*n*=4 the partial sum is 1+2+3+4=10

for

*n*=5 the partial sum is 1+2+3+4+5=15

etc.

If you draw the first five terms on a piece of paper, it looks like this

But you don't have to add all those numbers to calculate value at each point. The numbers, it so happens, are a sequence (1,3,6,10,15, . . .) that you can calculate with a simple formula called a generating function. In this case the generating function is

G(

*n*)=

*n*(

*n*+1)/2

To get a number at any point in the sequence, like say the 5th spot, just plug in 5 for

*n*, and you get the answer

for

*n*=5, G(5)=5(5+1)/2=15

As it turns out, you can also use the generating function to figure out what the values would be between whole numbers. Essentially, you replace the number

*n*with an

*x*, which can have any numerical value you like. If you plot the result, you get this, for positive

*x*.

The curve you see here goes up to infinity as

*x*goes to infinity. So far, so good. But if we're going to plot the graph between the various values of

*n*, we might as well look at negative values of

*x*too.

When you do that, you get a graph like this

There are three interesting areas in this graph. The area above the horizontal axis and to the right of the curve (let's call it A), the area above the horizontal axis and to the left of the curve (call it B), and the area trapped between the curve and the horizontal axis (which I call C).

A and B are large areas - infinite actually, as you extend the curves to infinity, but C is small. In fact, if you use calculus to determine it's size, it's 1/12.

And because it's below the axis, it's conventionally considered negative, so it's -1/12. It's an easy integral to do, but in case you're feeling lazy, I did it on Wolfram Alpha for you, just click here.

Interesting, isn't it? The curve generated using the partial values of the series 1+2+3+4+ . . . gives you a graph with a little region in it that has an area of -1/12. Hmmmmm.

This is how 1+2+3+4+ . . . and -1/12 are associated. They aren't equal (update 2-6-14: provided you are taking limits), but -1/12, in the form of area C, is a characteristic of the curve While 1+2+3+4+ . . is the series that generates the curve in the first place.

So, despite my previous, non-mathematical argument to the contrary

**. . .**

*Update 2-6-14: The limit as m goes to infinity of 1+2+3+4+ . . .+m does not equal -1/12.*As I see it, -1/12 is a kind of label for the curve that you can generate using partial sums of 1+2+3+4+ . . .

The same thing works for 1+2^3+3^3+4^3+ . . . , and 1+2^5+3^5+4^5+ . . . and so on for any odd power (i.e., zeta(-3), zeta (-5), etc). I used this method to calculate the associated values of the zeta function for powers up to 13. In each case, you get a specific value the area C that's associated with the zeta function that creates the curve.

Here's a list of the C areas I calculated for curves generated by several series

zeta(-1) = 1+2+3+4+ . . . ---> -1/12

zeta(-3) =1+2^3+3^3+4^3+ . . . ---> 1/120

zeta(-5) =1+2^5+^5+4^5+ . . . ---> -1/252

zeta(-7) =1+2^7+3^7+4^7+ . . . ---> 1/240

zeta(-9) =1+2^9+3^9+4^9+ . . . ---> -1/132

zeta(-11) =1+2^11+3^11+4^11+ . . . ---> 691/32760

zeta(-13) =1+2^13+3^13+4^13+ . . . ---> -1/12

They agree with the published values of the zeta function for negative integers listed on Wikipedia.

(Although Wikipedia stops at zeta(-7) and I go to zeta(-13). The fact that the number associated with zeta(-1) and zeta(-13) are the same looks like potential trouble, BTW - after all, how would you know if your -1/12 is associated with zeta(-1) or zeta(-13)? It also suggests that the Numberphiles could have shown that -1/12 = 1+2^13+3^13+4^13+ . . ., or that 1+2^13+3^13+4^13+ . . .= 1+2+3+4+ . . ., if they felt like it.)

This little procedure works for even powers too, except the answer is always zero. Here's what the C area looks for for the curve generated from zeta(-2)=1+2^2+3^2+4^2+ . . .

The parts above and below the axis cancel for all series of this type with even powers, and as a result the total area doesn't give you any information. As you can see for the integral of the curve that comes from1+2^2+3^2+4^2+ . . .

**How Could I Have Screwed Up So Badly?**

I'm going to blame my misadventure on the trouble with using words to describe mathematical ideas. Before working out this problem, there was no way I could understand what it means when someone says that the value of -1/12 "can be assigned to an infinite series." They sounded like gibberish,

Ramanujan is to blame a bit too. After all, how are we supposed to understand what he was trying to say here?

*"I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal."*

*-S. Ramanujan in a letter to G.H. Hardy*

Are we supposed to realize that "under my theory" means that "=" doesn't mean equal?

I haven't found his original work, but several people have reproduced a calculation by Euler that uses an equal sign in the same way. If the two sides aren't equal then, as I recall from second grade math, you can't use an equal sign.

At least one person I spoke to said that in order to understand it, you have to know what the ". . ." in the expression 1+2+3+4+. . . means. As you can see from the example above, the dots mean exactly what they always mean. If they didn't, then we'd be in almost as much trouble as having equal signs that don't mean equal.

Finally, there are the physicists that say they need the relation 1+2+3+4+ . . . = -1/12. Some of them seem to believe the equation is what it is, and that our failure to understand it shouldn't stand in the way of using it. That struck me as the most romantic view, and it was the one I latched onto at the end of the day. Now, I realize that this is a far too mysterious view. It's interesting, but I hope this post shows

While 1+2+3+4+ . . . = -1/12 is clearly not true

I hope there are other, comparably bizarre mathematical controversies out there. One thing is for sure, though, I'm not going to rely on words, even from the most decorated experts in their fields, to try to understand things like this. I'm going to get out a pencil, fire up Wolfram Alpha, and just do the math.

## 40 Comments:

Galilée007 said...

Une somme partielle s'arrête à n : Sn = 1 + 2 +... + n. On ne peut pas appliquer les propriétés de l'addition dans l'expression 1 + 2 + 3.... (sans fin) et c'est la raison pour laquelle on en donne une définition toute faite.

et tant qu'à faire, on choisit évidemment la plus intuitive, mais intuitive ou pas, c'est une définition et la cohérence interne des propriétés des séries numériques tient dans celle des propriétés des limite (quand n -> + infini), un point, c'est tout. Dès lors qu'une autre définition se présente et qu'elle produit un autre espace de raisonnement cohérent, elle est tout aussi valable dans son champ mathématique d'application. Tout prolongement analytique d'une fonction f initialement non définie en un point, qui attribue une valeur en ce point licitement en ce sens que la fonction n'interdit pas de le faire, est fondé. Attribuer la valeur (-1/12) ne pose aucun problème, c'est une définition aussi licite que celle qui consiste à dire que 1 + 2 +.... = lim Sn

(= + infini). Si les gens n'ont pas compris que l'addition est une loir de composition ''interne'' de R dans R ! et non de

R U {infini} dans R U {infini} ! et que par conséquent, toutes les propriétés de l'addition - l'associativité, l'existence d'un élément neutre ''0'', d'un symétrique (-x), la commutativité, la distributivité de la multiplication sur l'addition - n'ont plus de sens lorsque qu'on parle ''d'ajouter'' une infinité de termes, alors, c'est que les gens ne parlent pas de la même opération. Des propriétés dépendent des définitions que l'on donne aux choses, il ne faut croire que la chose a une essence bien définie avant qu'on la définisse nous-même !

Friday, August 19, 2016 at 2:03 AM

Andy Holland said...

In order to use negative numbers for the integral, doesn't that require the problem to be the sum of all natural numbers including the negative ones? The left hand integral area cannot cancel the right hand area otherwise. The sum of all natural numbers from -x to x in the limit as x approaches infinity is -1/12?

Monday, July 18, 2016 at 10:44 PM

alan doak said...

Every term of Zeta(-1) is larger than Zeta(1), except the first which are both 1. Since Zeta(1) equals infinity (proven by the comparison test) -> Zeta(-1) is also infinity, due to the comparison test. QED.

Thursday, June 23, 2016 at 9:38 PM

Ronnie Johansson said...

Please delete "Dr Ebute"'s fraudish nonsense.

Tuesday, May 3, 2016 at 6:05 AM

Merychris Calib-og said...

How can I find the following cardinalities?

|{{{1},{2{3,4}},Emptyset}}

Sunday, January 31, 2016 at 7:29 AM

Anonymous said...

The term above -143n^10 should be -143n^10/60. Inadvertently the denominator 60 was omitted.Also the term n^c should be +n^c with

a plus sign inserted. Integrating this corrected function from 0 to -1

will now render the result of -1/12 = RZ(-13) = RZ(-1). Alan Walter,Sydney.

Friday, January 29, 2016 at 9:00 PM

Anonymous said...

Thanks for your youtube email reference.

RZ(-c) is Riemann's Zeta value at -c = I(n=0to -1)[1^c+2^c+3^c+...+n^c]dn

Where I(n=0to-1) is the definite integral from 0 to -1(lower limit)

RZ(-1)=I(n=0to-1)[1+2+3+...+n]dn=I(n=0to-1)[(n/2)(n+1)]dn= -1/12evaluated.

RZ(-13)=I(n=0to-1)[1^13 +2^13 +3^13 +...+n^13]dn

RZ(-13)=I(n=0to-1)[n^14/14+n^13/2+13n^12/12-143n^10+429n^8/84-429n^6/60

+715n^4/132 -691n^2/420]dn= -1/12 on evaluation= -B(14)/14= -(7/6)(1/14)

Note1^c+2^c+3^c+...n^c for odd c values has factors of n(n+1)with zeros at 0,-1.

Anyone interested in an expression for the yet unsolved sum of the positive ODD

Zeta series Z(2n+1) in terms of π^(2n+1)? Alan Walter, Sydney.

Wednesday, January 27, 2016 at 8:39 PM

Anonymous said...

The author of this article is absolutely correct to point out that these supposedly mysterious values can be found by calculating the definite integral between 0 and -1 of the expression for the respective sum to the nth term (a.k.a. their partial sum expressions).

This is also mentioned in this interesting 'response' video (it responds to the claims made in the video that is the subject of this discussion): https://www.youtube.com/watch?v=BpfY8m2VLtc

It shows what happens if you do the series manipulations with rigorously (hint - you do not get -1/12) and it explains why other methods get this -1/12 result.

The response video claims this -1/12 result is the result of a mistake. The mistake is one of taking a function that applies to just positive whole numbers, manipulating it in ways that bring decimal numbers and negative numbers into play, and then interpreting the result as though it still relates to positive whole numbers.

Tuesday, January 26, 2016 at 7:02 PM

Zero said...

Let's assume I gave you zero fucks. How many fucks have I given you?

Friday, May 15, 2015 at 11:28 PM

Buzz said...

Interesting, except removing thingZ from bagA doesn't leave you with -thingZ in bagA. to do that you would have to remove 2*thingZ. Otherwise, taking thingZ out of bagA, and then replacing it would leave you with nothing in bagA. But clearly removing thingZ from bagA and replacing it shouldn't make thingZ dissappear.

Friday, April 3, 2015 at 9:39 AM

Anonymous said...

I understand how puzzling it can be. Is zero a quantity or not? Is a negative number a quantity or not? Talking about apples and pears is a little worn out, I think.

So, here:

There are two different bags; let's call them bagA and bagB. They are freshly made, and nothing has ever been put in them before.

Each bag has nothing in it. The bags are in an enclosed vacuum, so there is no air in them and no air outside of them.

There are two different things; let's call them thingZ and thingY.

You put thingZ into bagA, and thingY into bagB.

You then have a bag - bagA - containing thingZ and a bag - bagB - containing thingY.

If you remove thingZ from bagA and thingY from bagB, the bags will then be empty because the things that were in them have been removed.

Because there used to be a thingZ in bagA, and one was taken out, there is a shortage of one thingZ.

BagA therefore contains -1 thingZ.

Likewise, bagB therefore contains -1 thingY.

But all the time there was a thingZ in bagA, bagA also had no thingY.

Similarly, while there was a thingY in bagB, bagB also had no thingZ.

In fact, ever since the bags were made, there was also none of thingX, none of thingW, no tomatoes, onions, apples, bicycles, orang-utans or saxophones in them. None of any other thing, in fact, in either bag. And the same goes for what is outside of each bag either, because it was a vacuum.

If it is true that zero is something that does exist, then zero is a complete lack of stuff. But we know that it wasn't always like that, because there used to be a thingZ and thingY in the bags.

Regarding anything and everything else, there always was a complete lack of anything else in the bags, and everything outside the bags.

Because you cannot remove something that isn't there, bagA still contains zero thingY and bagB still contains zero thingZ - and zero everything else except as was stated in the paragraph before this one. And outside of the bags is still zero everything.

But the bags don't contain zero of everything, they contain -1 of one thing.

Zero is more than -1. The bags therefore collapse under the pressure of absolutely nothing.

Friday, April 3, 2015 at 2:30 AM

Anonymous said...

Hi .....

do you wants....idea from amateur...from me or not...

Easy...! Open Mind...

Zeta Functions not importtant ....Harmonic important more!

Zeta function that important Zeta(-1) and Zeta(-3)

I show you....develop zeta in term cartoon.....and blinding set and anti

and i predict somethings in Higher Dimension...You can help me proof

http://www.quora.com/Zeta-1-1-12-really-Sure-1

http://www.quora.com/What-is-the-solution-to-sqrt-1+2-sqrt-2+3-sqrt-3+4-sqrt-4+5-sqrt-5+6-sqrt-6+

sawat

Wednesday, December 3, 2014 at 11:13 PM

Anonymous said...

To be more specific, irrational mathematics('mathematics'), is not sustainable.

Monday, November 3, 2014 at 11:53 PM

Anonymous said...

I have posted a message explaining the dilemma.

What has happened here was inevitable, mathematics is not sustainable.

Here is the link to my message: http://marques.co.za/duke/news_win.htm

It will not surprise me if this comment is censored (Moderated)

Friday, October 31, 2014 at 5:13 AM

Anonymous said...

1+2+3+4+... = -1/12 (R) where (R) is the Rumanujan Summation. This is not a normal -1/12. It basically is a categorization of the series in question. It should be read, "the sum of one plus two plus . . . has a Rumanujan summantion of -1/12" (as opposed to "equals -1/12").

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

Tuesday, October 14, 2014 at 3:12 PM

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Saturday, August 9, 2014 at 1:31 AM

Anonymous said...

No, zero is not in the real world.

Imagine that you have two bags, in one there is one apple and in the other you have one pear. Then we remove the apple and the pear out of the bags, now in one bag you have zero apples and in the other zero pears, but zero apples and zero pears have the very same properties, therefore they must be the same thing, as we know, pears and apples are not equal to each other, so it must be that zero is nothing but a concept that does not exist in the real world.

Monday, June 2, 2014 at 1:58 AM

Anonymous said...

Area A = Area B since they both are infinitely large areas. That series diverges, you learned that in Cal2 or Math Physics of DiffEq. C'mon.

Thursday, May 29, 2014 at 10:47 AM

Richard Smart said...

This article is superb, I love it. That infinite series thing really perturbed me when I first saw it but I couldn't see how to examine it more effectively such that I could get to the point that I wasn't perturbed by it any more. Watching you do it above now makes me annoyed I didn't have the idea of doing that myself. But the fact is I didn't. So simple, so insightful, so satisfying. Thanks for that and if you could now just solve every other annoying problem for me I would be most grateful ;o)

Thursday, May 22, 2014 at 3:41 PM

Bernd Jantzen said...

Of course, the partial sums of 1+2+3+4+... are tending to infinity. Nobody claimed anything different. The whole discussion here is about how to assign a meaningful finite value even to a divergent sum like this one. And the discussion is about the question whether one may call this finite number the value of the divergent series or just a meaningful number that one may use in replacement of the series under some conditions.

Sunday, April 6, 2014 at 6:01 PM

Anonymous said...

1+2=3+3=6+4=10+5=15+6=21+7=28+8=36+9=45+10=55.......................................=infinity

Saturday, April 5, 2014 at 10:22 AM

Anonymous said...

1+2+3+4+5.................= infinity!!!

Saturday, April 5, 2014 at 10:20 AM

Anonymous said...

i think 1+2+3+4+.........=infinity

Saturday, April 5, 2014 at 10:18 AM

Anonymous said...

I possess no sheep, therefore I possess 0 sheep. 0 is very much in the real world.

Wednesday, March 19, 2014 at 11:08 AM