# 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.

(Update 2-5-14: I made a stupid error in this part, but I think the overall point is the same.)

In fact, it's easy to calculate the space between them out to any distance. Here's a Wolfram Alpha calculation of the difference that I performed by subtracting the area under B out to -100 from the area under A out to 100. If you keep taking the area calculation to larger and large distances, the difference increases. (I ignored a tiny portion of the area, but it doesn't matter in the long run because both numbers are going to infinity eventually and a tiny missing piece makes no difference.)

And of course, the difference between area A an B is itself infinity if you extend the curves out infinitely.

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+ . . .

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 on 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.

## 84 Comments:

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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.

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