# How to Drive a Mathematician Crazy. (With pies.)

In mathematics, you don’t understand things. You just get used to them. -Johann von Neumann

Sometimes, I have to deal with series: lots of numbers all added together. Some series clearly approach a limit, like the following:

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + …

I can visualize this in terms of pies. (What do you want? I’m hungry!) One = one whole pie. So that first number starts me out with one whole pie.

When I add the second number in my series, I’m clearly adding half a pie to that, for a total of a pie and a half.

When I add the third term, I’ve now got one-and-three-quarters pies, and by adding the fourth term, I’ve got one-and-seven-eighths pies.

If I add more and more terms in this series, I’m going to get closer and closer to two whole pies, but I’ll never quite reach it, always being just a tiny sliver away.

And that’s okay! It means that the sum of this series is two. The series has a limit, or — in math-speak — the series converges.

But not all series converge! Take a look at this one:

1 + 1 + 1 + 1 + 1 + 1 + …

The first term starts me out with a pie. Fine, so did the last series. But when I add the second term in, I realize I now have two pies!

In fact, if I add the third one in I have three pies, if I add the fourth in I get four pies, and so on! The more terms I add, the more pies I get.

This last part — that the series doesn’t approach a limit — is really important. It means that I can keep adding more terms, but when I do, the value of the series keeps changing; it doesn’t approach one single value. Therefore, this series does not converge, and the word we give to this is that the series diverges.

Or, to quote Lindsay Lohan (correctly) from Mean Girls, The Limit Does Not Exist!

But what if I were a tease about it? What if I gave you this series:

1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + …

After one term, there is a pie, but after two terms, there aren’t any pies; the second term takes the first pie away!

And it looks like the third term adds a pie, and the fourth one takes it away, and this continues. You’re adding pies (one-by-one) and taking them away (one-by-one) as you keep on adding terms.

So you might ask yourself, does the series converge, and if it does, what is its limit?

A mathematician is likely to tell you that the limit does not exist. Why not? Because for half of the terms, the sum is one full pie. But for the other half of the terms, the sum is zero pies. I could group these terms together to show you. For example, if I put my parentheses like so:

( 1 – 1 ) + ( 1 – 1 ) + ( 1 – 1 ) + ( 1 – 1 ) + ( 1 – 1 ) + …

it’s clear that this series sums up to zero. But what if I’m clever, and I organize my terms like this:

1 + ( -1 + 1 ) + ( -1 + 1 ) + ( -1 + 1) + ( -1 + 1) + …

I can see that my series sums up to one whole pie! So who’s right: is it zero pies, one pie, or does the solution simply not exist? To get it right, you have to ask yourself what happens on average.

Well, for half of the terms you have a pie, and for the other half of the terms you have no pies. On average? You have half a pie. Even though at no point do you actually have half-a-pie, this series sums to one-half. (For more rigor, you might want to read this.)

1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + … = 1/2.

This is a hard one to wrap your head around, and many mathematicians and physicists that I knew in graduate school were unable to do it. Want to try something more ambitious? Have a go at this one:

1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + …

(Answer here.) And if this isn’t up your alley, don’t worry. We’ll go back to the astrophysics on Wednesday.

1. #1 Benjamin Clegg
March 29, 2010

“1 + ( -1 + 1 ) + ( -1 + 1 ) + ( -1 + 1) + ( -1 + 1) + … ”

Associativity is not defined on an infinite series.

2. #2 ScentOfViolets
March 29, 2010

I thought associativity was defined, but only if the series converged. And as far as I know, this series does not converge to 1/2 or any other number – even on average. For example, you could rearrange the terms to be: 1+1-1+1+1-1+1+1-1+ . . . or 1-1-1+1-1-1-1+1-1-1-1-1+1- . . . so that this series can go to any integer you care to name. Or to no finite integer, for that matter.

3. #3 RogerTheGeek
March 29, 2010

I noticed that you didn’t use any meat pies in your analysis. Is there any significance to this? Do you think there is any possibility of a grant application?

4. #4 John Armstrong
March 29, 2010

[Evidently the comments here don’t like me linking to my own weblog. Go there yourself and search on “associativity in series”.]

The first couple of comments are close. I made two posts on the subject when I covered series.

However, even so: you won’t drive any mathematicians crazy. What you’re talking about here can be handled in two different ways.

First is “analytic continuation”. For example, the geometric series doesn’t converge at $x=-1$, but where it does converge it converges to the function $\frac{1}{1-x}$. This function can be evaluated at $x=-1$ to give the answer $\frac{1}{2}$.

Second is “Cesàro summation”. What we do here is replace the partial sum of the first $n$ terms with the average for the first $n$ partial sums. It turns out that if a series is summable, then it’s also Cesàro summable; but if it’s not summable it might still be Cesàro summable, and thus give us the right answer.

If the “many mathematicians and physicists” you knew really couldn’t wrap their heads around it, then they never passed a basic course in Fourier analysis.

5. #5 Egaeus
March 29, 2010

I’m no mathematician, but it doesn’t seem correct to me to say that the limit of the average of the nth partial sum of a series can be considered to be the same as sum of a series. If the series is convergent, then they do end up being equal, but that is a special case.

For example, consider the trivial series sum(1^n). The partial sums would be 1,2,3,4,5,…, and the Cesaro summation as defined in the Wikipedia article would be 1/1, 2/2, 3/3, 4/4, 5/5, … , or lim as n->infty(n/n) = 1 via L’Hopital’s rule. However, it’s a divergent series where lim as n->infty of s_n = infty, not an indeterminate value such as {1,-1}.

POTENTIAL SPOILER FOLLOWS

As far as your more ambitious question…. The Cesaro sum turns out oscillate between {1/2, -1/2} as n->infty. I haven’t looked at the solution, and can’t justify what I say next, but perhaps you could again apply the Cesaro sum to the oscillating terms of the partial sums and get 1/4?

6. #6 John Armstrong
March 29, 2010

Egaeus, You’ve made a mistake in your calculations. It’s not the average of the sequence up to that point, it’s the average of the partial sums. For example:

The sequence is
1, -1, 1, -1, 1, -1, 1, -1, …

The sequence of partial sums is
1, 0, 1, 0, 1, 0, 1, 0, …

The sequence of sums of partial sums is
1, 1, 2, 2, 3, 3, 4, 4, …

The sequence of averages of partial sums is

1/1, 1/2, 2/3, 2/4, 3/5, 3/6, 4/7, 4/8, …

Which converges to 1/2.

7. #7 Egaeus
March 29, 2010

My calculations were for 1^n, not -1^n.

8. #8 John Armstrong
March 29, 2010

Okay, let’s look at that one, Egaeus:

The sequence is
1, 1, 1, 1, 1, 1, 1, 1, …

The sequence of partial sums is
1, 2, 3, 4, 5, 6, 7, 8, …

The sequence of sums of partial sums is
1, 3, 6, 10, 15, 21, 28, 36, …

The sequence of averages of partial sums is

1/1, 3/2, 6/3, 10/4, 15/5, 21/6, 28/7, 36/8, …

Which diverges, just as it should. If a series diverges to infinity, it still does when Cesàro summation is used.

9. #9 Egaeus
March 30, 2010

Oh, duh, that’s right. I did it right later, but not on that one. Now the solution for the second problem makes more sense. I could see what needed to be done, but couldn’t justify it. It’s too late for math after a day of paper writing, I guess. I knew something didn’t look right in the first paragraph too.

But still, it doesn’t seem correct to equate s and the limit of 1/n*sum(s_k}.

10. #10 John Armstrong
March 30, 2010

It’s not too difficult to show that if the series converges then the sum and the Cesàro sum are equal. Yeah, it does seem to be sort of a cheat at first, but when you see how it makes the inverse Fourier transform work out you get used to it quick.

11. #11 Bee
March 30, 2010

You shouldn’t throw together different types of summability, at least not without explicit warning.

March 30, 2010

Although on average you have half a pie, the function remains without a limit although it is bounded. You simply have to accept a very large error in stating the value.

13. #13 Jérôme ^
March 30, 2010

If you wanted pies, You sould also have used the convergent series ∑(1/n²), whose sum is exactly π²/6; and also 1-1/3+1/5-1/7+… = π/4 (via power series expansion of arctangent), which was actually the formula used for centuries to compute π. By the way, this last one could even have a geometrical interpretation and be representable in pies, but I don’t see it right now.

And for the divergent geometric series ∑(-1)^n : as John Armstrong says, analytic continuation hints at a sum = 1/2 (though it converges neither in ℝ or ℚp), but you can also « cheat » by saying

S = 1 – 1 + 1 – 1 + 1 – 1 …

= 1 – (1 – 1 + 1 – 1 + …)

= 1 – S, thus S = 1/2.

(This is the exact same method used to compute the sum of a _convergent_ geometric series, but applied to the divergent case q=-1, it becomes confusing).

14. #14 ENT-TT
March 30, 2010

The correct answer is “Schrödinger’s Pie”.

15. #15 Sili
March 30, 2010

Good.

Now deal with how 1+2+3+4+5+ … = -1/12.

16. #16 Jérôme ^
March 30, 2010

Hey Sili, that’s amusing how (R) summation makes the familiar (1^3+2^3+3^3+…) = (1+2+3+…)^2 fail, considering that you have 1+2+3+4+5+… = -1/12 and 1^3+2^3+3^3+4^3+5^3+… = 1/120…

Oh, and a fun reference book on the subject is Hardy, _Divergent series_ (though, as all Hardy books, it is not easy reading!)

17. #17 ScentOfViolets
March 30, 2010

You shouldn’t throw together different types of summability, at least not without explicit warning.

Posted by: Bee

Especially when they are not the same thing; even Lebesgue integration and Riemannian integration are better equivalents. Such is the paucity of the mathematical vocabulary

18. #18 ScentOfViolets
March 30, 2010

(This is the exact same method used to compute the sum of a _convergent_ geometric series, but applied to the divergent case q=-1, it becomes confusing).

Posted by: Jérôme

This is also a way to show that 1/1+1/2+1/3+… diverges (somewhat apocryphal, but supposedly the work of a law student): Let S=1/1+1/2+1/3+… and split it into even and odd parts, S=(1/1+1/3+1/5+…)+(1/2+1/4+1/6+…). But this is just S=(1/1+1/3+1/5+…)+(1/2)S, so that (1/2)S=(1/1+1/3+1/5+…), and termwise, 1/1>1/2, 1/3>1/4, etc. So this is a contradiction unless both parts sum to infinity.

19. #19 Egaeus
March 30, 2010

I think that if you can show that 1+2+3+4+5+… = -1/12, then you might just overthrow the foundations of all mathematics….

20. #20 rob
March 30, 2010

Nigel Tufnel: The series all sum to eleven. Look, right across the board, eleven, eleven, eleven and…
Marty DiBergi: Oh, I see. And most series sum up to ten?
Nigel Tufnel: Exactly.
Marty DiBergi: Does that mean it’s convergent? Is it any more convergent?
Nigel Tufnel: Well, it’s Cesaro convergent, isn’t it? It’s not ten. You see, most blokes, you know, will be converging at ten. You’re on ten here, all the way converged, all the way converged, all the way converged, you’re on ten on your series. Where can you go from there? Where?
Marty DiBergi: I don’t know.
Nigel Tufnel: Nowhere. Exactly. What we do is, if we need that extra push over the cliff, you know what we do?
Marty DiBergi: Converged it up to eleven.
Nigel Tufnel: Eleven. Exactly. One bigger.
Marty DiBergi: Why don’t you just make ten more convergent and make ten be the top number and make that a little more convergent?
Nigel Tufnel: [pause] These converge to eleven.

21. #21 Mu
March 30, 2010

I just ate the pie, please recompute for N=2 to infinity.

22. #22 vagueofgodalming
March 30, 2010

Saying it’s ‘right’ that the sum is 1/2 is only justified if there’s a theorem showing that any summability method for divergent series that leads to a reasonable set of properties is equivalent to Cesàro summability (or that this particular series is a special case).

Otherwise you have the possibility of other ‘right’ answers under different sets of assumptions.

My knowledge maths isn’t up to that – but I haven’t heard that’s the case.

23. #23 Duncan Ivry
April 3, 2010

“In mathematics, you don’t understand things. You just get used to them. -Johann von Neumann”

This is a little bit funny, but as a mathematician, I can tell you: It is *not* true. And, well, no, today this not really funny any more, it is lame.

Ethan, don’t get it wrong: Handling series like the ones you showed here is basic mathematics, and has been handled during the first couple of semesters at university. Nobody was driven crazy by this.

24. #24 hipo
August 22, 2010

i love pie

25. #25 joshua
August 22, 2010

hihihihihiihihihihihihihihi do you like pizza

26. #26 poolover
August 22, 2010

do you like poo i most certainly do

27. #27 Recepti
February 3, 2012

Interesting math. Bravo for the idea!
http://kuhinjarecepti.com/