Riemann's paradox: pi = infinity minus infinity

Today I want to let you in on a really neat trick that I learned from a very famous mathematician. How to subtract infinity from infinity to get exactly pi. So for that I'll return to this blackboard here that I found in The Simpsons that I discussed in this video up there, and particularly that infinite series down there at the bottom. So it's 1 minus 1 half plus 1 third minus 1 fourth plus 1 fifth all the way to infinity, an infinite series with sum log 2. So that's about 0.69. What I want to do here is something that somebody discovered about 150 years ago. So just see what I do. It's a bit of a paradox. So what I do is I take a bit more of this and I really need the whole board. So we'll take out every fourth term and then what's left over has odd denominator and even denominator terms. So let's separate those out too. Now we're going to work our way in from the left, take those two guys and put them down there, take those two guys put them down in the gap, take those two guys and put them down in the next gap, and you keep on going like this. And obviously when you do this you use up all the terms that were there in the beginning and just rearrange them here and re-bracket them a little bit. Now we're going to work our way through the brackets. So first one here 1 minus 1 half is 1 half, 1 third minus 1 sixth is 1 sixth, and so on. So you can see how this works. It's a nice pattern here. That's what happens. So we've just rearranged the terms and we re-bracketed them a little bit and that should not change anything about the sum of this thing, right? So if you compare what we've got now to what we started with, well the bit that we started with is what? Log 2. But now we can compare term by term. So 1 half of 1 is 1 half, 1 half of 1 half is 1 fourth, and half of 1 third is 1 sixth, and so on. And so what you see is that term by term the bit at the bottom here is always 1 half of the bit at the top. So what that means is it should really be equal to 1 half log 2. Now if you put all this stuff together you get log 2 is equal to 1 half log 2, which really amounts to the same thing as saying that 1 is equal to 2. And whenever you come across something like this you start worrying. So there's really, you know, two different options here. First one is maths is broken or we made a mistake. What do you think is more likely here? Well, yes, we made a mistake. So where's the mistake? Let's have a close look. So there was three different identities. So here, here, and there. So two of these are correct and one of these is not correct. So the first bit, well I already said this and just believe me, this is equal to log 2. Then the bottom bit is also correct. In fact that bracketing here, and I just say this now, you can put in brackets any way you want. It's not going to change anything about this sum here. It's always going to add up to 1 half log 2. It's correct. So what's really messing things up is that we are rearranging the terms. So we're changing the order of the terms and that really creates an infinite series with a different sum than the one before. So basically when we're dealing with these infinite series, unlike with the finite ones, order matters. Rearranging the terms may change the sum. Okay, well, some people are getting really worried, oh no, maths is broken anyway. No, it's not broken, we just have to get used to this. I've got to be careful about this sort of stuff. So we just saw that we can rearrange this series into another series with a different sum. And actually there's a famous mathematician, Riemann, Bernard Riemann, who came up with a theorem, the Riemann rearrangement theorem, which says that, well, in this case of this particular series, you can rearrange this thing into series that add up to anything you want. For example, pi. And I'm going to show you how that is done, how he makes up these different sums. Now you've probably heard this name, the Riemann hypothesis is at the moment the most sought after thing that you can prove in mathematics. So eternal fame awaits if you can do this. And it's all about the Riemann zeta function. And there's also the Riemann integral, which you use all the time, and lots and lots of Riemann named bits in mathematics. So this guy is a mathematical superhero. Usually when it comes to these things that are named after him, they're quite difficult to explain properly. But that Riemann rearrangement theorem, and you can do like this with my animation magic. So let's just do it. But before we do it, I really have to remind you of what it actually means for one of those series to add up to a certain number. So what we do is we translate this series into a sequence of numbers, the partial sums, the sequence of partial sums. So the first partial sum is just the first term, which is 1. The second partial sum is just those two guys added up. That's 0.5 in this case. Then third partial sum, fourth partial sum. And so you get this sequence of numbers here. And in the case of this series, the sequence of numbers converges to a certain number, which is log2. And then we say that this series has sum that number, log2 in this case. So that's the definition of a sum of an infinite series. So this is an example of a convergent series, a series that actually has a sum, which is a finite number. Now there's also lots of series that don't converge, and they're called divergent. So for example, if we replace all the minuses in here by pluses, I already showed in that other video there, that this adds up to plus infinity. But there's other types of divergence. So for example, if we take this guy there, it's 1 minus 1 plus 1 minus 1. The partial sums here are 1, 0, 1, 0, 1, 0. They don't settle down to anything, not finite, not infinite. That's also divergent. So different sorts of divergent. To explain what's going on here, let's have a look at the positive terms and the negative terms separately. So if you just add up all the positive terms, you'll actually find you get infinity. And if you actually add up the negative terms, you get minus infinity. And actually, if you've watched the other video, you should be able to prove that yourself and maybe do it in the comments. So what we're really doing here is we're subtracting infinity from infinity in a controlled manner. What Riemann's rearrangement theorem now says is that if you've got a convergent series whose positive terms add up to infinity and whose negative terms add up to minus infinity, then you can rearrange the series into series that have any sum you want. So let me demonstrate this now for pi with this particular series. Okay, so we want pi. That's positive. So we're going to start with some positive terms. So the first term of my new series is going to be 1. So the first partial sum is also going to be 1. So partial sums I put up there. That's not enough, so we add a few more of those positive terms. Let's get close to pi. Now we actually have to add quite a few of them to get close to pi. We're still just under here. So let's add one more. That gets us just over. This string of terms we choose as the beginning bit of our new series. So we're just over. Now we're going to use some negative terms to get just under. Okay, so we use the, actually the minus one half is good enough. So that gets us under 2.6 and so on. All right, now let's go out over again. So for that we'll just use some positive terms. We actually need quite a few again, but we can be absolutely sure that we can get over because at any stage of this process the positive terms that are left over will add up to infinity and the negative terms that are left over will add up to infinity. So I can always be sure that no matter how far I'm under or how far I'm over, I can always take enough terms to get under or over. All right, now we can just keep on going like this. So then a negative term to get us under again, then a positive, positive terms again to get us over, and we go flip back and forth, and eventually we get to pi. And we can also be sure that we get to pi because the terms themselves, they get smaller and smaller in magnitude. So that means that as I overshoot and undershoot, I get closer and closer to pi. And actually I can rearrange this initial series in infinitely many ways, different ways, to get me pi. I can also rearrange it to get me an infinite sum, all sorts of other things. So maybe you also want to do this in the comments, figure out the details there. Now for other series you can actually have the situation where nothing changes no matter what you do. So there's lots of convergent series where the positive terms add up to a positive number and the negative terms add up to a negative number. So for example it could be, you know, 2, the positive terms, and minus 7, the negative terms. And then no matter how you reshuffle this series, the sum will always be 2 minus 7 is minus 5. Okay, now there's something else that we're actually doing here. We're actually bracketing. So we're bracketing. So we can bracket like this, or we can bracket like that, or maybe we don't do any brackets whatsoever. And actually as long as we're dealing with infinite series that converge, either this type or that type, it doesn't matter what, you can put in brackets any way you want and the sum won't change. So I mean that's a bit reassuring. So basically commutativity is broken but associativity, at least for convergent series, doesn't get mucked up. So it's quite nice. But there's at least one more surprise happening here. That's just for convergent series. So if you actually look at some divergent series, strange things can happen even when you bracket them. So for example this guy here, the 1 minus 1 plus 1 minus 1 divergent series, if you just bracket it right, you'll actually get a convergent series with a certain sum. And actually you can bracket this in many many different ways and get different sums out. And maybe that's also a puzzle for you to sort out. How can you re-bracket this thing to get different different sums out? Well there's really amazing things to be discovered yet. The first one is basically saying, and you can actually with what I showed you today, you can actually figure out the details yourself, is that if I give you an arbitrary infinite series there's three different cases. So you can ask in how many different sums can I rearrange this thing. So the first case is no matter how I rearrange this thing I'll never get a get a sum. Second case is it rearranges in all possible sums. And the third case is it just has one sum no matter how you rearrange it. Now this gets even freakier when you're considering infinite series of complex numbers or infinite series of vectors and you can do this in Banach space and all kinds of other things. But maybe that's enough for today.