▶︎ 0:00 Hi, I'm Joel David Hamkins, and I want to tell you about infinity in this series of lectures, the fascinating topic of infinity. I want to start with a classical thinker,
▶︎ 0:13 Zeno of Elea, who made a very interesting argument. Namely, Zeno argued, this was in about 450 BC, and Zeno argued that all motion is impossible. You cannot go from here to there. Let me just stand up. Of course, we know it's not true. You can go from here to there because I can walk from here to there. But Zeno argued that, "No, no, this is an illusion," and so it's no good to just say, "Well, we know the conclusion of Zeno's argument is false because we can go from here to there." Rather, we need to understand the argument itself and find the flaw in his reasoning.
▶︎ 0:54 And so Zeno argued like this. He said, "Well, suppose you want to go from here to there, from A to B. Then before you do that, before you go from A to B, you have to get halfway to B." That's something that you must do before you go from here to there. But before you get halfway there, you have to get halfway to the halfway point. So that's two things that you have to do before you go from here to there. And before you get halfway to the halfway point, you have to get halfway to that point and so on, and before you get halfway to the halfway to the halfway point you have to get halfway there and so on.
▶︎ 1:35 And so it seems impossible for you to move at all, because before you move at all, you would have already have to have gone halfway as far. And therefore, Zeno concluded that all motion is impossible. I find that quite interesting. It's a ridiculous conclusion, but maybe one struggles with the reasoning. Is it possible to do infinitely many things? Can you do infinitely many things? This seems to be the core problem that Zeno is identifying.
▶︎ 2:07 Let me tell you about another paradox that Zeno had talked about, Achilles and the tortoise. Achilles and the tortoise are going to have a race, and they're going to run around the stadium. They're both going to start here, and they're going to go all the way around. And the tortoise challenges Achilles, saying, "I'm going to win, but you have to give me a little bit of a head start." And how is it possible that the tortoise could win the race around the stadium against the great warrior Achilles?
▶︎ 2:38 Well, the tortoise argues like this. The tortoise gets a head start one-quarter of the way around, say at point A. So Achilles starts here and the tortoise starts at A, and they're off. And now the tortoise is moving while Achilles very rapidly gets to that point A. And the point is that, by the time Achilles gets to this point A, then the tortoise will have moved on to a point B. So the tortoise is still ahead.
▶︎ 3:14 And now, again, by the time Achilles gets to the point B, the tortoise will have also moved on because when Achilles arrived at A, the tortoise was already at B, and so they're both continuing from those points. And so there's a further point C such that, by the time Achilles gets to B, the tortoise has moved on to C, still ahead, and so on. So again, D and so on. So there's infinitely many points. Whenever the tortoise is at a place and Achilles is not yet there, then by the time Achilles gets there, the tortoise will have moved on. And so the tortoise argues that it's impossible for Achilles ever to catch up because he would need to do all those infinitely many things.
▶︎ 3:58 These puzzles are maybe related to the concept of supertasks, a task involving infinitely many steps or infinitely many actions. We'll have another lecture all about supertasks itself, but maybe Zeno's paradox is the origin of supertasks.
▶︎ 4:15 I want to talk about a slightly different way of understanding Zeno. Zeno had argued that all motion is impossible because before you get from here to there you need to get halfway, and before you do that you need to get halfway to that halfway point and so on. But then we can turn the argument around a little bit and argue like this. Namely, we cannot go from here to there because before going from here to there, we must have already gone halfway. So we must get to the halfway point. And then after that, we must get halfway of what remains. And then after that, we must get halfway of what remains again and so on. And so before arriving, we must have done those infinitely many things, which Zeno argues is impossible.
▶︎ 5:02 And so there's a way of understanding this argument maybe in a slightly more contemporary manner. If we think about the line segment from zero to one, then if we want to go from zero to one, we have to first get halfway there. And then now we're at the halfway point, and before we go from the halfway point to the end, we must get halfway to there, and there's another quarter there, so this is the three-quarter point. And then standing at the three-quarter point, we want to go halfway again. And so this is one-eighth of the total segment, and then one-sixteenth and so on.
▶︎ 5:45 Well, let's think about these numbers that we've written down. We can say the total length that we traverse is one-half, the initial one-half plus one-quarter plus one-eighth plus one-sixteenth and so on. And so we're adding up infinitely many numbers, and what does that really mean to add up infinitely many numbers? Well, it's quite clear, isn't it, that the total length of all of these segments will exhaust the original interval, and so this infinite series is adding up to one, the total original length. This is a kind of contemporary way of understanding what's going on in Zeno's paradox. So one-half plus one-quarter plus one-eighth plus one-sixteenth and so on adds up in total to one.
▶︎ 6:34 There's another way of arguing this. Let me show you how it goes. Let me show you another way of seeing this infinite summation. If we take the whole unit square, the one-by-one square. So this is a one-by-one square. The total area is one. And if I take half of it, like this, that's half of the total area. So there's half on this side, and I can take one-quarter of that. Half of what remains would be one-quarter. So therefore, this square is also one-quarter. And if I take half of that, it would be one-eighth. And this is also one-eighth, so half of it is one-sixteenth, and so on. I just keep chopping what remains in half.
▶︎ 7:23 And so the total area, which is one, can be thought of as one-half plus one-quarter plus one-eighth plus one-sixteenth, and so on. So it's another way of seeing that this infinite sum of numbers can nevertheless add up to a finite number, the number one.
▶︎ 7:53 Next I would like to tell you the most contested equation in middle school. Maybe some of you have heard of this equation, and maybe you've had arguments about it, and that's what I mean by it being the most contested equation in middle school. The equation is the following. We think about the number 0.9999 repeating, so the nines never stop. And the equation says that that number is in fact equal to one. It's the same as 1.000 forever. 0.999 repeating is the same as one. That's the most contested equation in middle school.
▶︎ 8:43 So let's give a couple of arguments for why this is true. Maybe the first argument is a quite common argument, which is, we want to understand, what is the value of this number? Sometimes people think it should be a little bit less than one, because it seems like there's all these nines, and so there should be something extra left over. But I want to argue that in fact that's not true, that in fact this equation is correct.
▶︎ 9:04 So if we call this number x, I'm going to let x be 0.999 repeating. And if I multiply x by 10, and we see 10x, whenever you multiply a number like this in decimal notation by 10, you just are moving the decimal point over. So when I multiply this number by 10, I get 9.999 repeating, with infinitely many nines. And then, if I set up a little subtraction, 10x here minus x, but x was 0.999 repeating, and if I just subtract 10x minus x on the left side, I've got 9x, and on the right-hand side, I've got 9.999 repeating minus 0.999 repeating, so all these nines are just going to cancel, and we just get nine. So 9x equals nine, and therefore, x must be one, just like we said. So therefore, 0.9 repeating is equal to one.
▶︎ 10:13 Sometimes mathematicians criticize this argument, because it presumes that 0.999 repeating is meaningful. I said let x be that, but that only makes sense if this expression does mean something. In fact, it does mean something, and we can give an argument for it, but we haven't given such an argument here.
▶︎ 10:32 And so I want to talk about, what is the meaning of a decimal expansion of this form? So let me give another argument for this contested equation though. Maybe it's less controversial to say that 0.3 repeating is equal to one-third. I think a lot of people know that. And if you divide one by three using the long division algorithm, then you're going to end up with this expression here. And now we can see, obviously 0.9 repeating is three times as big as 0.3 repeating, and therefore, whatever the value of this is should be three times as much as one-third, which is one. So it's another way of seeing that 0.9 repeating must be one.
▶︎ 11:27 Let's though get a little bit more basic and fundamental about what these things mean. When you have a number in decimal notation, say I write a number like 8547, 8,547, what it means is, this is a positional number system, so it means that we have eight thousands, and five hundreds, four tens, and seven ones. That's what the notation means. So we can write this out in scientific notation. 8 times 10 to the 3, plus 5 times 10 squared, plus 4 times 10 to the 1, plus 7 times 10 to the 0, which is really just one. So the meaning of our number notation means that it's the sum with one term for each digit.
▶︎ 12:32 When I write a number with digits after the decimal point, it means exactly the same kind of thing. So, for example, if I have a number like this, say, 3.14159265 and so on, this is the beginning of the number pi. What it means is, 3 plus, this 1 here means 1 times 10, oh, let me write it like this, 1 times one-tenth, plus 4 times 1/100 or 4 over 100, plus 1 over 1,000, plus 5 over 10,000 and so forth. So we get one digit here, corresponding to one term here multiplied by a power of 10, except now the powers of 10 have negative exponents. So the meaning of a number with infinitely many digits is precisely an infinite sum, where you have one term for each digit.
▶︎ 13:34 And so if you think about what it means to have the number 0.9 repeating, 0.999 repeating, what it means is nine tenths, that's the first nine, plus nine one-hundredths, that's the second nine, plus nine one-thousands, nine ten-thousands, and so on. So the very meaning of this expression is already an infinite sum of the kind we were looking at earlier. And this particular kind of sum has the property that each term in it is one-tenth as big as the previous term, and that's what's called a geometric series.
▶︎ 14:26 A geometric series is an infinite sum such that each term is a constant multiple of the previous term, in this case, one-tenth. So let's try to find the value of a general geometric series. Let me just look at this. For example, in our previous case, we had one-half plus one-fourth, plus one-eighth, plus one-sixteenth. This was the sum that comes up for Zeno, and it's a geometric series because each term is half as big as the previous one. We're multiplying by a half each time. That's a geometric series.
▶︎ 15:13 In general, for a geometric series, we start with a term A, and we multiply by this ratio, A sub r, A sub r-squared, A times r-cubed, and so on. Each term is multiplying by r times the previous term. So we want to know, what is, in general, the value of such a geometric series? This is exactly the kind of series that comes up with Zeno. So if the constant A, maybe it's easiest to just take A to be 1, so we're really considering 1 plus r, plus r-squared, plus r-cubed, and so on.
▶︎ 15:55 When we're trying to understand the value of an infinite sum with infinitely many terms, then what we should do is maybe think about what happens to the value if we take only finitely many of those sums. This is a kind of view of the meaning of the infinite sum which has a potentialist character. There's this distinction between potential infinity and actual infinity. Potential infinity is the idea that you never complete the infinite task. You only take more and more, finitely much of it, at a time. So the very meaning of an infinite sum has a potentialist character because we look at what happens to the values of the finite partial sums as we take more and more in those series.
▶︎ 16:40 So if I just take the numbers, say, r-squared, r-cubed, up to r sub n. I'm taking the terms all the way up to r sub n, and I want to know, what is the value of that part of the infinite series? This is what I want to call x. Let's call it x. So x is the number which is obtained by adding up the geometric series up to the nth power. So 1 plus r, plus r-squared, r-cubed, and so on. And let's think about what happens if we add the next term, x plus r to the n plus 1. That's the very next term. So this is 1 plus r, plus r-squared, and so 1 plus r to the n, and then one more term, r to the n plus 1.
▶︎ 17:35 And now, if I think about after the number 1 here, I can do a little bit of mathematics. All of these terms are multiples of r, so I can factor r out. So x plus r to the n plus 1 is equal to 1 plus r. Then this is r times 1, plus r, and so on, all the way up to r to the n. So I've taken the latter terms here, they're all multiples of r, and I factor an r out, so all of the exponents drop by one. So r times 1 is r, r times r is r-squared, r times r-squared is r-cubed, and so on, and the last one will be r times r to the n, which is r to the n plus 1. But now, the reason I did that, is that this is x again.
▶︎ 18:24 So what we see is that x plus r to the n is equal to 1 plus r times x. So let me write that equation up at the top here. What we have is x plus r to the n plus 1 is equal to 1 plus rx. That's what we've derived here. But this is an equation that we can solve for x. If I just do some algebra, I put this rx on the other side with a minus sign, so this is x minus rx is equal to 1 minus r to the n plus 1, and now these are both multiples of x, so I can factor 1 minus r out. So x times 1 minus r equals 1 minus r to the n plus 1.
▶︎ 19:07 And so altogether, this means that x equals 1 minus r to the n plus 1 over 1 minus r. So we found exactly the value of the finite partial sum, and now we can understand what happens with this as n becomes bigger and bigger. So remember, we have this idea of the meaning of the infinite sum. So maybe n is 100 and I'm taking 100 terms, and then I get this particular value. Or maybe n is 1,000 or 1 million, and so I'm going to let n get bigger and bigger. And this formula here tells me exactly the value of the finite partial sum, if I go out to that point.
▶︎ 19:51 Now, in the case, let's summarize what we have. We're interested in the sum 1 plus R plus R squared plus R cubed and so on, the infinite sum. But we've observed that if I just go out and take N terms, then what I get is this value here. So if R is bigger than 1, or even at least 1 or bigger, then this infinite sum will become infinite, because the finite partial sums will just get bigger and bigger. If R is big, then all these terms are going to be bigger than 1, and I'm just adding up more and more numbers that are all very big, and so the sum is just going to get bigger and bigger, and it will be infinite in the end.
▶︎ 20:49 But if R is less than 1, maybe the absolute value of R is less than 1, so maybe it's negative, then I can think about what happens. So that was what was true in the geometric series for Zeno where R was a half, remember each term was half as big as the previous one. So we're trying to understand how could it be that infinitely many numbers add up to a finite sum? That's the puzzle of Zeno. And so we're trying to explain exactly how that can happen.
▶︎ 21:23 If R is less than 1, then if I think about the finite partial sum, it's given by this expression, and the point is that as N becomes very large, I'm multiplying this number that's less than 1 by itself many, many times in this term here. So if I multiply a number less than 1 by itself many, many times, it just gets smaller and smaller and smaller. I can make it as small as I like. So the point is that as N becomes large, the value of the finite partial sum goes to just 1 over 1 minus R. That's all that's left.
▶︎ 21:59 So we can make the value of the finite partial sum as close to 1 over 1 minus R as we like, just by taking N large enough. By taking enough terms, we're going to make it as close to 1 over 1 minus R as we like. And therefore, what that means is that the total value of the geometric sum is 1 over 1 minus R. When we start with 1 plus R plus R squared, R cubed, and so on, we're going to get 1 over 1 minus R.
▶︎ 22:30 In general, in the case when A isn't 1, this same kind of reasoning, if we have a series A plus AR plus AR squared plus AR cubed and so on, we're going to get A over 1 minus R. This is the value of the geometric series. And what it means is that no matter how close you want to be to this value, then if you take enough terms, all the finite sums that have at least that many terms will be within that tolerance of the limit value. So that's the kind of potential, this understanding of the value of an infinite series.
▶︎ 23:13 So let's apply this formula to the Zeno case where we had one-half plus one-quarter plus one-eighth plus one-sixteenth and so on. So this is the case where A is a half, we started with a half, and R was also half because each time we were multiplying the previous term by R. So the total sum here will be A, which is one-half, over 1 minus one-half, and one-half over 1 minus a half, well, 1 minus a half is a half, so it's a half over a half, which is 1. And that's exactly what we said. This sum is exactly 1 in the case of Zeno.
▶︎ 23:55 And if we think about the most contested equation in middle school, the 0.9 repeating, well, this, as we said, is nine-tenths plus nine-one hundredths plus nine-one thousandths and so on, and so this is a case where A is nine-tenths, so we have A is nine-tenths, that's what we started with, and R is one-tenth because each one is one-tenth as big as the previous one. So therefore, the total sum is going to be A over 1 minus R, so nine-tenths over 1 minus one-tenth. But that's nine-tenths over nine-tenths, which is one.
▶︎ 24:42 So therefore, it's another way of seeing that 0.9 repeating, not only is, must be equal to one if it has any meaning at all, but furthermore, this argument shows that it does have a meaning, it does converge to one because we can make the value by taking more terms, more digits here, we can make the value be as close to one as we like. Okay, so maybe that's the most contested equation of middle school.
▶︎ 25:11 So let's look at some other interesting series. There's another very interesting case of this. You might say, the confusing thing about these infinite series is that it seems impossible maybe at first that you could add up infinitely many numbers to make a finite answer. Nevertheless, we argued that there's cases where you do get a finite answer, namely one-half plus one-quarter plus one-eighth, one-sixteenth, and so on, we add up all of those infinitely many numbers, and yet we still got a finite answer, equal to one. In that case, that's an instance of the geometric series.
▶︎ 25:49 Now obviously, we have other cases where we're adding up infinitely many numbers and we don't get a finite answer. If I add up one plus two plus three plus four and so on, then obviously, this is not going to be converging to any finite number. It's impossible for that to be a finite answer. But that's a case where the individual numbers are getting larger and larger. Even one at a time, they're getting larger, so when we add them up all together, of course they're not going to be finite.
▶︎ 26:15 What about a case like one plus one plus one plus one and so on forever? If I add up infinitely many ones, obviously I'm not going to get a finite answer. I can make that number bigger than any given bound by just taking enough terms. So maybe one might think, maybe if the individual terms in the sum are getting smaller, as they are in a geometric series, like one-half plus one-quarter plus one-eighth, the terms are getting smaller, and maybe that's what it takes for them to add up to a finite sum.
▶︎ 26:45 So let's consider another case that's like that. This is a famous series. It's called the harmonic series. The harmonic series is one plus one-half plus. So far, it looks like the geometric series. But instead of putting one-quarter, I'm going to do one-third here, and then one-quarter, and then one-fifth, one-sixth. I just take the next number in the denominator each time. That's the harmonic series.
▶︎ 27:35 And we want to know, what does it add up to? Does it have a sum? And the numbers individually are getting smaller, which is what we had said should be a necessary condition for it to have a finite value. But let's think a little bit more about it. Suppose I've added up many, many terms. Maybe I've got one plus a half plus a third plus a fourth and so on, and I've gone all the way out to one over n. So I've got the first n terms. Maybe n is a billion.
▶︎ 28:17 Then what I'm going to do is I'm going to think about doubling the number of terms. So the next term here is one over n plus one, plus one over n plus two, and I keep going. And if I have n terms, then the last one will be one over 2n, n plus n. So I had a billion terms, now I have two billion terms. And the question is, how much more did I add?
▶︎ 28:44 Audience: Yes?
▶︎ 28:45 And so let's think about what this extra bit is. Well, I've got n terms, and they're getting smaller. So in particular, each of them is at least as big as the last one. Each term is at least as big as one over 2n. So therefore, if I add up n numbers and each of them is at least that big, then the total is at least n times that bound, n times one over 2n, which is a half.
▶︎ 29:36 Let's think about that. So I said, no matter how many terms you add up, if you take that many terms again, then those terms, well, you'll have n of them, and each of them is at least as big as the last one, which has value one over 2n. So the total of this extra part I added was n times one over 2n, which is a half, because the ns cancel. So what I've said is, no matter how much you have, I can always add one half by doubling the number of terms.
▶︎ 30:10 Well, suppose I want to add another one half. Well, I just double again. And now I've added one to the total, and now, if I want to add another one-half, I just double it again, and double it again. I can always add an extra one-half. So if I keep doing that, I can add an extra 17 if I want. I just double it 34 times. So what we've shown is that no matter how many terms in the harmonic series you take, you can always add one-half more by just taking twice as many terms.
▶︎ 30:40 So therefore, it can't be that the harmonic series converges to a finite value, because for it to converge to a finite value would have to mean that if you take enough terms, you can get very close to that value, and no matter how many more terms you take, you'll still be very close to that value. But that won't be true for the harmonic series because we can always add one-half more, or 2 more, or 5 more, just by doubling the number of terms enough times. So therefore, the harmonic series is famous for being divergent. It does not converge to a finite value, even though the terms individually are going to zero. So, this is a totally different situation than the Zeno situation with the geometric series.
▶︎ 31:26 Let's twist things around a little bit. Maybe it's a little bit surprising why I'm talking about all this math and these series and so on, even though we started with a philosophical idea of Zeno's paradox, but the point I want to make about this is that, really, when you're looking at the philosophy of infinity, it blends into the mathematics so gradually and easily, and the way of using our mathematical knowledge to understand the nature of infinity is quite a natural thing to do. And so, one is pushed towards these mathematical ways of thinking.
▶︎ 32:04 Let's look at what is called the alternating harmonic series, and this is the series 1 minus 1/2 plus 1/3 minus 1/4 plus 1/5, and so on. So it's alternating because the signs vary, positive, negative, positive, negative and so on. So every other term is positive, and every other term is negative, and that's what makes it an alternating series. We saw that if they're all positive, then it doesn't converge to a finite number. It adds up to infinity. But what about this one? Because sometimes we're subtracting instead of adding.
▶︎ 32:49 Let's think about what happens if we try to do this. Let me make a graph. Here's the number of terms. So we're starting with 1, we have the value 1. But then we subtracted 1/2, so then we're going down to 1/2. But then we add in 1/3, so we go up to 1/3 on the next, so now we're at 1/2 plus 1/3, which is not quite 1. It's less than 1, because we add 1/2 plus 1/3, don't get all the way to the top. And now we subtract 1/4, which is going down, but it's going down by less than the amount that we just went up, because this was minus 1/2, and this was plus 1/3, and now we're going to go minus 1/4, and up 1/5, and so on. Minus 1/6, up 1/7, and so on.
▶︎ 33:46 So it has this zigzag notion, but the thing to observe is that every time we go up, then the next time we go down, we go down by less than we just went up. And every time we go down, then the next time we go up, but we go up by a smaller amount than we just went down by, and so what that means is that this upper envelope of the points above are going down, and the lower envelope, the ones that are below, right after we subtract, are going up each time. Because every time we go up, the next time we go down by less than that, and every time we go down, then the next time we go up, but by less than the amount that we just went down. So that's a way of seeing that these two curves are exactly, the upper one is descending and the lower one is ascending.
▶︎ 34:38 But furthermore, the distance between those curves is exactly the magnitude of the differences that we're zigzagging by. So therefore, those two curves are getting closer and closer together, as close as we like, and in fact, one can prove that they're converging exactly to the natural log of 2, which is about 0.69 and so on. So therefore, the alternating series is convergent, it has a finite value, and it's the natural log of 2. So this is a case where, when all the terms were positive, it added up to infinity, but if some of them are negative, then it added up to log 2.
▶︎ 35:30 Now I want to tell you about something amazing about this kind of series. This is called a conditionally convergent. It's a series that converges to a finite answer, but if you take the absolute value of each term, if you make them all positive, then it no longer converges to a finite value. So in particular, I can't view this series as first taking the positive numbers and then taking the negative numbers, and subtracting the two.
▶︎ 36:12 You might say, "Well, if I'm doing 1 minus 1/2 plus 1/3 minus 1/4 and so on, that should be the same thing as 1 plus 1/3 plus 1/5 plus 1/7 altogether, minus one-half plus one-fourth plus one sixth and so on, the negative terms." You can't group the positive terms and the negative terms together and then subtract them, because the positive terms add up to infinity and the negative terms also add up to minus infinity. And therefore, the subtraction doesn't make sense, even.
▶︎ 36:45 And there's a profound theorem about this kind of situation, and it's called Riemann. Oh my god, am I spelling it right? Is it E-I or I-E? I have to look up. It's I-E. So there's a profound theorem called the Riemann Rearrangement Theorem. Riemann Rearrangements. The Riemann Rearrangement Theorem says that if you have a conditionally convergent series like this, then you can rearrange the terms to make the value whatever you want, any value for any conditionally convergent series, not just this one, any conditionally convergent series. And any target value, then there's a way of rearranging the terms so that it now has that value.
▶︎ 37:39 So let me show you how to prove it. So if we do it with, say, the alternating harmonic series, this particular series. So suppose we want the value to be 1.4. Log2 is less than 1, it's 0.69, so I want it to be a little bit bigger. So this is going to be the target, the new target. And what I'm going to do is, well, I'm just going to keep taking the positive terms until I'm bigger than the target. So, 1 plus a third. That's 1.33 and so on, so it's not quite 1.4, but if I add 1 plus a third plus a fifth, then I think that's bigger than 1.4, oh my god. Let's see, this is 1.3, one-fifth is .2, so that would be, like, 1.53 and so on, so definitely bigger than 1.4.
▶︎ 38:33 Then once I've exceeded the target, then I start taking negative terms until I'm below the target, so minus a half. So this is already going to be below the target, so therefore, with one negative term I already got below the target, and now I start taking positive terms again. So the next positive term is going to be one-seventh, and so on. So the point is that if you have some target. Here's the original value of log2, and remember, there was that zigzag thing happening, but now I want a new target of 1.4, then what I should do is I should take the first value going up and the next one and the next one until I'm above the target, and then I take the value going down below, and then I take values going up until I'm above, yeah? And maybe I'm not quite reaching, so I need, whenever I'm still below after adding a positive term, then I take another positive term until I get above the target.
▶︎ 39:30 And the point is that you can always do that, precisely because the positive terms add up to infinity, so you can make, by taking finitely many, you can make it as big as you want. So you can always get above the target again by taking enough of the positive terms, and therefore you can always get below the target again by taking enough of the negative terms. And so if you just do that process, what you're doing is zeroing in on this target value, and so the rearranged form of the alternating harmonic series will add up to exactly your target.
▶︎ 40:03 I view that as quite profound, because what it means is that when you're adding up infinitely many numbers, it's just not true that you get the same answer when you rearrange them. That's not true. If the numbers that you're adding up come from a conditionally convergent series, then the order in which you're adding them can affect the numerical answer of the result, which I think is profound and surprising. And so all of these observations maybe shed light on what's going on with Zeno's paradox, in my view.
▶︎ 40:38 Zeno is troubled by the possibility of doing infinitely many things in a finite space of time. And so this is also connected with the concept of supertasks, which we're going to talk about in the next lecture. But one way of understanding specifically the idea of going from here to there, we were led to the concept of the geometric series of going halfway and then a quarter and then an eighth and so on. And adding up all those lengths and getting a finite answer, it's a way of understanding how it could be that an infinite sum nevertheless has a finite answer.
▶︎ 41:11 And so we were able to calculate using the geometric series that, in some cases, when we can describe exactly the geometric series, we can know exactly what the numerical answer is. And then there's the wrench in the works when we have alternating series that have sometimes negative terms and sometimes positive terms, because they can be convergent to an answer, but now the order in which you take those terms can affect the final answer, with the Riemann Rearrangement Theorem.
▶︎ 41:38 Well, I hope you enjoyed that story about Zeno's paradox, and it led us into considering geometric series and adding up infinitely many numbers to have a finite answer, and then there was the twist with the alternating harmonic series, and ultimately, we found out about Riemann's Rearrangement Theorem, a profound idea that, actually, when you're adding up infinitely many numbers, then the order in which you take those numbers can affect the results. So thank you very much.