Zeno's Paradox and Infinite Sums—Lectures on Infinity (lecture 1)
An ancient puzzle leads ultimately to a remarkable observation on the malleable nature of infinite sums.
Welcome to the Lectures on Infinity, a series of lectures exploring all my favorite paradoxes and conundrums.
In this first lecture, we introduce the ancient puzzle of Zeno’s paradox and follow the thread where it leads—to the very meaning of our number expressions, to infinite summations, to convergent and divergent series. We will discuss what I call the most contested equation in middle school, and by the end we shall reach the remarkable Riemann rearrangement theorem, concerning the malleable nature of infinite sums.
I shall be gradually sharing the individual lectures here on Infinitely More in the coming weeks and months.
Please enjoy!
Find the lectures here on Infinitely More in the lectures-on-infinity tag.
The lectures will appear on YouTube.
The whole lecture course is hosted at Ergo: Lectures on Infinity.
Find other philosophy lecture courses at Ergo.org.
This lecture is based on my essay Zeno’s Paradox.
The essay also appears in my new book, The Book of Infinity.
Zeno’s Paradox and Infinite Sums
A lightly edited transcript. Timestamps link to the video on the Ergo website.
Welcome to Infinity
I’m Joel David Hamkins, and in this series of lectures I want to tell you about infinity, that most fascinating of topics. I want to start with a classical thinker.
Zeno’s Paradox: Why All Motion Is Impossible
Zeno of Elea made a very interesting argument, around 450 BC, to the effect that all motion is impossible. You cannot go from here to there. Now, of course we know the conclusion is false, because we can simply walk from one place to another. But it is no good to simply say, “Well, we know Zeno’s conclusion 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.
Zeno argued as follows. Suppose you want to go from point A to point B. Before you can go from A to B, you must first get halfway to B. But before you can get halfway to B, you must first get halfway to the halfway point. And before you can get halfway to the halfway point, you must first get halfway to that point, and so on. Before you move at all, it seems you would already have to have gone halfway as far, which generates an infinite regress of tasks that must be completed before any motion can begin. Zeno concluded from this that all motion is impossible.
I find this quite interesting. The conclusion is of course absurd, but one may still struggle with the reasoning. The core problem Zeno is identifying seems to be this: is it possible to do infinitely many things?
Achilles vs the Tortoise
Let me tell you about another of Zeno’s paradoxes: Achilles and the tortoise. Achilles and the tortoise are going to have a race around the stadium, both starting from the same point and running all the way around. The tortoise challenges Achilles, claiming to be able to win, on the condition that it is given a small head start.
The tortoise’s argument runs as follows. Suppose the tortoise begins one-quarter of the way around the track, at a point we will call A, while Achilles starts from the beginning. They set off, and Achilles very rapidly reaches point A. But by the time Achilles arrives at A, the tortoise has moved on to a new point, B. So the tortoise is still ahead.
Now the same reasoning applies again. By the time Achilles reaches B, the tortoise has moved on to a further point C, and is still ahead. Then by the time Achilles reaches C, the tortoise has moved on to a point D, and so on. At every stage, whenever the tortoise occupies some point and Achilles has not yet arrived there, by the time Achilles does arrive, the tortoise will have moved on.
The tortoise therefore argues that it is impossible for Achilles ever to catch up, because doing so would require completing infinitely many such steps.
What Are Supertasks?
These puzzles are perhaps related to the concept of supertasks: a task involving infinitely many steps or infinitely many actions. We will have another lecture devoted to supertasks itself, but Zeno’s paradox may well be the origin of the concept.
Turning Zeno into an Infinite Sum
I want to discuss 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 we can turn the argument around and put it this way: we cannot go from here to there because, before going from here to there, we must have already gone halfway. Then after reaching the halfway point, we must get halfway of what remains, and then halfway of what remains again, and so on. Before arriving, we must have done infinitely many things, which Zeno argues is impossible.
There is a way of understanding this argument in a more contemporary manner by thinking about the line segment from zero to one. If we want to go from zero to one, we must first get halfway there. Now standing at the halfway point, before we go from there to the end, we must get halfway to there, which brings us to the three-quarter point. Standing at the three-quarter point, we go halfway again, covering one-eighth of the total segment, then one-sixteenth, and so on.
Consider the numbers we have written down. The total length traversed is one-half, the initial one-half, plus one-quarter, plus one-eighth, plus one-sixteenth, and so on. We are adding up infinitely many numbers, and what does it really mean to add up infinitely many numbers? It is quite clear that the total length of all of these segments will exhaust the original interval, and so this infinite series adds up to one, the total original length. This is a contemporary way of understanding what is going on in Zeno’s paradox: one-half plus one-quarter plus one-eighth plus one-sixteenth, and so on, adds up in total to one.
There is another way of seeing this infinite summation. Consider the unit square, a one-by-one square with total area one. If I take half of it, that portion has area one-half, and then half of what remains is one-quarter. Taking half of that gives one-eighth, half again gives one-sixteenth, and so on, each time chopping what remains in half. The total area, which is one, can therefore be thought of as one-half plus one-quarter plus one-eighth plus one-sixteenth, and so on. This is another way of seeing that an infinite sum of numbers can nevertheless add up to a finite number, namely one.
The most contested equation in middle school
Next I would like to tell you about the most contested equation in middle school. Perhaps some of you have heard of it, and perhaps you have had arguments about it, which is precisely what I mean by calling it the most contested equation in middle school. The equation concerns the number 0.9999 repeating, where the nines never stop, and it says that this number is in fact equal to one. That is, 0.999 repeating is the same as 1.000 forever. Let me give a couple of arguments for why this is true.
Sometimes people feel that 0.999 repeating should be a little bit less than one, since there seem to be all these nines with something extra left over. I want to argue that this intuition is mistaken and that the equation is correct. Let x be 0.999 repeating. If I multiply x by 10, I obtain 10x, and multiplying a decimal number by 10 simply moves the decimal point one place to the right, giving 9.999 repeating, with infinitely many nines after the decimal point.
Now consider the subtraction 10x minus x. On the left side this is simply 9x. On the right side, 9.999 repeating minus 0.999 repeating causes all the nines after the decimal point to cancel, leaving exactly 9. So 9x equals 9, and therefore x equals 1, just as claimed: 0.999 repeating is equal to one.
Mathematicians sometimes criticize this argument on the grounds that it presupposes 0.999 repeating is meaningful. When I said “let x be that number,” the step only makes sense if the expression actually denotes something. In fact it does denote something, and an argument can be given for that, but we have not given such an argument here.
What Decimal Notation Actually Means
Let me give another argument for this contested equation. Perhaps it is less controversial to say that 0.3 repeating is equal to one-third. Many people know this, and if you divide one by three using the long division algorithm, you arrive at exactly that expression. Now, 0.9 repeating is simply three times 0.3 repeating, and therefore its value must be three times one-third, which is one. So that is another way of seeing that 0.9 repeating must equal one.
Let us get a little more basic and fundamental about what these expressions actually mean. When you write a number in decimal notation, say 8,547, this is a positional number system, and the notation means that we have eight thousands, five hundreds, four tens, and seven ones. We can write this out as 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. The meaning of the notation is precisely a sum with one term for each digit.
The same principle applies when we write digits after the decimal point. Take a number like 3.14159265 and so on, the beginning of pi. What it means is 3, plus 1 times one-tenth, plus 4 times one-hundredth, plus 1 times one-thousandth, plus 5 times one ten-thousandth, and so forth. Each digit corresponds to one term in the sum, multiplied by a power of ten, except that now those powers of ten carry negative exponents.
The meaning of a number with infinitely many digits is therefore precisely an infinite sum, with one term for each digit. When we write 0.999 repeating, what it means is nine-tenths, plus nine one-hundredths, plus nine one-thousandths, plus nine ten-thousandths, and so on. The very meaning of that expression is already an infinite sum of the kind we were examining earlier, and this particular sum has the property that each term is one-tenth as large as the previous one. That is what is called a geometric series.
Deriving the Geometric Series Formula
A geometric series is an infinite sum in which each term is a constant multiple of the previous term. In our earlier example, drawn from Zeno, we had one-half plus one-fourth plus one-eighth plus one-sixteenth, and so on. Each term is half as large as the one before it, so we are multiplying by one-half each time. In general, a geometric series begins with a term a and proceeds through ar, ar-squared, ar-cubed, and so on, where each term is r times the previous one.
To find the value of such a series in full generality, it is easiest to take a equal to 1, so we are considering 1 plus r plus r-squared plus r-cubed, and so on. When we try to understand the value of an infinite sum, what we should do is think about what happens to the value when we take only finitely many of those terms. This approach has a distinctly potentialist character. The distinction between potential infinity and actual infinity is relevant here: potential infinity is the idea that one never completes the infinite task but instead takes more and more, finitely much at a time. The very meaning of an infinite sum carries this potentialist character, because we look at what happens to the values of the finite partial sums as we include more and more terms.
So let us call x the value obtained by summing the geometric series up to the nth power: that is, x equals 1 plus r plus r-squared plus r-cubed, all the way up to r to the n. Now consider what happens when we add the very next term, r to the n plus 1. We get x plus r to the n plus 1, which equals 1 plus r plus r-squared, and so on up to r to the n, and then one further term, r to the n plus 1. After the leading 1, every remaining term is a multiple of r, so we can factor r out of those terms, reducing each exponent by one. The result is that x plus r to the n plus 1 equals 1 plus r times the quantity 1 plus r plus r-squared, and so on up to r to the n, which is simply 1 plus r times x.
We have therefore derived the equation x plus r to the n plus 1 equals 1 plus rx, and this is an equation we can solve for x. Moving the rx term to the left-hand side gives x minus rx equals 1 minus r to the n plus 1. Factoring the left-hand side yields x times 1 minus r equals 1 minus r to the n plus 1, and so altogether we find that x equals 1 minus r to the n plus 1, divided by 1 minus r. This gives us the exact value of the finite partial sum out to that point.
With this formula in hand, we can now understand what happens as n becomes larger and larger. Perhaps n is 100, or 1,000, or 1 million. The formula tells us precisely the value of the finite partial sum at each such stage, and by letting n grow without bound we can grasp the meaning of the infinite sum itself, in keeping with that potentialist conception of infinity.
When Do Infinite Sums Converge?
Let us summarize what we have. We are interested in the infinite sum 1 plus R plus R squared plus R cubed and so on. We have observed that if we take N terms, we obtain a specific finite value. If R is greater than or equal to 1, then this infinite sum will become infinite, because the finite partial sums will simply grow without bound. All the terms will be at least as large as 1, and adding more and more such numbers causes the sum to grow indefinitely.
But if the absolute value of R is less than 1, including the possibility that R is negative, then we can reason carefully about what happens. This is precisely the situation in Zeno’s geometric series, where R was one-half and each term was half as large as the previous one. The puzzle of Zeno is exactly this: how can infinitely many numbers add up to a finite sum? That is what we are trying to explain.
When the absolute value of R is less than 1, the finite partial sum is given by a certain expression, and the key point is that as N becomes very large, we are multiplying a number less than 1 by itself many, many times. A number less than 1 raised to higher and higher powers gets smaller and smaller, and we can make it as small as we like. Therefore, as N grows large, the value of the finite partial sum approaches 1 over 1 minus R, which is all that remains in the expression.
By taking enough terms, we can make the finite partial sum as close to 1 over 1 minus R as we like. This means that the total value of the geometric series 1 plus R plus R squared plus R cubed and so on is equal to 1 over 1 minus R. More generally, if the series begins with a leading coefficient A, so that we have A plus AR plus AR squared plus AR cubed and so on, the same reasoning gives us A over 1 minus R as the value of the geometric series.
What this means precisely is that no matter how close you wish to be to this limiting value, if you take enough terms, every finite partial sum with at least that many terms will fall within your chosen tolerance of the limit. That is the proper understanding of what it means for an infinite series to have a value.
Applying the Formula to Zeno and 0.999...
Let us apply this formula to the Zeno case, where we had one-half plus one-quarter plus one-eighth plus one-sixteenth and so on. This is the case where a is one-half, since that is the term we started with, and r is also one-half, because each term is obtained by multiplying the previous one by r. The total sum is therefore a over 1 minus one-half, and one-half over one-half is simply 1. That is exactly what we said: this sum is exactly 1 in the case of Zeno.
Now consider what is perhaps the most contested equation in middle school: 0.9 repeating. As we noted, this equals nine-tenths plus nine-hundredths plus nine-thousandths and so on. This is a case where a is nine-tenths, since that is the first term, and r is one-tenth, since each term is one-tenth as large as the previous one. The total sum is therefore a over 1 minus r, which is nine-tenths over 1 minus one-tenth, and that is nine-tenths over nine-tenths, which equals 1.
This gives us another way of seeing that 0.9 repeating must equal 1. Not only must it equal 1 if it has any meaning at all, but this argument shows that it does have a meaning: it converges to 1, because by taking more and more terms, more digits, we can make the value as close to 1 as we like. That is perhaps the most contested equation of middle school.
The Harmonic Series Blows Up
Let us consider some other interesting series. The confusing thing about infinite series is that it can seem impossible, at first, that one could add up infinitely many numbers and arrive at a finite answer. Nevertheless, we argued that there are cases where a finite answer does result: one-half plus one-quarter plus one-eighth plus one-sixteenth, and so on. We add up all of those infinitely many numbers and still obtain a finite answer, equal to one. That is an instance of the geometric series.
Obviously, there are also cases where adding up infinitely many numbers does not yield a finite answer. If I add one plus two plus three plus four and so on, this cannot converge to any finite number. But that is a case where the individual terms are getting larger, so of course their sum will not be finite. Similarly, if I add up infinitely many ones, one plus one plus one plus one forever, I will never reach a finite answer; I can make that sum exceed any given bound simply by taking enough terms.
One might therefore think: perhaps what it takes for an infinite sum to be finite is that the individual terms must be getting smaller, as they do in the geometric series, where the terms one-half, one-quarter, one-eighth, and so on decrease toward zero. Let us consider another famous series of exactly that kind. It is called the harmonic series, and it goes: one plus one-half plus one-third plus one-quarter plus one-fifth plus one-sixth, and so on, taking the reciprocal of each successive integer. The individual terms are indeed getting smaller, which we said should be a necessary condition for the sum to be finite. But let us think more carefully about whether the series actually converges.
Suppose we have added up many terms, say the first n terms: one plus one-half plus one-third plus one-quarter, all the way out to one over n. Perhaps n is a billion. Now consider doubling the number of terms. The additional terms are one over n plus one, plus one over n plus two, and so on, up to one over 2n. That is exactly n new terms. Each of these new terms is at least as large as the last one, which is one over 2n. So the total contribution of this additional block is at least n times one over 2n, which equals one-half, since the n’s cancel.
What this shows is that no matter how many terms you have already summed, you can always add at least one-half more to the total simply by doubling the number of terms. If you want to add another one-half on top of that, double again. If you want to add yet another one-half, double again. You can add an extra 17 to the sum, if you like, just by doubling 34 times. There is no ceiling.
Therefore, the harmonic series cannot converge to a finite value. For convergence to a finite value would require that once you have taken enough terms, you are very close to that value and remain close no matter how many additional terms you take. But that is precisely what fails here: we can always add one-half more, or two more, or five more, simply by doubling the number of terms enough times. The harmonic series is thus famous for being divergent. It does not converge to a finite value, even though its individual terms go to zero. This is an entirely different situation from the Zeno case and the geometric series.
The Alternating Harmonic Series
Let me twist things around a little. Perhaps it seems surprising that I have been talking about all this mathematics, about series and sums, when we started with the philosophical idea of Zeno’s paradox. But the point I want to make is that, when you are looking at the philosophy of infinity, it blends into mathematics so gradually and naturally that one is simply pushed toward these mathematical ways of thinking. Our mathematical knowledge is a quite natural tool for understanding the nature of infinity.
So let us look at what is called the alternating harmonic series. This is the series 1 minus 1/2 plus 1/3 minus 1/4 plus 1/5, and so on. It is called alternating because the signs vary: positive, negative, positive, negative, and so on. We saw earlier that if all the terms are positive, the harmonic series does not converge to a finite number; it adds up to infinity. But what about this one, where we are sometimes subtracting instead of adding?
Consider a graph with the number of terms along one axis and the running total along the other. We start at 1. Then we subtract 1/2, dropping to 1/2. Then we add 1/3, rising to 1/2 plus 1/3, which is less than 1. Then we subtract 1/4, going down, but by less than we just went up. Then up by 1/5, down by 1/6, up by 1/7, and so on. The series has this zigzag character, but with a crucial feature: every time we go up, the subsequent downward step is smaller than the upward step that preceded it, and every time we go down, the subsequent upward step is smaller than the downward step that preceded it.
This means that the upper envelope of the zigzag, the values recorded just after each upward step, is descending, while the lower envelope, the values recorded just after each downward step, is ascending. Moreover, the distance between these two envelopes equals the magnitude of the steps we are zigzagging by, and since those steps are shrinking, the two envelopes are drawing closer and closer together. One can prove that they converge to exactly the natural log of 2, which is approximately 0.69. The alternating harmonic series therefore converges; it has a finite value, and that value is the natural log of 2.
So here we have a striking contrast. When all the terms are positive, the harmonic series adds up to infinity. But when the terms alternate in sign, the series adds up to log 2. The mere introduction of subtraction transforms a divergent sum into a convergent one.
Rearranging Terms Changes the Answer
There is something remarkable about this kind of series. It is called a conditionally convergent series: a series that converges to a finite answer, but if you take the absolute value of each term, making them all positive, it no longer converges to a finite value. In particular, you cannot view such a series as first summing the positive terms and then summing the negative terms and subtracting the two. You might think that 1 minus 1/2 plus 1/3 minus 1/4 and so on should be the same as taking all the positive terms, 1 plus 1/3 plus 1/5 plus 1/7 and so on, and subtracting all the negative terms, 1/2 plus 1/4 plus 1/6 and so on. But you cannot group them that way, because the positive terms add up to infinity and the negative terms also add up to infinity, and so the subtraction does not make sense at all.
There is a profound theorem that addresses exactly this kind of situation: the Riemann Rearrangement Theorem. It states that if you have a conditionally convergent series, you can rearrange its terms to make the sum equal to whatever target value you like. For any conditionally convergent series, not just this one, and for any target value, there is a rearrangement of the terms that converges to that target.
Let me show you how the proof works, using the alternating harmonic series as our example. Suppose we want the new sum to equal 1.4. Since log 2 is approximately 0.69, that target is comfortably above the original value. The strategy is straightforward: keep adding positive terms until you exceed the target. For instance, 1 plus 1/3 is about 1.33, which is not yet above 1.4, but 1 plus 1/3 plus 1/5 is already above 1.4, so we stop there. Once we have exceeded the target, we begin adding negative terms until we fall below it. Subtracting 1/2 already brings us below 1.4, so with just one negative term we are back under the target. Then we resume adding positive terms, starting with 1/7, and continue the process.
The key point is that this zigzagging procedure always works. Because the positive terms alone sum to infinity, you can always take enough of them to climb back above the target. Because the negative terms alone also sum to infinity in magnitude, you can always take enough of them to fall back below the target. By repeating this process, you zero in on the target value, and the rearranged series converges to exactly that value.
I find this genuinely profound. What it means is that when you are adding up infinitely many numbers, you do not necessarily get the same answer when you rearrange them. If the terms come from a conditionally convergent series, the order in which you add them can affect the numerical result. That is surprising and, I think, deeply illuminating. And all of these observations shed light, in my view, on what is really going on with Zeno’s paradox.
Recap and What’s Next: Supertasks
Zeno is troubled by the possibility of doing infinitely many things in a finite space of time. This concern connects directly with the concept of supertasks, which we will take up in the next lecture. To understand specifically the idea of moving from one place to another, we were led to the concept of the geometric series: going halfway, then a quarter, then an eighth, and so on. Adding up all those lengths and arriving at a finite answer gives us a way of understanding how an infinite sum can nevertheless have a finite result.
Using the geometric series, we found that in cases where we can describe the series exactly, we can determine precisely what the numerical answer is. Then came the complication introduced by alternating series, which have sometimes negative and sometimes positive terms. Such a series can converge to an answer, and yet the order in which you take those terms can affect the final result, as the Riemann Rearrangement Theorem shows.
I hope you enjoyed the story of Zeno’s paradox and the path it opened up into geometric series, into the possibility of adding up infinitely many numbers to obtain a finite answer, and then into the twist provided by the alternating harmonic series. Ultimately, we arrived at Riemann’s Rearrangement Theorem, a profound idea: when you are adding up infinitely many numbers, the order in which you take those numbers can affect the result.
The other lectures will appear in the lectures-on-infinity tag. The full collection of essays is available at The Book of Infinity. And the book is now available:
These look great can't wait to watch them all!