I'm Joel David Hamkins, and I want to tell you about infinity, that most fascinating of topics, in this series of lectures. I want to begin with a classical thinker,
Zeno of Elea made a very interesting argument, around 450 BC, that all motion is impossible, that you cannot go from here to there. Of course, we know the conclusion is false; one can simply walk from one place to another. But it is no good to merely assert that the conclusion is false. Rather, we need to understand the argument itself and find the flaw in his reasoning.
Zeno argued as follows. Suppose you want to travel from point A to point B. Before you can reach B, you must first get halfway there. But before you can reach that halfway point, you must first get halfway to it, and before that, halfway again, and so on indefinitely. There are thus infinitely many things you must accomplish before you can move at all, which Zeno took to be impossible. He therefore concluded that all motion is an illusion.
The conclusion is plainly absurd, yet the reasoning is not so easy to dismiss. The core problem Zeno is identifying seems to be this: can one do infinitely many things? That question is what gives the paradox its force.
Consider another of Zeno's paradoxes: Achilles and the tortoise. Achilles and the tortoise are to have a race around a stadium, and the tortoise challenges Achilles by saying, "I will win, but you must give me a small head start." How could the tortoise possibly defeat the great warrior Achilles?
The tortoise's argument runs as follows. Suppose the tortoise is given a head start of one-quarter of the way around the track, starting at a point A, while Achilles begins at the starting line. They set off, and Achilles very rapidly reaches point A. But by the time Achilles arrives at A, the tortoise has already moved on to a further point B, so the tortoise is still ahead.
Now, by the time Achilles reaches B, the tortoise has moved on yet again to a point C, and then from C to a point D, and so on. At every stage, whenever Achilles arrives at the position the tortoise previously occupied, the tortoise has already advanced to a new position. There are thus infinitely many such points, and the tortoise is ahead at every one of them.
The tortoise therefore argues that it is impossible for Achilles ever to catch up, because doing so would require him to complete infinitely many such steps.
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 in its own right, but Zeno's paradox may well be their origin.
Consider a slightly different way of understanding Zeno's paradox. 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. We can turn the argument around, however, and frame it like this: we cannot go from here to there because, before arriving, we must already have gone halfway, then halfway of what remains, then halfway of what remains again, and so on. Before reaching the destination, we must have completed infinitely many such steps, 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. To travel from zero to one, we must first reach the halfway point at one-half. From there, before completing the remaining distance, we must reach the halfway point of what remains, which is the three-quarter point. Standing at three-quarters, we must again go halfway, covering one-eighth of the total segment, then one-sixteenth, and so on.
Now consider the numbers we have identified. The total length traversed is one-half, plus one-quarter, plus one-eighth, plus one-sixteenth, and so on, an infinite sum. Yet it is quite clear that all of these segments together exactly exhaust the original interval, and so this infinite series sums to one, the total original length. This gives us a contemporary way of understanding what is at stake in Zeno's paradox: one-half plus one-quarter plus one-eighth plus one-sixteenth, continued indefinitely, adds up in total to one.
Here is another way of seeing this infinite summation. Consider the unit square: a one-by-one square with total area one. If we take half of it, that portion has area one-half, and half of what remains is one-quarter. Taking half of that gives one-eighth, and half again gives one-sixteenth, and so on; we simply keep chopping what remains in half. The total area, which is one, can therefore be expressed 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.
Next, I would like to present what may be the most contested equation in middle school, one that many of you have likely heard of, and perhaps argued about. The equation concerns the number 0.999…, where the nines never stop, and it states that this number is in fact equal to one. That is, 0.999… repeating is exactly the same as 1.000… forever. Let us consider a couple of arguments for why this is true.
Some people feel that 0.999… should be a little bit less than one, as though all those nines leave something extra unaccounted for. The following argument shows that this intuition is mistaken. Let x equal 0.999… repeating. When we multiply this number by ten, we simply shift the decimal point, giving us 10x = 9.999… repeating, with infinitely many nines. Subtracting the original equation from this one, we get 10x − x = 9.999… − 0.999…, and since all the nines cancel on the right-hand side, we are left with 9x = 9. Therefore x = 1, confirming that 0.999… repeating equals one.
Mathematicians sometimes raise a subtle objection to this argument: it presupposes that 0.999… is a meaningful expression in the first place. When we write "let x be that number," the step only makes sense if the expression actually denotes something well-defined. In fact it does, and a proper argument can be given for this, but we have not supplied that argument here.
Let me give another argument for this contested equation. It is perhaps less controversial to say that 0.3 repeating is equal to one-third. Many people accept that readily, and indeed if you divide one by three using the long division algorithm, you obtain exactly that repeating decimal. Now, 0.9 repeating is simply three times 0.3 repeating, and therefore its value must be three times one-third, which is one. That is another way of seeing that 0.9 repeating must equal one.
Let us get more fundamental about what decimal notation actually means. When you write a number such as 8,547, this is a positional number system, meaning we have eight thousands, five hundreds, four tens, and seven ones. We can express this in scientific notation as 8 × 10³ + 5 × 10² + 4 × 10¹ + 7 × 10⁰. The notation means precisely that the number is a sum with one term for each digit, each term being that digit multiplied by the appropriate power of ten.
Exactly the same principle applies to digits after the decimal point. A number like 3.14159265…, the opening digits of π, means 3 + 1 × (1/10) + 4 × (1/100) + 1 × (1/1,000) + 5 × (1/10,000), and so on. Each digit corresponds to one term, multiplied by a power of ten with a negative exponent.
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…, that expression means 9/10 + 9/100 + 9/1,000 + 9/10,000 + ⋯. The very meaning of the notation is already the infinite sum we were examining earlier. This particular sum has the property that each term is one-tenth as large as the previous one, which is exactly what is called a geometric series.
A geometric series is an infinite sum in which each term is a constant multiple of the previous one. To find the value of a general geometric series, consider first the example from Zeno: one-half plus one-fourth plus one-eighth plus one-sixteenth, and so on. This is a geometric series because each term is half the previous one; we multiply by one-half at every step. In general, a geometric series begins with a term A and proceeds through Ar, Ar², Ar³, and so on, where each term is obtained by multiplying the previous one by the ratio r.
Taking A = 1 for simplicity, we are considering the series 1 + r + r² + r³ + ⋯. When trying to understand the value of an infinite sum, it is natural to examine what happens when we take only finitely many terms. This approach has a distinctly potentialist character, reflecting the classical distinction between potential infinity and actual infinity. Potential infinity is the idea that one never completes an infinite task, but instead takes more and more, finitely much at a time. The very meaning of an infinite sum embodies this character, since we look at the values of the finite partial sums as we include more and more terms.
Let x denote the finite partial sum 1 + r + r² + ⋯ + rn. Now consider what happens when we add the very next term: x + rn+1 = 1 + r + r² + ⋯ + rn + rn+1. All terms after the leading 1 are multiples of r, so we can factor r out of them: the result is 1 + r(1 + r + r² + ⋯ + rn), which is simply 1 + rx. We therefore have the equation x + rn+1 = 1 + rx.
This equation can be solved for x by straightforward algebra. Bringing the rx term to the left gives x − rx = 1 − rn+1, and factoring the left-hand side yields x(1 − r) = 1 − rn+1. Dividing both sides by (1 − r), we obtain the exact value of the finite partial sum:
x = (1 − rn+1) / (1 − r).
This formula tells us precisely the value of the partial sum out to the nth power, and it now allows us to understand what happens as n grows larger and larger. Whether we take 100 terms, 1,000 terms, or 1 million terms, this expression gives the exact value at each stage, and from it we can determine the limiting behavior that defines the value of the full infinite series.
Let us summarize what we have. We are interested in the sum 1 + R + R² + R³ + ···, the infinite series. 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 diverge to infinity, because the finite partial sums grow without bound. When R is large, every term is greater than 1, and adding more and more such terms causes the sum to increase indefinitely.
But if the absolute value of R is less than 1, including the case where R is negative, the situation is quite different. This is precisely what occurred 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 |R| < 1, the finite partial sum is given by a closed expression, and the key observation is that as N grows very large, we are multiplying a number less than 1 by itself many times over. A number less than 1, raised to higher and higher powers, becomes smaller and smaller without limit. Consequently, as N grows large, the finite partial sum approaches 1 / (1 − R), and we can make it as close to that value as we wish simply by taking enough terms.
Therefore, the total value of the geometric series 1 + R + R² + R³ + ··· is 1 / (1 − R). More generally, when the leading term is A rather than 1, the series A + AR + AR² + AR³ + ··· has the value A / (1 − R). This is 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 sufficiently many terms, every partial sum beyond that point will lie within that prescribed tolerance of the limit. This is the rigorous sense in which we understand the value of an infinite series.
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 first term, and r is also one-half, since each term is obtained by multiplying the previous one by r. The total sum is therefore a over one minus r, which is one-half over one minus one-half. Since one minus one-half is one-half, we get one-half over one-half, which equals one, exactly the result we stated for Zeno's sum.
Now consider what is perhaps the most contested equation in middle school: 0.999… As we noted, this equals nine-tenths plus nine-hundredths plus nine-thousandths, and so on. Here a is nine-tenths, the first term, and r is one-tenth, since each successive term is one-tenth as large as the previous one. Applying the formula, the total sum is a over one minus r, which is nine-tenths over one minus one-tenth, that is, nine-tenths over nine-tenths, which equals one.
This argument does more than confirm that 0.999… must equal one if it has any meaning at all; it also shows that it does have a meaning. The series genuinely converges to one, because by taking more and more terms, more digits, we can make the partial sums as close to one as we like. That, perhaps, is the most contested equation of middle school.
Consider some other interesting series. The confusing thing about infinite series is that it seems impossible, at first, that you could add up infinitely many numbers and arrive at a finite answer. Nevertheless, we argued that there are cases where you do get a finite answer: 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.
There are obviously other cases where adding up infinitely many numbers does not yield a finite answer. If we add one plus two plus three plus four and so on, this cannot converge to any finite number. That case, however, involves individual terms that are growing larger, so it is no surprise the sum is infinite. Similarly, if we add up infinitely many ones, one plus one plus one plus one forever, we will never reach a finite answer, since we can make the total exceed any given bound simply by taking enough terms.
One might therefore think: perhaps if the individual terms in the sum are getting smaller, as they are in the geometric series, that is what it takes for the sum to be finite. Let us examine another famous series with exactly that property. The harmonic series is one plus one-half plus one-third plus one-quarter plus one-fifth plus one-sixth, and so on. At each step, the denominator simply increases by one. The individual terms are getting smaller, which we said is a necessary condition for convergence. But let us think more carefully about whether the series actually converges.
Suppose we have added up many terms: one plus one-half plus one-third plus one-quarter, all the way out to one over n, so we have the first n terms. Now consider doubling the number of terms. The additional terms are one over n plus one, plus one over n plus two, continuing all the way to one over 2n, that is, exactly n new terms. How much do these extra terms contribute?
Each of these n new terms is at least as large as the last one, which has value one over 2n. Therefore, their total is at least n times one over 2n, which equals one-half, since the n's cancel. In other words, no matter how many terms we have already summed, doubling the number of terms always adds at least one-half to the total. If we want to add another one-half, we double again. We can add an extra seventeen to the sum simply by doubling thirty-four times.
This shows that the harmonic series cannot converge to a finite value. For convergence to a finite value would require that, once enough terms are taken, adding further terms changes the total by only a negligible amount. But that is never true here: we can always add at least one-half more, or two more, or five more, simply by doubling the number of terms sufficiently many times. The harmonic series is therefore famous for being divergent — it does not converge to a finite value, even though its individual terms tend to zero. This is an entirely different situation from the Zeno-style geometric series.
Perhaps it seems surprising that, having started with the philosophical idea of Zeno's paradox, we find ourselves deep in mathematics and the theory of series. But the point is that when one looks seriously at the philosophy of infinity, it blends into mathematics so gradually and naturally that one is almost inevitably pushed toward these mathematical ways of thinking. Mathematical knowledge is simply the most powerful tool we have for understanding the nature of infinity.
With that in mind, let us examine what is called the alternating harmonic series: the series 1 − 1/2 + 1/3 − 1/4 + 1/5 − ⋯. It is called alternating because the signs vary, positive, negative, positive, negative, and so on. We already saw that when all the terms are positive, the harmonic series does not converge to a finite number; it adds up to infinity. But here, because we are sometimes subtracting instead of adding, the situation may be different.
To see what happens, consider a graph of the partial sums plotted against the number of terms. We begin at 1. We then subtract 1/2, dropping to 1/2. We add 1/3, rising to 5/6, not quite back to 1. We subtract 1/4, falling again, but by less than we just rose. Then we add 1/5, rising again, but by less than we just fell. The series has a zigzag character, but with a crucial feature: every upward step is followed by a downward step of strictly smaller magnitude, and every downward step is followed by an upward step of strictly smaller magnitude.
This means that the upper envelope (the values of the partial sums just after each addition) is strictly decreasing, while the lower envelope (the values just after each subtraction) is strictly increasing. Moreover, the gap between these two envelopes equals the magnitude of the most recent step, which tends to zero. The two envelopes therefore converge toward each other, and one can prove that they meet at exactly the natural logarithm of 2, which is approximately 0.693.
The alternating harmonic series is therefore convergent; it has a finite value, namely ln 2. This is a striking contrast: when all the terms are positive, the series diverges to infinity, but when the signs alternate, the series converges to ln 2.
There is something remarkable about this kind of series. It is called conditionally convergent: it converges to a finite answer, but if you take the absolute value of each term, if you make them all positive, it no longer converges to a finite value. You might think that computing 1 − 1/2 + 1/3 − 1/4 + ⋯ should be the same as grouping all the positive terms 1 + 1/3 + 1/5 + 1/7 + ⋯ and subtracting all the negative terms 1/2 + 1/4 + 1/6 + ⋯. But you cannot do that, because the positive terms sum to infinity and the negative terms also sum to infinity, and so the subtraction is undefined from the outset.
There is a profound theorem about precisely this situation: the Riemann Rearrangement Theorem. It states that if you have any conditionally convergent series, you can rearrange its terms to make the sum equal to whatever target value you wish. This applies to any conditionally convergent series, not just the alternating harmonic series, and for any target value at all.
Let me show how the proof works using the alternating harmonic series as an example. Suppose we want the rearranged series to converge to 1.4, which is larger than log 2 ≈ 0.69. The strategy is to keep adding positive terms until the partial sum exceeds the target, then add negative terms until it falls below the target, and then repeat. For instance, 1 + 1/3 ≈ 1.33, which is still below 1.4, but 1 + 1/3 + 1/5 ≈ 1.53, which exceeds it. At that point we subtract 1/2, dropping below the target, and then resume adding positive terms beginning with 1/7, and so on.
The process works in general precisely because the positive terms alone sum to infinity and the negative terms alone sum to negative infinity. No matter how far the partial sum has drifted, you can always return above the target by taking sufficiently many positive terms, and always return below it by taking sufficiently many negative terms. By repeating this procedure, the partial sums zero in on the target, and the rearranged series converges to exactly that value.
I find this genuinely profound. It means that when you are adding up infinitely many numbers, rearranging them need not give the same answer. If the terms come from a conditionally convergent series, the order in which you add them can change the numerical result, which is surprising and, I think, sheds real light on what is going on in Zeno's paradox.
Zeno is troubled by the possibility of doing infinitely many things in a finite space of time, and this connects directly with the concept of supertasks, which we will take up in the next lecture. One way of understanding the specific idea of traversing a distance, going from here to there, led us to the concept of the geometric series: going halfway, then a quarter, then an eighth, and so on. Summing all those lengths and arriving at a finite answer gives us a way of understanding how an infinite sum can nevertheless yield a finite result. In cases where we can identify the series precisely as geometric, we can determine the exact numerical answer.
A complication arises with alternating series, which have terms that are sometimes negative and sometimes positive. Such a series can converge to a finite answer, yet the order in which the terms are taken can affect that answer, a fact made precise by the Riemann Rearrangement Theorem. This is a profound result: when adding up infinitely many numbers, the order in which those numbers are summed can change the outcome.
We began with Zeno's paradox, which led us into the study of geometric series and the possibility of infinitely many numbers summing to a finite value. The story then took a striking turn with the alternating harmonic series and culminated in Riemann's Rearrangement Theorem, a deep reminder that infinite addition does not always behave the way finite addition does.