# Primer on Infinite Series

## Example One: Zeno Crosses the Room

One of the origin stories for infinite series is attributed to the Greek Philosopher Zeno. It’s a good place to start motivating why you end up dealing with infinite series and also a place from which you can begin to appreciate the need for a way of thinking about infinity that doesn’t lead to pure nonsense. Although he had many paradoxes, they all boil down to certain weirdnesses that come up when you imagine space or time being infinitely divisible. Things that are obvious and simple from the point of view of everyday experience are contradicted by their analysis from the point of view of infinite divisibility. Consider the following scenario.

You want to cross the room.

That is you wish to cover one room’s worth of distance. Easy enough right? You don’t even need to stretch. But…(Zeno’s voice) to cross the room, first you need to get halfway there. Before you cross the 2nd half of the room, you need to cross half of the 2nd half of the room. And so forth. At each stage you subdivide the remaining distance in half and cover that distance.

Zeno’s problem was that this sequence of instants is infinite. If each turn took say 1 second, it would take infinity seconds to cross the room. From this perspective any motion—not just room crossing—appears impossible. Zeno’s conclusion was that bringing infinity into the discussion, in this case dividing a given distance into infinitely many pieces, leads to nonsense and should be avoided.

noteFor this and other similar reasons, the official use of infinity in mathematical arguments was postponed until the innovation of Calculus. Essentially, due to the simultaneous creation of mechanics and all the practical uses to which it could be put, the problems with reasoning with infinite sums were swept under the rug for a while. Calculus uses infinity all the time and doesn’t apologize for it. Newton, Leibniz, etc. didn’t figure out how to solve Zeno’s paradox, they just ignored it. It wasn’t until a couple centuries later that mathematicians really confronted the problem.

Zeno’s room crossing can be written using numbers and symbols instead of sentences, where it might look a bit more familiar as an “infinite sum”.

$$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\ldots$$

And we can think of its sum either from the statement of the problem (where it is clear that the sum should be 1, a whole room) or by a direct calculation of the distance remaining to cover and noting where its limit is.

Step Distance Covered Distance Remaining
0 0 1
1 $$\frac{1}{2}$$ $$\frac{1}{2}$$
2 $$\frac{1}{2}+\frac{1}{4}$$ $$\frac{3}{4}$$
3 $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$$ $$\frac{7}{8}$$
4 $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$$ $$\frac{15}{16}$$
5 $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$$ $$\frac{31}{32}$$

So, at this point, we have a good reason to want to be able to write

$$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\ldots = 1$$

Metaphysical or existential doubts notwithstanding, it just looks like we should end up with one if this process of making up half the remaining distance were to continue. In this and many other situations, its clear that adding together infinitely many things should work out to give something finite. The way we would read these symbols in words is

If you keep adding halves of halves ($\frac{1}{2}$ + \frac{1{4}$and so on), you get closer and closer to 1. If somehow you could add forever, you’d have exactly one. ## Exact and Approximate At this point we should clear up one thing about exact versus approximate. You might be thinking that, regardless of the “infinite” part of the sum, this is a clear case of “good enough for all practical purposes”. At some point you get close enough to touch the wall at the end of the room. So yes, we can think of the sum as approximating one. And these infinite sums were originally very useful because they gave ways of approximating difficult to calculate things with very easy to calculate things. But there is some subtlety here in the words we use. When we say, $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\ldots = 1$$ We do not mean $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\ldots \approx \text{ (is approximately equal to) }1$$ It would be correct to say $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \approx 1$$ or also $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32} \approx 1$$ but once we have infinitely many terms, once we are asking what would happen if the process were allowed to continue indefinitely, we are no longer taking about approximately, we are talking about exactly equal. And this is where people, not just young students but the many mathematical thinkers through history, have had trouble. We are used to the things related by equal signs both being simple numbers. Finite things. Not infinite processes. So what does it mean for an infinite process to be exactly equal to a finite number? So far in this discussion we’re not yet sure. But we have a clue. We have exactly one case where we have an infinite process that has an obvious limit. We’d like to equate them. Next, let’s look at a case where the infinite process, instead of telling us something we already know, helps us see something new. ## Example 2: Archie and the Area of a Circle You may have been asked to remember formulas for the circumference and area of a circle in school. $$\text{Circumference }(C) = \pi \cdot \text{Diameter }(D) = 2\pi \cdot \text{Radius }(r)$$ $$\text{Area of a Circle}(A) = \pi \cdot r^2$$ But did you ever wonder why these formulas are true or why they both happen to have $$\pi$$ in them? Did you ever wonder how we figured out the area of a round shape like the circle in the first place? Well, Archimedes did. And the way he figured it out was by creating an infinite sum. The formula for circumference is easy. It’s just a definition. Ancient Greeks and many others noticed that $$\frac{C}{D}$$ was the same ratio for any circle (WHY?). They didn’t call this $$\pi$$ or even consider it a number, but we do. By definition, $$\frac{C}{D} = \pi$$, so of course $$C=\pi D$$. The real mystery is with the area formula. And this is what Archimedes did. He imagined a regular polygon inscribed within a circle. Here’s a picture. ![](some pic) Now, in my picture, there are 8 sides. But there’s nothing special about 8. It could be 6, or 12, or a million. But…the more sides there are, the more the polygon looks like a circle, the better approximation its area is for the area of the circle. And, as it turns out, the area of the polygon is easy to compute. Moreover, computing the area of the pylon, we can see how it changes with the number of sides and how it stays the same, and we can see what happens to this area if we somehow had infinitely many sides. Let’s look: First, we notice that a polygon can be split into a bunch of triangles, each of which is the same. How many? The same number as we have sides of the polygon. Just like slices of pie. Add up the area of all the slices to get the area of the original polygon. $$\text{Area of polygon with }n \text{ sides} = \text{sum of } n \text{ triangles}$$ Of course, each triangle has the same area, and any triangle has area $$= \frac{1}{2} \cdot \text{base}\cdot \text{height}$$. Here, the base is the side of a polygon (let’s call it $$s$$) and the height we can call $$h$$. Our formula for the area of a polygon becomes: $$\text{Area of polygon with }n \text{ sides} = \underbrace{\frac{1}{2}sh+\frac{1}{2}sh+\ldots+\frac{1}{2}sh}_{n \text{ times}}$$ Now, factor the $$\frac{1}{2}h$$ out of the above expression to get something even nicer. $$\text{Area of polygon with }n \text{ sides} = \frac{1}{2}h(\underbrace{s+s+\ldots+s}_{n \text{ times}})$$ And look, $$n$$ sides added together is nothing other than the perimeter ($P\$) of the polygon.

$$\text{Area of polygon} = \frac{1}{2}hP$$

This formula is true regardless of how many sides we have. All we have to do now to recover the formula for the area of the circle is to imagine what happens to all of the parts when the number of sides, the extent of the approximation, the number of terms in the underlying sum, gets larger.

$$\text{Area of polygon} \rightarrow \text{Area of a Circle}(A)$$

$$h \rightarrow r$$

$$P \rightarrow C$$

So our formula becomes

$$A = \frac{1}{2}rC = \frac{1}{2}r(\pi D) = \frac{1}{2}r(\pi(2r))=\pi r^2$$

Amazing, no?

There are tons of other situations where you can answer a question by putting together infinitely many pieces, situations where we would like to be able to formally write down something of the sort:

$$\text{infinite sum} = \text{finite number}$$

but there are also obvious and not so obvious situations where doing so is nonsense. We would like to go so far as to extend our rules for algebra to be able to operate on these infinite sums (things like multiplying both sides of an equation by the same number or grouping terms), but the truth is we don’t have enough information to know when that does and doesn’t work out. Next we will take a look at this more general question.

Parents:

• Mathematics

Mathematics is the study of numbers and other ideal objects that can be described by axioms.