Primer on Infinite Series

What does it mean to add things to­gether for­ever? While many get in­tro­duced to in­finite se­ries in Calcu­lus class, it seems that ev­ery­one skips ahead to mem­o­riz­ing ter­minol­ogy with­out first mo­ti­vat­ing the sub­ject, or con­sid­er­ing that it’s sorta weird. In re­al­ity, the long his­tory of in­finite se­ries would sug­gest that the sub­ject is sub­tle and strange. But by steam­ing ahead, mem­o­riz­ing the vo­cab­u­lary, ex­am­ples, and rules for con­ver­gence, learn­ers miss a chance to re­ally think about what these in­finite sums re­ally are or could be. They end up be­ing able to more or less ac­cu­rately par­rot those rules, but if you speak with them a bit or over­hear a con­ver­sa­tion with oth­ers about the topic, you re­al­ize their un­der­stand­ing is a hol­low shell. This ar­ti­cle is a look at in­finite se­ries, adding things to­gether for­ever, from more or less first prin­ci­ples.

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.