A problem that’s always bugged me a little

Note – just to be clear from the start – this post is only half serious, and I understand that I’m playing a little fast and loose with a few things – especially the definition of a limit – but it is true that the concept I’ve outlined below has always bugged me.

I started reading (actually listening to) Eugenia Cheng’s How to Bake Pi today. It is a fun book so far, and certainly made working out at the Harvard stadium this morning much more enjoyable than usual.

One of her statements that got me thinking was the idea that nearly everyone hits a wall in mathematical abstraction at some point. I may not have Cheng’s exact words exactly right, but the idea was pretty interesting to me. It made me wonder about some fairly abstract ideas that I’d never completely reconciled in my mind.

What came to mind is strange property of the harmonic series (1 + 1/2 + 1/3 + 1/4 + 1/5 + . . .). Nearly everyone I’ve shared this idea with thinks I’ve completely lost my mind, so with that warning (and with the extra note at the start!), I guess, here we go 🙂

A standard proof that the harmonic series diverges goes something like this:

Step 0: 1 is bigger than 1/2
Step 1: 1/2 + 1/3 is bigger than 1/2
Step 2: 1/4 + 1/5 + 1/6 + 1/7 is bigger than 1/2
Step 3: 1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 is bigger than 1/2

The reason that the sum in step 3 is larger than 1/2 is that all 8 terms are larger than 1/16, so the sum is greater than 8/16 which is 1/2.

Step n: 1/2^n + 1 / (2^n + 1) + \ldots + 1/(2^{n + 1} - 1) is greater than 1/2

The reason that this sum is greater than 1/2 is more or less the same reason given for step 3. Each of the 2^n terms is larger than 1/(2^{n+1}) so the sum is larger than 2^n / 2^{n+1} which is 1/2.

At each step m, you have a sum with 2^m terms which is greater than 1/2, and that tells you that the finite sum 1 + 1/2 + 1/3 + \ldots 1/2^m is greater than m/2 and so as you let m approach infinity, the sum also goes to infinity.

Totally seperately, when you learn about set theory you learn about some of Cantor’s theorems about power sets. The “power set” of a set is the set of all subsets of the set, including the null set and the original set itself. The notation that I learned for the power set of a set S was 2^S. That notation, I assume, comes from the fact that for a finite set S with n elements, the number of elements of the power set of S is 2^n.

Cantor proved that the cardinality of the power set of a set S is always greater than the cardinality of the set itself. In particular the cardinality of the set of positive integers – the “smallest” infinity and usually denoted by \aleph_0, is larger than \aleph_o. In fact, the cardinality of the power set of the set of positive integers is equal to the cardinality of the real numbers. If we call the cardinality of the real numbers R, what I’ve written above can be summarized by saying:

R = 2^{\aleph_0} > \aleph_0

Loosely speaking, that equations tells you that there are “more” real numbers than there are integers.

Ok, so back to the standard proof that the harmonic series diverges. The way that I’ve constructed the proof above, at every step m we show that the sum of 2^{m+1} terms starting at 1/2^m is greater than 1/2.

So, as m goes to infinity, the number of terms we add in each additional step – namely 2^{m+1} terms – also gets *really* large. In fact, so large that it is actually a different infinity! Right?? That means that the “last” sum that is greater than 1/2 in the proof above – namely the step that “finally” gets you to infinity requires adding up more terms than there are integers!

A little crazy, I know, but at least I felt less crazy about being puzzled by this idea after learning that 1 + 2 + 3 + \ldots = -1/12 🙂

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