Infinity is not an intuitive concept for most people. My favorite example is the magician with the inifinite hat:

Boffo the Impossible has an amazing trick. Starting at 11:00 am, Boffo puts two balls into a hat, and pulls one out. Every time he gets half-way closer to noon, he puts two balls in, and takes one out. So, at 11:30, he puts two in and takes one out. At 11:45, he puts two in and takes one out. At 11:52:30, he puts two in and takes one out. And so on — he becomes an flurry of activity as he gets closer to noon. At noon, he stops.

The question is: how many balls are left in his hat at 12:01 pm?

The obvious answer is an infinite number of balls, but this is not necessarily correct. The real answer is: any positive whole number of balls between zero and infinity.

Let’s say that Boffo numbers the balls as he puts them in. So that the first two balls (at 11:00 am) are numbers 1 and 2. The second two (at 11:30 am) are numbers 3 and 4. So on, and so forth. For every positive whole number (1, 2, 3, 4, 5, …) you can point to the exact time that he put it in.

Now the question becomes: In what order does Boffo take the balls out of the hat?

The natural assumption is when he puts in 1 and 2, he takes out #1. When he puts in 3 and 4, he takes out #3. When he puts in 5 and 6, he takes out #5. When you look at the problem this way, you clearly have an infinite number of balls (as you have all of the even numbered balls left.)

But… let’s say that Boffo *doesn’t* take the balls out of the hat that way. Let’s say that when he puts 1 and 2 in, he takes out #1. When he puts in 3 and 4, he takes out #2. When he puts in 5 and 6, he takes out #3. Now the problem is changed: *for every ball that Boffo has put in, you can name the time that he takes it out.* Whoa there! This means that you can’t name a single ball that’s left in the hat! Since you can’t name any ball that is left in the hat, there must be no balls left in the hat. (I triple dog dare you to name a single ball left in the hat.)

Now, if Boffo takes out the balls in the order #1, #3, #4, etc. then he has one ball left in the hat. You can construct a simular sequence to leave any number of balls in the hat that you want.

What’s going on here? This is an example of what mathematicians call a *conditionally divergent series* (or a conditionally convergent series depending on your perspective.) Each series like this has what’s referred to as a natural ordering.

For Boffo’s hat trick the natural ordering is: (2 – 1) + (2 – 1) + (2 – 1) + … which is clearly 1 + 1 + 1 + …, which is infinite.

Now, let’s say you reorder the terms, so that you have (2 – 1 – 1) + (2 – 1 – 1) + (2 – 1 – 1) + … You can do this since you have an infinite number of twos and minus ones. This ordering of the sequence is clearly 0 + 0 + 0 + … and so forth. This sums to zero.

The “conditional” part of conditional convergence/divergence is so-called because the result you get is entirely dependent on the sequence in which you sum up the terms.

And now you can say that you’ve had exposure to one of the key parts of advanced calculus. Thanks, Boffo!