The Mathematical Con Artist

Suppose that, one day, walking along the street, you meet a man who shows you a series: 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}....  He tells you that this converges to \ln{2}\approx 0.69 and indeed, as you try a few terms, it seems to be true.  He then bets you £100 that it will always converge to \ln{2} no matter how you write it.  Should you believe him?

To answer this, I welcome you once again…

…to Let’s Do Real Maths.

LDRM is a recurring segment in which I take a great paper from mathematical history and rewrite it so that it’s easy to understand.  Today, I’m doing part of Riemann’s Ueber die Darstellbarkeit einer Functiondurch eine trigonometrische Reihe.  (Confession time: I haven’t been able to get hold of a copy of this in English, and I don’t speak German, but I’ve got a fairly good idea of what it says from various secondary sources.)

Riemann managed to show that, for any series that followed certain rules, you can rearrange the terms to make it converge to literally any number you like.  Specifically, the sequence has to converge (it must have a well-defined sum), but it must not absolutely converge.  That’s to say, the version of the sequence where you replace minus signs with plus signs, in this case 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}... must not converge.  In fact, it doesn’t, which makes this a prime candidate for rearrangement.  (If you don’t believe me, start adding terms.  It grows extremely slowly, but it does grow.)

Now, let’s denote our sequence by S=a_1+a_2+a_3+a_4...+a_n+..., where an infinite number of a_n are positive and an infinite number are negative.  A little thought will show that that’s the only way that the sequence can converge but not absolutely converge.  Divide the sequence up into two sequences, S^+ and S^-, containing the positive and negative terms of S respectively.  Obviously, neither S^+ nor S^- converge.

We now need to realise something.  For a sequence to converge, its members must become closer to zero as we go further along.  It’s no use trying to add numbers that keep getting bigger; obviously you’ll get an infinite sum.  So, a_n\rightarrow0 as n\rightarrow\infty, and that means that the members of S^+ and S^- get smaller as you go along as well.

We’re ready for our proof.  Pick a number, any number, and call it M.  Start adding together terms of S^+, from the first onwards, until the answer is as small as possible whilst still being bigger than M.  If S^+ were 6+5+4+3... and M were 12, you would add 6+5+4=15 because adding any more would make the result needlessly large but adding any fewer would make the result less than M.  Now, keep adding numbers, but with a catch — these ones come from S^- instead which, you’ll remember, contains only negative numbers.  Add just as many of these as you need to make the result less than M.

You can probably see where we’re going with this.  Add a few more terms of S^+ to get something bigger than M, add a few more terms of S^- to make it smaller again, and keep going.  But, since terms of S get smaller as you go on, the distance between M and the series of partial results must also get smaller.  Think of it as trying to park a car in a particular space.  If you start by moving it six feet at a time, you’re going to overshoot the parking space by a lot.  If, on the other hand, you only move it six inches before stopping to check where you are, you’re going to be a lot closer to actually making it into the parking space.

Keep doing this forever, and you’ll have added all the terms from S^+ and S^- together in some order.  Wait a minute.  That’s all the terms from S!  Thus, there must be some arrangement of the terms in S which converges to M.

Just as an example, let’s go back to the sequence we were shown earlier.  We’ll start by saying that 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...=\ln{2}.  Now, rearrange the sequence, so that for every addition, you make two subtractions.  Don’t change the numbers, just the order.  This gives you (1-\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{6}-\frac{1}{8})....  I’ve put them in groups for reasons that will become apparent in a minute.  Tidy up the groups a little bit by subtracting the second number from the first and leaving the third, to get (\frac{1}{2}-\frac{1}{4})+(\frac{1}{6}-\frac{1}{8})....  This is equal to \frac{1}{2}(1-\frac{1}{2})+\frac{1}{2}(\frac{1}{3}-\frac{1}{4})....  Remind you of anything?  In fact, this is exactly half of our original series, and so converges to \frac{\ln{2}}{2}.

Please don’t go trying to show that \ln{2}=\frac{\ln{2}}{2}.  That would be missing the point entirely.  The Riemann Rearrangement Theorem, as this is known, shows that the idea of a “sum” doesn’t really work for a lot of infinite series.  Obviously, some you can sum (1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...=2 no matter which way you look at it), but some you can’t.  Our man on the street is quite clearly a mathematical con artist!

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