### Infinite Series – A Health and Safety Warning

18th January, 2014

There’s been a bit of a rumpus recently about this video from *Numberphile* which purports to show that \[1+2+3+\ldots = -\frac{1}{12}\]

It got posted on Slate with an accompanying blogpost by Phil Plait, which then got shot to pieces by various incensed bloggers, and Phil now has a much better follow-up post.

I’m not going to rehearse the story again – the short version is that the equation above is a really bad and misleading way of communicating a really interesting and surprising result. But I thought I’d share another titbit which has a similar moral, namely:

Infinite series often behave in very weird ways.

This in turn has two corollaries, both amply illustrated in the story above:

Unexpected and cool things can happen.

but at the same time

It’s very easy to start talking nonsense if you’re notextremelycareful.

The video is about the series \(1+2+3+4+5+\ldots \) This is what we call a *divergent* series, meaning that as you add up more and more terms, the series grows without limit. An example of a *convergent* series is \(0.9+0.09+0.009+0.0009+\ldots\) Here, as you add up more and more terms, you get ever closer to the fixed value 1. This is the *limit* of the series. Convergent series are generally friendlier beasts than divergent series, but the purpose of this post is to illustrate that even convergent series can do shocking things.

Humans are not used to infinite series. But we are used to adding up finite collections of numbers. If we have three numbers \(A,B,C\) we know that the order in which we add them up does not matter: \(A+B+C=B+A+C=A+C+B\) etc.. This sort of thinking is so natural and automatic that it is all too easy to transfer it to the realm of infinite series, where they may not hold – not even (necessarily) for convergent series.

This post is about the following series: \[S: \ \ 1 – \frac{1}{2} + \frac{1}{3} – \frac{1}{4} + \frac{1}{5} – \frac{1}{6} + \frac{1}{7} – \frac{1}{8} + \frac{1}{9} \ldots\]

The first obvious question is: does \(S\) converge? And the answer is yes. It turns out to converge to a number near 0.69. (More precisely, but unexpectedly and coolly, the limit is^{[1]} the natural logarithm of 2 or \( \ln 2 \). But you don’t have to know what that means to follow the rest of the post.)

Now, let’s reorder the elements of \(S\). Notice that the numbers on the bottom alternate between being odd and even. Let’s mix things up so that they go ‘odd even even odd even even’ instead, but being careful to keep the sign of every element the same at it was before:

\[ 1 – \frac{1}{2} – \frac{1}{4} + \frac{1}{3} – \frac{1}{6} – \frac{1}{8} + \frac{1}{5} – \frac{1}{10} – \frac{1}{12} + \ldots \]

To reiterate, we have not added or removed any elements here, just reordered the elements of \(S\). But if you look carefully at this new series, you’ll see that the positive terms are all followed by the negative of their halves – I’ll group these together in brackets to make it clearer:

\[ \left(1 – \frac{1}{2} \right)- \frac{1}{4} + \left( \frac{1}{3} – \frac{1}{6} \right) – \frac{1}{8} + \left( \frac{1}{5} – \frac{1}{10} \right) – \frac{1}{12} + \ldots \]

Now let’s compute the bits in the brackets and give a name to the new series:

\[ T: \ \ \frac{1}{2} – \frac{1}{4} + \frac{1}{6} – \frac{1}{8} + \frac{1}{10} – \frac{1}{12} + \ldots \]

But what’s this? If you compare a term of \(T\) to the term in the corresponding position in the original series \(S\), you’ll see that it’s exactly half. And indeed, the limit of \(T\) is exactly half the limit of \(S\) – around 0.34 (more preciely, \(\frac{1}{2} \ln 2 \)).

So just by rearranging the terms of \(S\) we have completely changed its limit! That might seem surprising, but Riemann’s rearrangement theorem tells us that this is the tip of the iceberg. It is actually possible to rearrange the terms of \(S\) so that it converges to any number you care to name – or so that the series diverges!

So, be very careful when handling infinite series! Here endeth the lessen. But to continue with the mathematics a little: this happens because \(S\) is what is termed *conditionally convergent*, that means that if you replace all its negative terms with their positives:

\[H: \ \ 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} \ldots\]

you get a series which doesn’t converge. \(H\) is called the harmonic series; that it diverges was one of the first truly surprising facts that mathematicians discovered about infinite series. *Absolutely convergent series* (ones where the corresponding entirely positive series converges, as 0.9 + 0.09 + 0.009 +… does) are much better behaved, and Riemann’s theorem doesn’t apply there.

**Finally, a shameless plug:** All this material is covered in my book Maths 1001.

[1] This is the case \(x=1\) of the series expansion for \( \ln (1+x)\) which is given by the *Mercator series:* \(1 – x + \frac{x^2}{2} – \frac{x^3}{3} + \ldots \)