It’s usually quite simple to spot the error in ‘proofs’ that $1=2$: either someone’s divided by 0 or glossed over inverting a multi-valued function (conveniently forgetting the second square root, for example). You sometimes (as with the sum of natural numbers being $-\frac{1}{12}$, if you throw out all good sense) hit problems with an infinite series that doesn’t converge.

This one, via Futility Closet, is something a bit different - $S$ does converge!

Let $S = \frac11 - \frac 12 + \frac 13 - \frac 14 + \frac 15 - …$.

Now consider $2S = \frac21 - \frac 22 + \frac 23 - \frac 24 + \frac 25 - …$

Arrange the terms of $2S$ in three columns, so everything that has an odd denominator is in the first column, everything with a denominator double an odd number (i.e. even numbers that aren’t multiples of 4) in the second column, and denominators which are multiples of four in the third column, like this:

$2S=\frac 21 - \frac 22 - \frac 24 \\

+\frac 23 - \frac 26 - \frac 28 \\

+\frac 25 - \frac 2{10} - \frac 2{12} …$

However, if you combine the first pair of terms in each row and simplify the last, you get:

$2S = 1 - \frac 12 \\

+ \frac 13 - \frac 14 \\

+ \frac 15 - \frac 16 … = S$

So, $2S = S$, meaning either $S = 0$ (which it doesn’t) or $2=1$ (which it most certainly doesn't!) Where’s the problem?

Analysis? Who needs analysis? Oh, yeah, I need analysis.

The glib answer, by which I mean ‘it’s technically correct but doesn’t tell you much’ is that while $S$ does converge to a limit ($S = \ln(2)$) it’s *conditionally convergent* - and you’re not allowed to rearrange the order of terms in a conditionally convergent series.1

Conditionally convergent means that, even though the series as written converges, the sum of the absolute values doesn’t. The absolute values of $S$ give the harmonic series $1 + \frac 12 + \frac 13 + …$, which diverges (and, nicely, is a special case of the Riemann zeta function).

So it’s proved that it doesn’t work, and that’s very nice, but doesn’t really give any insight into why. My best answer is that $S$ is made of two divergent series: the negative terms are $\frac 12 + \frac 14 + \frac 16 + … = \sum \frac 1{2k}$, which is half of the harmonic series, which diverges; the positive terms are $1 + \frac 13 + \frac 15 + … = \sum \frac 1{2k-1} > \sum \frac 1{2k}$, so that also diverges. If you add and subtract two divergent series, you can get whatever you like out of it depending on how you arrange the terms.

I’m not 100% satisfied with my answer, but I’d be curious to hear if anyone has a better one!

* Edited 2014-12-15 to fix a LaTeX error.

* Edited 2015-01-14 to add a link.

- That’s a result due to Riemann:

The bulk of the theorems I learnt

Were proved by a German named Berndt

His results analytical

All proved to be critical

Apart from a couple that weren’t. [↩]

## Nathan Briggs

1 = 2, because George Osborne is running the blackboard

## Nathan Briggs

BTW, didn’t understand a word of your proof (this may be why I failed P3 😉 )

## Nathan Briggs

Nathan Briggs liked this on Facebook.

## srcav

Lovely post from @icecolbeveridge http://t.co/eS7sAsNjbX

## MathbloggingAll

A “Proof” that 1 = 2 http://t.co/F5sPlLfrbD

## Paul Coombes

THank you for an explanation that has convinced me. I must admit that I had not come across a Conditionally convergent series before.

## Colin

Thanks, Paul – it was a new one on me, too!