# Why I don't buy that $1 + 2 + 3 + ... = -\frac{1}{12}$

*Thanks to Robert Anderson for the question.*

I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal.

- Srinivasa Ramanujan

A recent Numberphile video made the astonishing claim that $1+2+3+4+… = -\frac1{12}$, and went on to ‘prove’ it as follows:

$S_1 = 1 + 1 - 1 + 1 - 1 + … = \frac12,$ on average. $S_2 = 1 - 2 + 3 - 4 + 5 - … = \frac 14,$ with a bit of manipulation.

Let $S = 1 + 2 + 3 + 4 + …$ and consider $S-S_2$; you get:

$4 + 8 + 12 + … = 4(1 + 2 + 3 + …) = 4S$

Therefore $S - S_2 = 4S$, or $S = -\frac 13 S_2 = -\frac{1}{12}$.

Q, apparently, ED.

Before I go on, I’d like to make clear: Numberphile is brilliant, thought-provoking, and features people I want to be like when I grow up. I’m taking a contrary position on this video partly because I don’t understand analytic continuation, but also partly to put the ‘traditional’ analysis case.

### In the words of the greatest scientist who ever live, I do not buy it

I’m not the greatest mathematician who ever live, though; that would probably be Ramanujan ((or, possibly, the Mathematical Ninja)) - who’s responsible for this abomination.

There’s a problem in the first line of the ‘proof’: the sum $1 - 1 + 1 - 1 + …$ is not well-defined. Simply averaging the two answers is the kind of thing a physicist might get away with, but around here, the Mathematical Ninja tends to chop off any hands waved in the justification of a result. For a sum to have a well-defined sum, there needs to be a point beyond which the terms get smaller in magnitude. (That’s necessary, but not sufficient – there are series, such as $1 + \frac12 + \frac13 + …,$ which have smaller and smaller terms but whose sum can be shown to exceed any number you pick). You’re probably more familiar with the infinite sums of geometric series you get in C2 - for example, $1 + \frac12 + \frac14 + \frac18 + …$ gets as close to 2 as you care to make it.

People like Cauchy spent a lot of time in the 1800s figuring out when it made sense to talk about infinite sums and when it didn’t. Sadly, neither $S_1$, $S_2$ or $S$ qualify – their sums are undefined.

### But wait, there’s more

*However*, there is a certain sense in which $S_1=\frac 12$ and $S_2=\frac14$ make… a certain sense.

Consider the binomial expansion of $(1+x)^{-1}$. That’s $B_1 = 1 - x + x^2 - x^3 + x^4 - …$ Look familiar? Well, if you just plonk $x=1$ in there, you get $2^{-1} = 1 - 1 + 1 - …$ just like the man said!

Unfortunately, that doesn’t quite work, either: the binomial expansion is only valid for $|x| < 1$, so you just did something illegal. The maths police will be here any minute. However, you *can* take a limit! The binomial expansion is defined for $x$ just a tiny bit smaller than 1, and the closer you make $x$ to 1, the closer the answer would get to $\frac 12$. So, *taking the limit from below*, $S_1$ does indeed approach a half.

Similarly, you can consider $B_2 = (1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + …$, which approaches $S_2$ as $x$ gets closer to 1 as well. That, just like the video says, approaches a quarter (when you take the limit as $x$ approaches 1 from below.)

This breaks down when you think about $S$, though, for several reasons. We could try the same trick directly, and consider $B = (1-x)^{-2} = 1 +2x + 3x^2 + 4x^3 + …$ – however, the limit as $x$ approaches 1 is undefined (the closer you get, the larger the sum gets, because you’re dividing 1 by a tiny number).

We could also try the bit of mathematical legerdemain ((good word)) those good people at Numberphile used, and take the difference between $B$ and $B_2$, which would give us $B-B_2 = 4x + 8x^3 + 12x^5 + … = 4x(1 + 2x^2 + 3x^4 + …)$. That bracket… I don’t believe that bracket is the same thing as $B$. It’s $(1-x^2)^{-2}$, and I’m not at all convinced that you can combine it with $B$, even in the limit as $x$ approaches 1.

### That’s as far as my skills in analysis go

As far as the real number system goes, the sum Tony and Ed discuss in the video is undefined. I gather there’s some way of extending the system so it makes sense - but I’d be very wary of simply asserting that an undefined sum has a specific value without explaining that extension.

If you have any thoughts on this, I’d love to hear them - pop them in the comments below!