Friday, January 17, 2014

Why the sum of all natural numbers is not negative one twelfth

This video posted by Numberphile has been making the rounds lately, supposedly proving that the sum of all natural numbers all the way to infinity is equal to negative one twelfth. The guy in the video, Tony, uses pretty basic math to derive this formula and at first glance it may not be immediately clear what is wrong with his derivation. To add to the mystery, he flashes a page from a String Theory book which quotes the formula, albeit in a very different context.

Despite a lot of controversy this video seems to be generating lately, I don't see a good explanation of what exactly is wrong with this proof among the top Google results, so I'm going to try to make it clear here.

Main points

The derivation uses "..." to denote something different in the case of computing S1 and something else in the case of S2, hence the entire derivation is invalid.

There is no interpretation of "..." which, if used consistently alongside sums of natural numbers, would lead to S = 1 + 2 + 3 + ... = -1/12.

Now, mathematical models may exist which use similar notation to the one used for integers and sums, but model phenomena other than natural numbers. Such models may abide by different rules and you may get results which don't carry over to the natural number world. This is probably what the String Theory book is doing (you can check it out at Google Books here). But these mathematical models would still have to be internally consistent in order to be useful, unlike the Numberphile framework which could be used to derive contradictory results, hence has no application other than entertainment.


The first part or the proof concludes that:

S1 = 1 - 1 + 1 - 1 + 1 - 1 ... = 1/2

The argument behind this is that if the sum were cut short at any given point: if this point were odd, the sum would be 1 and if it were even, the sum would be 0. Tony goes on to say: "Do we stop at an odd or even point? We don't know, so we take the average of the two."

Note that this whole argument relies on the assumption that we do stop the series at some point. We just don't know if this point is odd or even, but not stopping at all is not an option. This is an essential distinction.

The traditional mathematical definition of infinite sums, assumes that you never stop. Since you never stop, you never stop at a point which is odd or even, hence the sum is undefined (we'd call this a divergent series).

So what Tony is really computing here, is not the infinite sum, but the expected value of a finite sum stopped at a random place. In more verbose notation we would write what Tony said as:

S1 = 1 - 1 + 1 - 1 + 1 - 1 ... (n times)
S1 = {0 when n is even; 1 when n is odd}
P(n is even) = 1/2
P(n is odd) = 1/2
E(S1) = 1/2 * 0 + 1/2 * 1 = 1/2

There would be nothing wrong with using the infinite sum notation to denote the expected value of a finite sum. In fact, the above definition is called the Cesàro sum and is a well-known mathematical concept. The problem is that having chosen the Cesàro sum as the interpretation of "...", Tony would have to stick to this definition for the remainder of the proof. But he does not do that. In fact, for the rest of the proof, he uses the same notation to mean the actual infinite sum, which never stops.

This is clear, for example, in his derivation of S2. The derivation of S2 is based on adding the series to a version of itself shifted by one, like so:

S2 = 1 - 2 + 3 - 4 + 5 - 6 ...
2 * S2 = 1 - 2 + 3 - 4 + 5 - 6 ...
+ 1 - 2 + 3 - 4 + 5 ...

This is a perfectly valid operation when you are dealing with actual infinite series, because those go on forever. So even if you shift them, they can still align perfectly all the way to infinity. But it does not work under the expected value definition, since the expected value definition assumes that the series stops at some point, in which case the shifted version of S2 would have an extra term at the end which would not align with the un-shifted version.

With all this in mind, here is the video again: