No, $1+2+3+\dots$ isn't equal to $-1/12$. But $1+2+4+8+\dots = -1$.

A while ago the numberphile team released a video (apparently viewed more than 1.5 million times in one month), purporting to give a simple argument for the bizarre equation \[1+2+3+\dots = -1/12.\]

Obviously something fishy is going on here, because an infinite sum of larger and larger positive terms can't really have a finite value, let alone negative. Still, Tony Padilla and Ed Copeland claim in the video that the equation makes sense and is important for string theory in 26-dimensional space and what not.

In the discussions on the web, the video has been more or less burned at the stake. Mathematicians have criticized it for being misleading and for failing to properly discuss the concept of sum of a divergent series. Several commenters have dismissed it on the grounds that the whole thing is obviously nonsense. Among the critics are Blake Stacey and Evelyn Lamb. Quoting the blogger Skulls in the stars,

"a depressingly large portion of the population automatically assumes that mathematics is some nonintuitive, bizarre wizardry that only the super-intelligent can possibly fathom. Showing such a crazy result without qualification only reinforces that view, and in my opinion does a disservice to mathematics."

To some extent I agree with this. But if the wizardry is something a school kid can try and show to their friends, it might do a great service to mathematics even if it seems crazy and nonintuitive. The real problem with the video is

If kids try this at home, it won't work.

And this goes for kids of all ages and levels of education. Including myself. I know that $\zeta(-1)=-1/12$ but I still can't make sense of the derivation in the video.

So let's debunk it. Suppose \[1+2+3+4+\dots = S.\] Then, using the same type of calculation as in the video, we can shift the series by adding one or more zeros in front: \[S = 0+1+2+3+\dots = 0+0+1+2+3+\dots.\]
Now let's take the original series, subtract two copies of it shifted one step, and then add a copy shifted two steps:
\[
\begin{eqnarray}
S =& 1+2+3+4+5+\dots \\
-S =& 0-1-2-3-4-\dots\\
-S =& 0-1-2-3-4-\dots\\
+S =& 0+0+1+2+3+\dots
\end{eqnarray}
\]
Adding up columnwise, we get
\[ 0 = 1+0+0+0+\dots = 1.\]

That's not just weird, that's a contradiction. So the series can't sum to $-1/12$ or to any other number.

My main misgiving about contradictions isn't that they are wrong or anything. It's that they are boring. If somebody plays around on their own with this and similar series, perhaps trying to sum all even or all odd numbers, or all squares, they will just end up with a mess. If they ask about it in school, all the poor teacher can say is "you can't manipulate infinite sums the way you do with finite sums because you run into contradictions". And the gap between school mathematics and where the fun is will just appear larger.

But the thing is you can manipulate infinite sums, even divergent ones, in a lot of fun ways without running into contradictions. It's just that zeta regularization and Ramanujan summation is a bad first example. (For those who want to learn about that stuff, I recommend the excellent Wikipedia article 1+2+3+4+…).

So here's what I suggest instead:

Listen to the teacher's warning, and obey the following simple rules: (1) When you play with an infinite series, you can apply all the ordinary rules of arithmetic, as long as you only mess with a finite number of terms. (2) You can add (or subtract) two infinite series term by term. You may also multiply all the terms of a series by a fixed number, and the sum will be multiplied by the same number. And keep in mind that not all series have sums. By the way, rule (1) is called stability in the theory of divergent series.

With these rules, it is easy to sum a geometric series. For example, let $A=1+1/2+1/4+1/8+\dots$. Then
\[
\begin{eqnarray}
2A =& 2+1+1/2+1/4+1/8+\dots \\
-A =& 0-1-1/2-1/4-1/8-\dots\\
\end{eqnarray}
\]
Adding up, we get
\[ A = 2.\]
Amazing, and a little weird if you haven't seen it before. Now lets sum the powers of 2. Yes, positive powers. Let $B=1+2+4+8+16+\dots$. With the same trick,
\[
\begin{eqnarray}
2B =& 0+2+4+8+16+\dots \\
-B =& -1-2-4-8-16-\dots\\
\end{eqnarray}
\]
And adding up, $B = -1$.

How weird, the sum of a series of larger and larger positive numbers is negative! Well, strictly speaking we have only demonstrated that if the series can be assigned a value, that value has to be $-1$. But in fact assigning it this value is consistent with the rules I gave you.

I will not try to defend the pedagogical merits of summing divergent series. But at least it might serve to start a discussion of why it is okay to say that $A=2$, and whether it could somehow make sense that $B=-1$.

And what about the derivation in the video? First, Grandi's series (called $S_1$ in the video) must have the value $1/2$, because
\[
\begin{eqnarray}
S_1 =& 1-1+1-1+\dots \\
S_1 =& 0+1-1+1-\dots\\
\end{eqnarray}
\]
So $2S_1 = 1$.

Then let $S_2 = 1-2+3-4+\dots$, and just as in the video,
\[
\begin{eqnarray}
S_2 =& 1-2+3-4+5-\dots \\
S_2 =& 0+1-2+3-4+\dots\\
\end{eqnarray}
\]
So $2S_2 = S_1 = 1/2$, and therefore $S_2 = 1/4$. So far so good. Then they let $S=1+2+3+4+\dots$, and compute $S-S_2$:
\[
\begin{eqnarray}
S =& 1+2+3+4+5+6+\dots \\
-S_2 =& -1+2-3+4-5+6-\dots\\
\end{eqnarray}
\]
So $S-S_2 = 0+4+0+8+0+12+\dots$. But this is NOT the same thing as $4S$ (as is claimed in the video). To get from $0+4+0+8+0+12+\dots$ to $4+8+12+\dots$, you would have to delete an infinite number of zeros, and there is no rule that allows you to do that. That's where they go wrong. Up to that point, what they did was actually consistent.

I haven't mentioned Cesàro summation or Abel summation, and I haven't discussed the theory of stable, regular and linear summation methods. The point is that while equations like $1+2+4+8+\dots = -1$ may be bizarre and unintuitive, at least they are bizarre and unintuitive in a consistent way, and that gives a hint that there is something real going on behind the scenes. You don't have to sit passively and watch somebody pull rabbits out of a hat, you can actually try it yourself.

Which brings us finally to a little exercise. The Fibonacci numbers start 1, 1, and after that, each number in the sequence is the sum of the two previous numbers. So it goes $1,1,2,3,5,8,13,\dots$. Now compute
\[ F = 1+1+2+3+5+8+13+\dots!\]