I want to have a look at a very famous (and bizarre) result, the infinite sum

$\sum_{n=1}^\infty n = 1+2+3+4+5+... = -\frac{1}{12}$

At first, it doesn’t seem to make any sense. If you add all these numbers up, the result keeps getting larger, so how does it approach a finite number – and a negative one for that matter?

The short answer is, it doesn’t – the equals sign in the above statement is misleading, and hides a lot of complex ideas – essentially, it changes from meaning ‘equals to’ to meaning ‘is linked with’. But the result is still in some sense correct, and meaningful.

Let’s start with the “proof” of this statement, (If you see flaws in this, don’t worry – the whole thing is ridiculous, there are more rigorous and meaningful mathematical explanations.)

## A combination of the Numberphile and Mathologer proofs

We start with this infinite series:

1-1+1-1+1-1+1-1+1-1…=?

What does this equal? Well, the sum alternates between 1 and 0, so it doesn’t seem to approach any single value as it heads off to infinity.

Let’s look at it a different way – what’s the sum of this infinite series:

$1 + r + r^2 + r^3 + r^4 +...$

Well, let’s assign it a value and see if we can figure it out:

$L = 1 + r + r^2 + r^3 + r^4 +...$

If we multiply both sides by r:

$Lr = r + r^2 + r^3 + r^4 +r^5... = L-1$

but we see that, by multiplying by r, we have almost the same infinite sum as before, just without the “1” at the front.

$Lr -L=-1$

$L(r -1)=-1$

$L=-\frac{1}{r-1} = \frac{1}{1-r}$

so now we have collapsed our infinite sum into a single value. What happens if r=-1?

$1 + r + r^2 + r^3 + r^4 +... = 1 - 1 + 1 - 1 + 1 - 1 +...$

and if we plug r=-1 into our formula:

$L= \frac{1}{1-r} = \frac{1}{1--1}=\frac{1}{2}$

right, so we have now assigned a value to our sum:

$1-1+1-1+1-1+1-1+1-1...=\frac{1}{2}$

As an aside, you can also probably see that if you averaged out 0 and 1, the two values the sum fluctuates between, you get a half as well.

Next, we jump to another infinite series:

R = 1-2+3-4+5-6+7-8+….

What on earth is this sum equal to? Well, we can add it to itself:

Which gives us the strange result

$2R = 1-1+1-1+1-... = \frac{1}{2}$

$R = \frac{1}{4}$

Now we can get back to our main series:
$S = 1 + 2 + 3 + 4 + ...$

What happens if we subtract R from it?

S-R

Fixing up the brackets:

We see that our result is just four times (1+2+3+4+…), or 4S. So:

$S - R = 4S$

$S = -\frac{R}{3}$

But we already found that R (the sum 1-2+3-4…) is one quarter:

$S = -\frac{1}{12}$

## There must be some hidden flaws here, right?

As you can probably guess, this proof does a bunch of things you normally aren’t allowed to do with series.

### Dodgy thing #1

We found the limit of 1-1+1-1+1… by using the formula for geometric series (that 1+r+r2+… series), even though that formula is only supposed to apply for |r|<1 (the size of r is less than 1), and normally doesn’t work outside of that range.

### Dodgy thing #2

The next thing that we did was take the sum 1-2+3-4+5…, and add it to itself in a specific fashion, to get us back to that previous series. Except this series is divergent (it heads off to infinity), so it doesn’t make sense to give it any value; it seems meaningless to call the value of the sum “R” and then try doing algebra with it.

### Dodgy thing #3

When you have infinite series that have a mix of positive and negative terms, you can’t move the terms around in the sum, or they change the value of the sum (Here is a video on this topic). But we add 1-2+3-4+5… to itself by shifting the second copy along one, so the numbers line up the way we want them to.

## So the proof is wrong?

Well, not totally. Again, while this identity is not literally correct, it is still in some sense meaningful. There are far more complex and insightful proofs and interpretations of this identity, but they are a little hard to follow. Essentially, what we are doing is saying “I know the result is infinite – but that isn’t useful, so i am going to treat it as if it is finite, and see if I can come up with a result anyway.” And as it turns out, this is the only meaningful finite value for this series.

It is also known the result is meaningful because it appears in physics, particularly quantum physics and string theory. This is a good confirmation that something is right about the identity.

Essentially, the takeaway is that 1+2+3+4+5… does not equal -1/12, but that the two values are associated in some sense, and -1/12 can be substituted as the answer to the sum in some places.

## The Riemann Zeta function

This sum doesn’t come from nowhere – it is from an area of maths called analysis, sort of an extension of calculus. In fact, it is only one of a family of bizarre infinite sums. This specific one comes from something called the Riemann Zeta function, which looks like this:

$\zeta (s) = \sum_{n=1}^\infty \frac{1}{n^s} = \frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+...$

You’ll notice that, if you make s=-1:

$\zeta (-1) = \sum_{n=1}^\infty \frac{1}{n^{-1}} = \frac{1}{1^{-1}}+\frac{1}{2^{-1}}+\frac{1}{3^{-1}}+... = 1+2+3+4...$

So the value of this function at -1 is our strange series. This function is only really defined for values of s greater than 1, because otherwise the function heads off to infinity. However, through a very clever process called analytic continuation, the rest of the function can be assigned values, even though all the points are in reality infinite. This is part of why the sum 1+2+3+4… is so important – this function, the Riemann zeta function, is very important in mathematics (see the Riemann hypothesis.)

## The moral of the story is

There are a hundred different ways of looking at this result – most of them aren’t entirely correct, but each of them helps bring a little understanding to this bizarre mathematical creature. I seriously recommend going and doing more research on it, because it is an entry into a fascinating area of mathematics.

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