If you add up 1 and 2, what do you get? The answer is 3. Ok. Let’s go one step further. What if you add up 1 and 2 and 3? What do you get now? Now the answer is 6. Now 1 plus 2 plus 3 plus 4. That sums to 10. Now what if I do this for ever, in other words add up all the natural numbers right to infinity? What do I get? Most people say it is infinity. Mathematicians often say that there is no sum because technically you can’t sum a ‘divergent series’, as opposed to a series which converges to a number (such as 1+1/2+1/4+1/8+…, which converges to 2).

But let’s be ambitious and see where we get.

Let’s start simple and add up the following series:

1-1+1-1+1-1+1-1+1-1+… to infinity.

What is this?

If you stop at an odd step in the series, such as the first or third or fifth step, the series sums to 1. But if you stop at an even step, say the second or fourth or sixth, the series sums to 0. Both are equally likely, so it is intuitively obvious that we can take the average of 1 and 0, which is 0.5 as the solution of this equation.

For those who aren’t convinced by the obvious, however, we can show it a little more rigorously like this:

Let S = 1-1+1-1+1-1 …….

So, 1-S = 1 – (1-1+1-1+1-1 …) = 1-1+1-1-1-1…

So, 1-S = S

So, 2S=1

Therefore, S = ½.

We can also show it by the method of averaging partial sums, which I’ve added in an appendix to this post, as the third method.

So there are three different ways to demonstrate that the series: 1-1+1+1-1+1-… equals 0.5.

Now that we have established this, the task of calculating the solution to:

1+2+3+4+5… becomes quite straightforward.

So we have established that 1-1+1-1+1-1+… = ½

We’ll call this series S1.

But what if we want to calculate S2, which is the series 1-2+3-4 + ….?

The way to do this is to add it to itself, to get 2.S2

1-2+3-4 +5- … + (1-2+3-4+5…)

The easiest way to do this is to move the second series one step along, which is fine as it is an infinite series. The start with 1 and now add up each pair of the remaining terms.

So we get:

1 + (-2+1) + (3-2) + (-4+3) + (5-4) + … = 1-1+1-1+1 ………….

But we have seen this series before. It is S1, and is equal to ½.

So, 2.S2 = ½

Therefore, S2 = ¼

Now what we are trying to sum is 1+2+3+4+5+6+…….

Let us call this S.

So, S – S2 = 1+2+3+4+5+6+… – (1-2+3-4+5-6…)

So, S-S2 = 0+4+0+8+0+12+…

This series is identical to: 4+8+12+16+20+24+…

This is 4 x (1+2+3+4+5+6+…)

In other words, S-S2 = 4S

We know already that S2 = ¼

Therefore, S-1/4=4S

So, 3S = -1/4

S= -1/12

And that is the proof that the sum of all the natural numbers up to infinity equals -1/12.

Who said it’s infinity? Who said you can’t sum divergent series? It’s got a solution and it’s the only meaningful one. Add up all the positive integers up to infinity and you get a negative number, -1/12.

There is no mathematical sleight of hand here. It is a properly derived solution, and we know from everything we understand about the laws of modern physics that it works in explaining the real world.

My next question is to ask you whether infinity is odd or even. What happens if I press the number 1 after 1 minute, then zero after a further 30 seconds, then 1 again after a further 15 seconds, then zero after a further 7.5 seconds and so on. What number am I pressing at the precise end of two minutes? Am I pressing 1 or zero, or both simultaneously, or neither. Imagine this was a magic light bulb that never blew. At the end of precisely 2 minutes, would it be on or off or both on and off? Or would it be -1/12 on and +1/12 off, or vice-versa.

Makes you think!

Thanks, by the way, to the excellent people on the Numberphile channel on YouTube, who inspired my interest in this.

References:

http://www.nottingham.ac.uk/~ppzap4/response.html

Appendix:

In the series, 1-1+1-1+1-1 …

The first term = 1.

The sum of the first and second terms = 1-1 = 0

The sum of the first, second and third terms = 1-1+1=1

The sum of the first, second, third and four terms = 1-1+1-1 =0

And so on.

So the series of these sums (known as partial sums) = 1,0,1,0 …

Now, averaging these partial sums gives the following series:

1 divided by 1, 1+0 divided by 2, 1+0+1 divided by 3, 1+0+1+0 divided by 4, etc.

This works out as:

1, ½, 2/3, ½, 3/5, ½, 4/7 …

If we continue this series, we end up with ½, as the odd numbered terms get ever smaller, and eventually vanishingly so, leaving us just with 1/2.

This method of averaging partial sums to derive the total sum is well-established and can be used, for example, to calculate the sum of 1+1/2+1/4+1/8+ …

This can be shown to converge on 2 using the same method.

From → Mathematics, Puzzles