1+2+3+4+5+6+7+8+…= -1/12

Since the Knight’s Atari editors, and a number of its contributors, have a few doctorates in mathematics between them, it is probably fitting that we increase the amount of mathematics featured on Knight’s Atari – especially if we’re still trotting out the “editor of an award-winning publication featuring mathematical puzzles” waffle in job applications. To that end, we will attempt to include a little mathematics on a regular basis. Nothing too academic – just stuff which might crop up in everyday life.

Last night, for example, I attended the 30th birthday party of Dr. PÓC, one of Knight’s Atari’s contributors. His rather impressive cake* was adorned with small icing objects which bore some relevance to his life – a piano, a study desk, and so on. Of interest to us here, though, is the equation which was embossed in the copious icing around the edge of the cake:

1+2+3+4+5+6+7+8+… = -1/12

which anyone with any sort of grasp of mathematics will tell you is ridiculous.

Some of our readers might think that perhaps the cake artist simply made a grievous error in her calculations – summing infinite series is, after all, a difficult task – and that Dr. PÓC chose to let it slide. Those readers clearly do not know Dr. PÓC very well.

Let’s have a look then. Suppose you have the infinite series

1-1+1-1+1-1+1-1+… .

Let’s call this S1. What does this sum to? What if we added in a pair of brackets and rewrote the series as

1-(1-1+1-1+1-1+1-… )?

We can call this S2, but it is still the same thing, yes? We’ve just added a pair of brackets and adjusted the signs accordingly, but the sum is still the same. So even though we call them different things – hopefully for clarity – we still have that S2=S1.
Now – does the series inside the brackets in S2 look familiar? It should do. It’s our series S1 again.
So S2=1-S1. And if S2=S1, that means S1=1-S1, so the only solution is S1=1/2.

Let’s have a look at another series – let’s call it S3:

1-2+3-4+5-6+7-8+…

What does this one sum to? Well, we can see that if we take two copies of S3 and add them together as seen below, we find they equal our previous series:

1  – 2 + 3 – 4 + 5 – 6 + 7 – 8 + …
+        1 – 2 + 3 – 4 + 5 – 6 + 7 – …
=  1 – 1 + 1  –  1 + 1 – 1 + 1 – 1 + …

which is S1!

So if S1=1/2 and 2*S3=S1, then S3=1/4.

Now, back to our original series, which we should probably name at this point. Let’s call it SX. So we have SX= 1+2+3+4+5+6+7+8+…

Then SX-S3 is

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + …
– ( 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + …)

or, removing brackets,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + …
– 1 + 2 – 3 + 4 – 5 + 6 – 7 + 8 – …
= 0 + 4 + 0 + 8 + 0 + 12 + 0 + 16 + …
=4+8+12+16+…
=4*(1+2+3+4+5+6+…)
=4*SX

So SX-S3=4*SX and S3=1/4, so SX-1/4=4*SX and SX=-1/12!

So there you have it: 1+2+3+4+5+6+7+8+… = -1/12 .

Math – not even once.

Since it may seem like I am talking out my arse, I include here a few references, but do feel free to complain or even offer constructive criticism in the comments. It could be worth your while having a look at these links relating to the above piece of mathematics – not only for your erudition, but also in case you spot something new about the related zeta function which is an integral part of the Riemann Hypothesis, a proof of which would garner you a cool million dollars.

I also include a relevant quote from Niels Abel, after whom Mathematics’ infamous Abel Prize is named:
“The divergent series are the invention of the devil, and it is a shame to base on them any demonstration whatsoever. By using them, one may draw any conclusion he pleases and that is why these series have produced so many fallacies and so many paradoxes.”

*I will include a picture if one of our readers sends it in.