0=1–1/2–1/3–1/5+1/6–1/7+1/10-…

The series presented in the title is a very complicated summation, but with a very simple resolution.

It is what one could call “The Dirichlet Series for s=1 of the Möbius function”, which just means it is a infinite series like this:

And this letter μ represents the Möbius function in this situation.

Now the most peculiar is the origin of this function — if you do not already know it — , which is quite random and can be summarized mostly by:

But, with only this definition it does not match exactly the series written in the title of this post — if you notice that there is a missing 1/4 there — , but there woudn’t with only this definition. The Möbius funciton is also 0 when there are repeating prime factors on n.

Now after the undersating of this function— of which I describe much better in the beginning of the post “Why the Möbius Function is very polynomial-like” — , we can start to understand the reason for why — in sigma nottation:

To get the reason for this we must first understand what was Euler’s method to find a infinite product over the prime numbers which will represent the Harmonic Series.

The way to do this is quite ingenius and inovative and it can be sketched first by just writing our Harmonic Series as:

And now, if we were to take the half of our x, we would then have only the even fractions:

And then we took this representation of the half of the Harmonic Series out of the actual Harmonic Series, we would then have:

And then we did this again, for the next denominator after 1, we would then have:

And as we keep going the next number after 1 will obviously not be a multiple of any of the other prime numbers that came before it and therefore is as well prime and then, by extent if one would keep repeating this proccess for ever we’d have:

And then, by necessity, if one wrote these fractions more clearly we would say that:

Which is quite pretty on it’s own and deserves better understanding, but I won’t expalin more of it for now.

Consider now that product that is written as a coefficient of the Harmonic Series. That product’s expansion — at least as is written with 1–1/2 for example — will be forming the values of the Möbius function believe it or not, because each term it generates will, first, not have repating factors, and then, will have all numbers that can be formed with a multiplication of prime numbers — therefore all numbers — but without numbers with repeating factors, and then they will alternate just like defined on the Möbius function. More evidence of this can be found once we start expanding like this:

And, as you can see, they really do form the numerators as the Möbius function.

Now, the product in this form does not show obviously that it goes to 0, but written as I wrote before will reveal that the numerator is infinitely larger than the numerator in that fraction, and therefore, it must be 0, because there are infinitely many prime numbers. And therefore, in other words:

And there we have it!

Soli Deo Gloria.