# Number Theory Series Problem

First note that the operation * denotes the Dirichlet product, and µ denotes the Möbius function.

Ok so here is the problem:

Let f(x) be defined for all rational x in 0≤x≤1 and let $$F(n)=\sum_{k=1}^nf(\frac{k}{n})\;\;\;\;\; and \;\;\;\;\;\; G(n)=\sum_{k=1,\;(k,n)=1}^nf(\frac{k}{n}).$$

Prove that G=µ * F.

Formally here is what I've got: Set $$n=p_{1}^{e_1}...p_t^{e_t}.$$ than
$$\mu * F = \sum_{d|n}[\;\mu (d)\sum_{k=1}^{\frac{n}{d}}f(\frac{dk}{n})\;]$$$$=\sum_{k=1}^{n}f(\frac{k}{n})-[\;\sum_{k=1}^{\frac{n}{p_1}}f(\frac{p_1k}{n})+...+ \sum_{k=1}^{\frac{n}{p_t}}f(\frac{p_tk}{n})\;]+\sum_{k=1,\;i≠j}^{\frac{n}{p_ip_j}}f(\frac{p_ip_jk}{n})-...+f(\frac{n}{n}).$$

So this is probably a lot to take in but I'm pretty sure it's correct, just remember that the Möbius function is zero whenever d contains multiple copies of a prime in its factorization and thus many terms get zeroed out. Now, after working with some toy examples, the way I see this equaling G is that the initial series, which is just F, gets all its terms satisfying (k,n)>1 canceled out by terms coming from those series inside the brackets. But there are some terms in those series which have doubled up, tripled up, etc. and those terms will be canceled out by the later series(plural) which range over i,j and i,j,l and so on and so forth depending on the size of n. Now! The question is how the hell can I show this rigorously? It's just such a mess that there must be some way of simplifying the situation but I can't see it. Thanks.

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