So I am trying to prove the above stated formula. I am completely new to number theory so I thank you in advance for baring with me. I start off by using the definition:

$\frac{\phi(n)}{n} = \sum_{d|n} \frac{\mu(d)}{d}$

Where $\phi$ is Eulers function, and $\mu$ is the Mobius mu function. From here, I use the Mobius inversion formula to give:

$\frac{\mu(n)}{n} = \sum_{d|n}d\frac{\phi(\frac{n}{d})}{n}\mu(d)$

Now summing $n$ up to $x$:

$\sum_{n \leq x}\frac{\mu(n)}{n} = \sum_{n \leq x}\sum_{d|n}d\frac{\phi(\frac{n}{d})}{n}\mu(d)$

Now I'm not fully sure on how I can rearrange things inside the two summations on the right hand side, so leaving things as they are; using that $\sum_{d|n} \mu(d) = 1$, if and only if $n =1$. Hence:

$\sum_{n \leq x}\frac{\mu(n)}{n} = \sum_{d|1}d\phi(\frac{1}{d})\mu(d)$

Thus $d=1$. Subbing this in:

$\sum_{n \leq x}\frac{\mu(n)}{n} = \phi(1) = 1$

I am not sure what I have missed (or done wrong) as to have an equality and not a triangle relation. Any suggestions or ideas would be greatly appreciated

  • $\begingroup$ This is proved at pp. 66-67 of Apostol (Intro. to Analytic Number Thy.) Different approach. $\endgroup$ – daniel Dec 31 '15 at 19:10

Hint you don't need to use the Möbius inversion formula , you can directly use : $$1 = \sum_{d \leq x} \mu(d) \left\lfloor \frac{x}{d}\right \rfloor = \sum_{d \leq x} \mu(d) \frac{x}{d} - \sum_{d<x} \mu(d) \left \{ \frac{x}{d} \right\} \ \ \ \ \ (1)$$

and use the fact that $\left|\mu(d) \left \{ \frac{x}{d} \right\}\right|\leq 1$ and you're done (the $d<x$ in the second summation is very important)

  • $\begingroup$ in my opinion, from $\sum \mu(d) \lfloor x /d \rfloor$ you get $\sum_{d \le x} \frac{\mu(d)}{d} = \frac{1}{x} \left(1 + \sum_{d \le x} \mu(d) \{x/d\} \right)$ which is the solution to the question. and that $\sum_{d \le x} \mu(d) \{x/d\} = \mathcal{O}(x^{1/2+\epsilon})$ would imply $\sum_{d \le x} \frac{\mu(d)}{d} = \mathcal{O}(x^{-1/2+\epsilon})$ and thus the Riemann hypothesis. $\endgroup$ – reuns Dec 31 '15 at 19:46
  • $\begingroup$ yes that's what I mean by my hint, For the implication you're absolutely right but we don't know how to prove that $\displaystyle\sum_{d \le x} \mu(d) \{x/d\} = \mathcal{O}(x^{1/2+\epsilon})$ or the other equivalence $\displaystyle \sum_{n \leq x} \mu(n) = O(x^{1/2+o(1)})$ $\endgroup$ – Elaqqad Dec 31 '15 at 22:39
  • $\begingroup$ This result was not relegated to exercises in Apostol and OP say s/he is new to number theory. So I wonder if this is enough. $\endgroup$ – daniel Jan 1 '16 at 17:49
  • $\begingroup$ @daniel that's exactly why (S)he is allowed to leave comments if it's not enough $\endgroup$ – Elaqqad Jan 1 '16 at 19:36

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