1. ## Problem

Hi--

Let $\displaystyle \phi_{1}(n)= n \sum\limits_{d \mid n} |\mu(d)|/d$, then prove that $\displaystyle \phi_{1}(n)= n \prod\limits_{d \mid n} (1+ \frac{1}{p})$

2. I’m not sure but I think you can use the Möbius inversion formula.

3. Originally Posted by Chandru1
Hi--

Let $\displaystyle \phi_{1}(n)= n \sum\limits_{d \mid n} |\mu(d)|/d$, then prove that $\displaystyle \phi_{1}(n)= n \prod\limits_{d \mid n} (1+ \frac{1}{p})$
Theorem: if $\displaystyle f(n)$ is a multiplicative arithmetic function, then so is $\displaystyle \sum_{d\mid n}f(d)$. Let me know if you would like to see the proof of this.
$\displaystyle |\mu(n)|$ is the square indicator function, so one can easily deduce it's multiplicative. Since $\displaystyle \frac{1}{n}$ is too, so is $\displaystyle f(n)=\frac{|\mu(n)|}{n}$.
Therefore if $\displaystyle n=p_1^{\alpha_1}p_2^{\alpha_2}\cdot\cdot\cdot p_k^{\alpha_k}$, then $\displaystyle \phi_1(n)= n\prod_{i=1}^k \phi_1(p_i^{\alpha_i}) = \prod_{i=1}^k p_i^{\alpha_i}\sum_{d\mid p_i^{\alpha_i}} \frac{|\mu(d)|}{d} = n\prod_{p\mid n}\left(1+\frac{1}{p}\right)$.