Estimate for a specific series

Question

For a positive integer $m$ define $$ a_m=\prod_{p\mid m}(1-p), $$ where the product is taken over all prime divisors of $m$, and $$ S_n=\sum_{m=1}^n a_n. $$ I am interested in an estimate for $|S_n|$. Any references, hints, ideas, etc., will be appreciated.

Answer 1

Let $a_n = \prod_{p \mid n} (1 - p)$. As $a_n$ is a multiplicative function, we have that \[\sum_{n = 1}^{\infty} \frac{a_n}{n^s} = \prod_p \left(1 + \sum_{k = 1}^{\infty} \frac{(1 - p)}{p^{ks}}\right) = \prod_p \left(1 + \frac{1 -p}{p^s(1 - p^{-s})}\right) = \prod_p \left(\frac{1 - p^{-(s - 1)}}{1 - p^{-s}}\right).\] So \[\sum_{n = 1}^{\infty} \frac{a_n}{n^s} = \frac{\zeta(s)}{\zeta(s - 1)}.\] The right-hand side defines a meromorphic function on $\mathbb{C}$, and by the standard zero-free region for $\zeta(s)$, it has no poles in the region $\Re(s) > 2 - 1/\log(|\Im(s)| + 2)$. Using this and the same sort of methods used to prove the prime number theorem, one can therefore show that \[\sum_{n \leq x} a_n = O\left(x^2 e^{-c\sqrt{\log x}}\right).\] Assuming the Riemann hypothesis, this can be strengthened to \[\sum_{n \leq x} a_n = O_{\varepsilon}\left(x^{3/2 + \varepsilon}\right).\] One can probably also prove this by first showing that \[a_n = \sum_{d \mid n} d\mu(d),\] and then estimating $\sum_{n \leq x} a_n$ and using the fact that $\sum_{n \leq x} \mu(n) = O(xe^{-c\sqrt{\log x}})$ by the prime number theorem.

Answer 2

As Oussama Boussif notice, using the Euler product for the totient function $$\phi\left(m\right)=m\prod_{p\mid m}\left(1-\frac{1}{p}\right) $$ we have $$a_{m}=\prod_{p\mid m}\left(1-p\right)=\left(-1\right)^{\omega\left(m\right)}\frac{\phi\left(m\right)}{m}\prod_{p\mid m}p $$ where $\omega\left(m\right) $ is the number of distinct prime factors of $m $, so we have using the fact that $\prod_{p\mid m}p\leq m $ (equality holds if $m $ is a squarefree number) $$\left|\sum_{m=1}^{n}a_{m}\right|\leq\sum_{m=1}^{n}\phi\left(m\right)=\frac{3}{\pi^{2}}n^{2}+O\left(n\log\left(n\right)\right). $$