Is anything known about the asymptotics of the sum $$ \sum_{d \mid n} \frac{\mu(d)}{d^2} $$ as $n$ tends to infinity? I am particularly interested in $\liminf_{n \to \infty}$. Here, $\mu$ is the Möbius function.
2
-
9$\begingroup$ This factorizes as $\prod_{p\mid n}(1-1/p^2)$, whose liminf is $\prod_{p}(1-1/p^2)=1/\zeta(2)$ (and limsup is $1$). $\endgroup$– Ofir GorodetskyCommented Nov 27 at 9:46
-
2$\begingroup$ $$\sum_{d\mid n} \frac{\mu(d)}{d^2}=\frac{J_2(n)}{n^2}$$ where $J_k(n)$ is the Jordan's totient function with average order $$J_k(n)\approx \frac{n^k}{\zeta(k+1)}$$ and consequently the average order of your sum is $$\sum_{d\mid n} \frac{\mu(d)}{d^2}=\frac{J_2(n)}{n^2}\approx\frac{1}{\zeta(3)}.$$ $\endgroup$– Steven ClarkCommented Nov 27 at 17:02
Add a comment
|