Question about Mobius function.

Question

Let $N \in \mathbb{N}.$ I would know if is it true that $$-\underset{k\mid N}{\sum}\mu\left(k\right)\log\left(k\right)>0.$$I know that $$-\mu\left(k\right)\log\left(k\right)=\underset{r\mid k}{\sum}\mu\left(r\right)\Lambda\left(\frac{k}{r}\right)$$ but I don't think is useful.

Answer 1

Without loss of generality, suppose that $N = p_1 \cdots p_\ell$, where $p_i \neq p_j$ are primes. We can identify a factor of $N$ with a subset of $\{1,\ldots,\ell\}$, say through a mapping $\varphi$. Then $\mu(\varphi(A)) = (-1)^{|A|}$ and $\log\varphi(A) = \sum_{i \in A} \log p_i$. Therefore $$ \begin{align*} \sum_{k|N} \mu(k) \log k &= \sum_{A \subseteq \{1,\ldots,\ell\}} (-1)^{|A|} \sum_{i \in A} \log p_i \\ &= \sum_{i=1}^\ell \log p_i \sum_{B \subseteq \{1,\ldots,\ell\} \setminus \{i\}} (-1)^{|B|+1} \\ &= \begin{cases} -\log p_1 & \text{if }\ell = 1, \\ 0 & \text{if }\ell \neq 1. \end{cases} \end{align*} $$

Answer 2

In general, if $m\ge 1$, and $N$ has more than $m$ distinct prime factors, we have $$ \sum_{k\mid N}\mu(k)\log^m(k)=0. $$ This can be easily seen by induction.