Motivation. Let $\mu(n)$ the Möbius function and we denote with $$\operatorname{rad}(n)=\prod_{p\mid n}p$$ the radical of an integer $n\geq 1$, see this Wikipedia.
Fact. Using that $\mu(n)=0$ iff $\operatorname{rad}(n)<n$, and thus $$\sum_{n=1}^\infty\frac{\mu(n)}{n}\log\frac{n}{\operatorname{rad}(n)}=0+0+\ldots=0,$$ or well seeing the identity $(11)$ from this MathWorld, (by cases 1) $\mu(n)=0$, 2) $\operatorname{rad}(n)=n$ with $\mu(n)=1$ and 3)$\operatorname{rad}(n)=n$ with $\mu(n)=-1$ ) one can to prove $$\sum_{n=1}^\infty\frac{\mu(n)}{n}\log\operatorname{rad}(n)=-1.$$
Since series involving $\operatorname{rad}(n)$ are interestings, I've thought this exercise:
Question. We know Chebyshev's result about the asymptotic behaviour of $\sum_{p\leq n}\frac{\log p}{p}$ and since the series $$\sum_{n=1}^\infty\frac{\log\operatorname{rad}(n)}{n}$$ has positive terms, I know that it diverges. But, is it possible to deduce some more precisely about the asymptotic behaviour of the partial sums $$\sum_{n=1}^N\frac{\log\operatorname{rad}(n)}{n}$$ as $N\to\infty?$ Provide hints, or a detailed answer, as you prefer. Or references if it is well known. Many thanks.