What about the asymptotic behaviour of $\sum_{n=1}^N\frac{\log\operatorname{rad}(n)}{n}$ as $N\to\infty$? 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.

 A: In the following, $p$ always runs over primes, and $n,k$ over positive integers. By changing the order of summation, we obtain
\begin{align}
\sum_{n \leqslant x} \frac{\log (\operatorname{rad} n)}{n}
&= \sum_{n \leqslant x} \sum_{p\mid n} \frac{\log p}{n} \\
&= \sum_{p \leqslant x} \sum_{\substack{n \leqslant x \\ p \mid n}} \frac{\log p}{n} \\
&= \sum_{p \leqslant x} \frac{\log p}{p}\sum_{k \leqslant x/p} \frac{1}{k} \\
&= \sum_{p\leqslant x} \frac{\log p}{p}\biggl(\log \frac{x}{p} + \gamma + O\biggl(\frac{p}{x}\biggr)\biggr) \\
&= \log x\sum_{p\leqslant x} \frac{\log p}{p} - \sum_{p\leqslant x} \frac{(\log p)^2}{p} + \gamma\sum_{p\leqslant x} \frac{\log p}{p} + O\Biggl(\frac{1}{x}\sum_{p\leqslant x}\log p\Biggr).
\end{align}
From the prime number theorem, we have the strengthening of Mertens' first theorem that there is a constant $C$ such that
$$\sum_{p\leqslant x} \frac{\log p}{p} = \log x + C + o(1).\tag{1}$$
Also, $\vartheta(x) = \sum_{p \leqslant x} \log p \sim x$, so it remains to find the behaviour of
\begin{align}
\sum_{p \leqslant x} \frac{(\log p)^2}{p}
&= \bigl(\log x + C + o(1)\bigr)\log x - \int_2^x \frac{\log t + C + o(1)}{t}\,dt \\
&= (\log x)^2 + C\log x + o(\log x) - \frac{1}{2}\bigl((\log x)^2 - (\log 2)^2\bigr) - C\log x + C\log 2 - o(\log x) \\
&= \frac{1}{2}(\log x)^2 + o(\log x)
\end{align}
to overall get
$$\sum_{n \leqslant x} \frac{\log (\operatorname{rad} n)}{n} = \frac{1}{2}(\log x)^2 + (C + \gamma)\log x + o(\log x).$$
