# 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.

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

$$\sum_{n \leqslant x} \frac{\log (\operatorname{rad} n)}{n} = \frac{1}{2}(\log x)^2 + (C + \gamma)\log x + o(\log x).$$

• Nice! The term $o(\log x)$ is in fact $\frac12 (\log 2)^2 +C\log2+ C\gamma + 1+ O(e^{-c\sqrt{\log x}})$. – Sungjin Kim May 17 '17 at 22:12
• You are a number theorist! Many thanks for all details, and also thanks to @i707107 – user243301 May 18 '17 at 7:30