Asymptotics on sum of n/rad(n) [duplicate]

Question

I'm interested in asymptotics for the sum $$\sum_{n\le x}\frac{n}{\text{rad}(n)}.$$ For my research I only need to know whether or not this is $\mathcal{O}(x)$, but I would appreciate more precise asymptotics. Additionally, a reference would be much appreciated. The sum of $\frac{1}{\text{rad}(n)}$ has been discussed on the site, but I cannot find anything on my sum.

Answer 1

One has \begin{align*} \sum_{n \leq x}\frac{n}{\operatorname{rad}n} & = (1+o(1)) \, x \sqrt{\frac{2}{\log x \log \log x}} \exp\left((1+o(1))\sqrt{\frac{8\log x}{\log \log x}}\right) \ (x \to \infty), \\ & = x\exp\left((1+o(1))\sqrt{\frac{8\log x}{\log \log x}}\right) \ (x \to \infty) \\ & \neq O(x\, (\log x)^A) \ (x\to \infty) \end{align*}

for any $A \in \mathbb{R}$, where the first estimate holds according to user "Ofir Gorodetsky"'s detailed answer to a similar question to yours at Asymptotic behavior of a "strange" arithmetic function.

Answer 2

Here is a quick way to show that $\sum_{n\leq x}f(n)$ is not $O(x)$. For more precise results, see Ofir Gorodetsky's response here.

The arithmetic function $f(n):=n/\mathrm{rad}(n)$ is multiplicative, with values at prime powers given by $f(p^k)=p^{k-1}$, hence its Dirichlet generating series equals $$F(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}=\prod_p\left(1+\sum_{k=1}^\infty\frac{p^{k-1}}{p^{ks}}\right)=\prod_p\left(1+\frac{1}{p^s-p}\right).$$ It follows that the ratio $$\frac{F(s)}{\zeta(s)}=\prod_p\left(1+\frac{p-1}{p^{2s}-p^{s+1}}\right)$$ tends to infinity as $s\to 1+$ along the real axis. So $F(s)$ is not $O(1/(s-1))$ for $s>1$, which implies by partial summation that $\sum_{n\leq x}f(n)$ is not $O(x)$.

If we divide $F(s)$ not by $\zeta(s)$ but by $\zeta(s)^{A+1}$, we can conclude in a similar fashion that $\sum_{n\leq x}f(n)$ is not $O(x(\log x)^A)$.

Added. The OP asked in a comment below this post why $$\sum_{n\leq x}f(n)=O(x)\qquad\Longrightarrow\qquad F(s)=O\left(\frac{1}{s-1}\right).$$ Here are the details. Denote the sum on the LHS by $S(x)$, and assume that $S(x)=O(x)$. Then, for $s>1$, we have that $$F(s)=\int_{1-}^\infty\frac{dS(x)}{x^s}=\left[\frac{S(x)}{x^s}\right]_{1-}^\infty+s\int_1^\infty\frac{S(x)}{x^{s+1}}dx.$$ On the right-hand side, the first term vanishes by $S(x)=O(x)$. Therefore, assuming $1<s<2$, $$F(s)=\int_1^\infty\frac{O(x)}{x^{s+1}}dx=O\left(\frac{1}{s-1}\right).$$