Can we have $\sum_{n\leq [x]}e^{-\sqrt{\frac{\log x}{r}}}\ll \frac{x}{e^{c \sqrt{\log x}}}$ for some constant $c>0$, where $x>1.$

Question

Let positive interger $n$ is square-free, that is $n=p_1p_2\cdots p_r$ some $r$. Can we have

$$\sum_{n\leq [x]}e^{-\sqrt{\frac{\log x}{r}}}\ll \frac{x}{e^{c \sqrt{\log x}}}$$ for some constant $c>0$, where $x>1.$

Answer 1

No.$$\frac{x}{e^{\sqrt{\log x/\log\log x}}}\ll\sum_{n\le x}e^{-\sqrt{\log x/r}}$$holds. This can easily be seen by the Hardy-Ramanujan theorem.

$|w(n)-\log\log n|<(\log\log n)^{1/2+\epsilon}$ for all $n\le x$ except $o(x)$ numbers by the Hardy-Ramanujan theorem. And the number of square-free numbers less than $x$ are $\displaystyle\frac{6}{\pi^2}x+O(\sqrt{x})$. Thus $$\sum_{n\le x}e^{-\sqrt{\log x/r}}>\sum_{x/3\le n\le x/2}|\mu(n)|e^{-\sqrt{\log x/w(n)}}>\left(\frac{1}{\pi^2}-o(1)\right)\sum_{x/3\le n\le x/2}e^{-\sqrt{\log x/\log\log x}}\gg\frac{x}{e^{\sqrt{\log x/\log\log x}}}$$ for all sufficiently large $x$.