How can we use (as suggested by Tenenbaum's book on Introduction to Analytic and Probabilistic Number Theory) $$\lim_{X\rightarrow \infty}\sum_{\sqrt{X}\le p\le X}\dfrac{1}{p}=\log 2,$$
to prove that the following set has positive density?
$$\mathcal A_{1/2}:=\left\{n\le X:p\ge \sqrt{n} \text{ where $p$ is the largest prime divisor of $n$ }\right\}.$$
I think the following method is promising to conclude that $\mathcal A_{1/2}$ has indeed positive density, but it doesn't include the mentioned identity.
We have that the number of elements in $\mathcal A_{1/2}$, denoted $\#\mathcal A_{1/2}$, is at least (where we fixed $n$ along to be less than or equal to $X^{1/2}$ and combine it with a prime greater than or equal to $X^{1/2}$ to produce such number in the set. They are all different as $np=n'p'$ means $p|n'p'$ and $p|p'$ since $p>n$ so $p=p'$ and thus $n=n'$)
$$\sum_{n\le X^{1/2}}\sum_{\substack{p\le X/n \\ p\ge X^{1/2}}} 1=\sum_{n\le X^{1/2}}\left(\sum_{\substack{p\le X/n}} 1-\sum_{\substack{p\le X/n\\ p<X^{1/2}}}1\right),$$
and the last term is $o(X)$ where the term before it is (as $n\le X^{1/2}$ we have for large $X$, $\pi(X/n)>(X/n)/\log(X/n)$)
$$\sum_{n\le X^{1/2}}\sum_{p\le X/n}1>\sum_{n\le X^{1/2}}\dfrac{X}{n(\log X-\log n)}>\sum_{n\le X^{1/2}}\dfrac{X}{n\log X}\gg \dfrac{X}{\log X}\cdot \log X=X.$$
Therefore, $\#\mathcal A_{1/2}\gg X,$ or has positive density.
It seems that with the same method we might have $\#\mathcal A_{\epsilon}\gg X$ for as small $\epsilon>0$ as we please, where
$$\mathcal A_{\epsilon}:=\left\{n\le X:p\ge n^{1-\epsilon}\text{ where $p$ is the largest prime factor of $n$ }\right\}.$$
If this is true, it is pretty interesting that when $\epsilon=0$ the positiveness vanishes. I am not sure with the arguments and (as I asked first) how can we use such identity to prove this thing?
Best Answer
Let $n=mp$, with $1\leq m\leq X/p$ and $\sqrt{X}\leq p\leq X$ with $p$ prime. Then $n\in\mathcal{A}_{1/2}(X)$. Furthermore, since $p$ is the largest prime divisor of $n$, it follows that if $mp= m'p'$, then $p=p'$, and hence $m=m'$, so $n$ is unique for each $(m,p)$. Thus, $$|\mathcal{A}_{1/2}(X)|\geq\sum_{\sqrt{X}\leq p \leq X}\frac{X}{p}$$ Dividing by $X$ and taking the limit as $X\rightarrow\infty$ gives our desired result: $$\lim_{X\rightarrow\infty}\frac{|\mathcal{A}_{1/2}(X)|}{X}\geq \ln 2$$