An upper bound on $\sum_{n^{1/3}<p,q\leq n^{1/2}} \frac{n}{pq}-\lfloor \frac{n}{pq}\rfloor$

Question

I would like to ask if there is a good upper bound on the difference $$D_2(n)=\sum_{n^{1/3}<p,q\leq n^{1/2}} \left(\frac{n}{pq}-\left\lfloor \frac{n}{pq}\right\rfloor\right)\quad (1) $$where $p$ and $q$ range over primes in the given interval. I believe that the ratio $$R_2(n)=\frac{\sum_{n^{1/3}<p,q\leq n^{1/2}} \left\lfloor \frac{n}{pq}\right\rfloor}{\sum_{n^{1/3}<p,q\leq n^{1/2}} \frac{n}{pq} }\quad (2)$$ approaches 1, as $n\rightarrow \infty$ from below.

Question 1: Is the claimed limit for (2) correct? How would it be proved rigorously?

Question 2: Is there a tight upper bound on $D_2(n)$ as $n$ goes to infinity?

More generally, define

$$D_k(n)=\sum_{n^{1/(k+1)}<p_1,\ldots,p_k\leq n^{1/k}} \left(\frac{n}{p_1 p_2 \cdots p_k}-\left\lfloor \frac{n}{p_1 p_2 \cdots p_k}\right\rfloor\right)$$ where the $p_i$ range over primes in the interval given, and define $R_k(n)$ analogously.

Can Questions 1 and 2 be answered for these quantities? Note that I am intersted in both the case of $k$ fixed, as well as $k$ slowly growing, but satisfying $k\leq c \lfloor \log \log n \rfloor,$ with $c<2$ a constant. My goal is to approximate the sum with the floor functions by means of the sum without the floor functions.

Answer 1

Based on @GerhardPaseman and @alpoge's remarks I have written this answer, for $k=2.$ Consider $$ D_2(n)=\sum_{n^{1/3}<p,q\leq n^{1/2}} \left(\frac{n}{pq}-\left\lfloor \frac{n}{pq}\right\rfloor\right)\quad (1) $$ where $p,q$ are primes, and note that each term in (1) is positive, and in $[0,1),$ which gives $$ D_2(n)<\#\{p \in (n^{1/3},n^{1/2}]: p~~a~~prime~\}^2=[\pi(n^{1/2})-\pi(n^{1/3})]^2=$$ $$ \approx \left[\frac{2n^{1/2}}{\log n}-\frac{3n^{1/3}}{\log n}\right]^2\ll \frac{n}{\log^2 n}.$$ Now consider the approximating sum $$ S_2'(n)=\sum_{n^{1/3}<p,q\leq n^{1/2}} \frac{n}{pq}=\left[\sum_{n^{1/3}<p\leq n^{1/2}}\frac{\sqrt{n}}{p}\right]^2=n\left[\sum_{n^{1/3}<p\leq n^{1/2}}p^{-1}\right]^2\asymp $$ $$ \asymp n [\log\log(n^{1/2})-\log\log(n^{1/3})]^2=\log^2(3/2) n =c~ n,\quad(2) $$ with $c \in (0,1).$ From (1), the sum I am interested in satisfies $$ S_2(n)=\sum_{n^{1/3}<p,q\leq n^{1/2}} \left\lfloor \frac{n}{pq}\right\rfloor> c~n -\frac{n}{\log^2 n} \gg c~n$$ for $n$ large enough.

For $k$ larger and fixed, one can argue essentially the same way, and the implied constant $c$ in (2) becomes $\log^2((k+1)/k)$ so increasingly smaller.

The open question is, what can be done, if anything, to lower bound $$ S_k(n)=\sum_{n^{1/(k+1)}<p_1,\ldots,p_k\leq n^{1/k}} \left\lfloor \frac{n}{p_1 p_2 \cdots p_k}\right\rfloor $$ for $k=\alpha \log\log n,$ and $0<\alpha<2.$