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.

1more comment