Let $$ S(n)= \sum_{p \leq n \\ k > 0} \left\{\frac{n}{(p-1)p^k}\right\},$$ where sum is taken over the primes $p$ and $n$ is natural. If $k > \log_2 n$, then the remainder term is \begin{align*} \sum_{p \leq n, \, k > \log_2 n} \left\{\frac{n}{(p-1)p^k}\right\} &= \sum_{p \leq n, \, k > \log_2 n} \frac{n}{(p-1)p^k}- \sum_{p \leq n, \, k > \log_2 n} \left \lfloor\frac{n}{(p-1)p^k}\right\rfloor\\ &= \sum_{p \leq n} \frac{n}{(p-1)} \sum_{k \geq \lfloor \log_2 n \rfloor + 1} \frac{1}{p^k} \\ &= \sum_{p \leq n} \frac{n}{p^{\lfloor \log_2 n \rfloor}(p-1)^2}, \end{align*} for which an analytic expression can be obtained via the Abel summation formula. Thus, we have \begin{align*} S(n)= \sum_{p \leq n \\ k \leq \lfloor \log_2 n \rfloor} \left\{\frac{n}{(p-1)p^k}\right\} + n\sum_{p \leq n} \frac{1}{p^{\lfloor \log_2 n \rfloor}(p-1)^2}. \end{align*} The remainder term is expected to approach $0$ for large $n$ (as a matter of fact, it appears to decay like $1/\ln n$); now, the summation for positive $k \leq \log_2 n$ remains. We may bound this term by $$\sum_{p \leq n \\ k \leq \lfloor \log_2 n \rfloor} \left\{\frac{n}{(p-1)p^k}\right\} \leq \sum_{p \leq n \\ k \leq \lfloor \log_2 n \rfloor}1 = \pi(n)\lfloor \log_2 n \rfloor,$$ which implies $S(n)$ behaves like $n$ for $n$ large enough (as a matter of fact, $o(n)$). However, I believe we can actually obtain a closed form though I'm not sure how.
Question: Is there a closed form analytic expression for the sum $$\sum_{p \leq n \\ k \leq \lfloor \log_2 n \rfloor} \left\{\frac{n}{(p-1)p^k}\right\} ?$$