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I am attempting to prove/disprove convergence of the following sum
$$ \lim_{n \to \infty} \frac{1}{n} \sum_{p \leq n} \sum_{k=0}^\infty \ln p \left\{\frac{n}{(p-1)p^k} \right\}$$
where $\{ x\}$ denotes the *fractional part* of $x$.

Let $\epsilon_n$ denote the above double sum. It's a trivial fact $\epsilon_n > 0$, since the summand contains positive terms. On the other hand, it can be shown $\epsilon_n \geq 1$. Using the facts $\{x\} + \{y\} \geq \{x+y\}$, $\{\frac{m}{n}\} \leq 1-\frac{1}{|n|}$, and $\sum_{p \leq n} \ln p \sim n$, we have

$$\epsilon_n = \frac{1}{n} \sum_{p \leq n} \ln p \sum_{k=0}^\infty \left\{\frac{n}{(p-1)p^k} \right\} \geq \frac{1}{n} \sum_{p \leq n} \ln p \left\{\frac{n}{p-1}\sum_{k=0}^\infty \frac{1}{p^k} \right\} $$

$$= \frac{1}{n} \sum_{p \leq n} \ln p \left\{\frac{np}{(p-1)^2} \right\} \geq \left(\frac{1}{n} \sum_{p \leq n} \frac{-\ln p}{(p-1)^2} +\ln p \right) \sim 1-\frac{C_n}{n} $$

where $C_n = \sum_{p \leq n} \frac{\ln p}{(p-1)^2}$. Now, the limit $C_n$ is known to be convergent (by the limit comparison test to the derivative of the prime zeta function $P(s)$, which is known to be convergent for $\mathrm{Re}(s) > 1$), so $\lim_{n \to \infty} \frac{C_n}{n} \to 0$. Thus $\epsilon_n \geq 1$. Perhaps this may not be of help, as I experience difficulty establishing an upper bound.