# Asymptotics on a double sum over primes

I am attemping to find asymptotics of

$$\sum_{p \leq n}\ln p \left( \sum_{k=1}^\infty \left(\left\{\frac{n}{p^k} \right\} - \left\{\frac{n-1}{(p-1)p^k} \right\} \right) - \left\{\frac{n-1}{p-1} \right\} \right)$$ where $$\{ x\}$$ denotes the fractional part of $$x$$ and $$p$$ denotes prime. On January 2016, Fedor Petrov consequently has shown

$$\sum_{p \leq n}\ln p\sum_{k=0}^\infty \left\{\frac{n-1}{(p-1)p^k} \right\} = (1-\gamma)n+O(n\epsilon_n),$$ where $$\gamma$$ denotes the Euler-Mascheroni constant and $$\epsilon_n \to 0$$. This leaves us with determining asymptotics for $$\sum_{p \leq n}\ln p\sum_{k=1}^\infty \left\{\frac{n}{p^k} \right\},$$ which I suspect is determined somewhere in literature since this is the fractional part discussed in Legendre's Theorem.

• I think the asymptotic for your third sum is the same as the second sum (namely $(1-\gamma)n+o(n)$), so to get an asymptotic overall you will at least need to know the second order terms in more detail than you give. You don't mention the third sum - do you know the asymptotics for $\sum_{p\leq n} \ln p \{ \frac{n-1}{p-1}\}$? – Thomas Bloom Jun 13 '20 at 13:09
• @Thomas Bloom I should have specified where the asymptotics for the second and third sum come from. To be exact, Petrov has shown that \begin{align} \sum_{p \leq n} \ln p \sum_{k \geq 1} \left\{ \frac{n-1}{(p-1)p^k}\right\} &= o(n) \\ \sum_{p \leq n} \ln p \left\{ \frac{n-1}{p-1}\right\} &= (1-\gamma)n + o(n). \end{align} – Maiyu Jun 13 '20 at 18:48
• Beyond the current terms, I'm not sure how I would go about finding higher order terms. If you would like more information about the current estimates, here is the link: mathoverflow.net/q/228738/70508 – Maiyu Jun 13 '20 at 19:20
• Please always include a top-level tag like "nt.number-theory" – GH from MO Jun 14 '20 at 1:34