Order of Sum over Primes of Logarithm of p over p

Theorem

 * $\ds \sum_{p \le x} \frac {\log p} p = \log x + \map \OO 1$

Proof
We have that:


 * $\ds \map \Lambda m = \begin{cases}\ln p & m = p^k \text { for some prime } p \text { and } k \in \N \\ 0 & \text{otherwise}\end{cases}$

so:

that is:


 * $\ds \sum_{m \le x} \frac {\map \Lambda m} m - \sum_{p \le x} \frac {\log p} p = \sum_{k \ge 2} \sum_{p^k \le x} \frac {\log p} {p^k}$

We have:

Note that we have:


 * $\ds \sum_{p \le x} \frac 1 {p^{3/2} } \le \sum_{n \le x} \frac 1 {n^{3/2} }$

From Convergence of P-Series:


 * $\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^{3/2} }$ converges.

So from Convergent Real Sequence is Bounded, there exists a real number $C$ such that:


 * $\ds \sum_{n \le x} \frac 1 {n^{3/2} } \le C$

for all $x$.

Then:


 * $\ds 0 \le \sum_{p \le x} \sum_{k \ge 2, \, p^k \le x} \frac {\log p} {p^k} \le 4 C$

for all $x$.

So:


 * $\ds \sum_{p \le x} \sum_{k \ge 2, \, p^k \le x} \frac {\log p} {p^k} = \map \OO 1$

Therefore:


 * $\ds \sum_{p \le x} \frac {\log p} p = \sum_{m \le x} \frac {\map \Lambda m} m + \map \OO 1$

so:


 * $\ds \sum_{p \le x} \frac {\log p} p = \log x + \map \OO 1$