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
From the definition of the Von Mangoldt function, 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:


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

Consider the sum:


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

The sum runs over the pairs of natural numbers $\tuple {k, p}$ for which $p$ is a prime with $p^k \le x$ and $k \ge 2$.

Note that if $p^k \le x$, then $p \le x$.

So, alternatively, for each prime number $p$ we can sum over the natural numbers $k \ge 2$ for which $p^k \le x$.

That is:


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

Since $p \ge 2$, we have $\log p > 0$, so we only increase the sum by adding more terms.

So:

We have:

Note that for $p \ge 2$, we have:


 * $\ds \frac 1 2 p \le p - 1$

giving:


 * $\ds \frac 1 {p \paren {p - 1} } \le \frac 2 {p^2}$

So:


 * $\ds \sum_{p \le x} \log p \paren {\sum_{k \ge 2} \frac 1 {p^k} } \le 2 \sum_{p \le x} \frac {\log p} {p^2}$

From Order of Natural Logarithm Function, we have:


 * $\ds \frac {\log p} {p^2} \le \frac {2 \sqrt p} {p^2} = \frac 2 {p^{3/2} }$

for $p \ge 2$.

So:


 * $\ds 2 \sum_{p \le x} \frac {\log p} {p^2} \le 4 \sum_{p \le x} \frac 1 {p^{3/2} }$

Note that we have:


 * $\ds 4 \sum_{p \le x} \frac 1 {p^{3/2} } \le 4 \sum_{n \le x} \frac 1 {n^{3/2} }$

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 $C > 0$ such that:


 * $\ds \sum_{n \le x} \frac 1 {n^{3/2} } \le 4 C$

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:


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

That is:


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

Therefore:


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

so:


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