Order of Sum of Von Mangoldt Function of n over n

Theorem

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

Proof
We have:


 * $\ds \sum_{t \le x} \sum_{m \divides t} \map \Lambda m = x \log x - x + \map \OO {\map \log {x + 1} }$

Looking to the, we have:

From Second Chebyshev Function is Big-Theta of x, we have:

So we have:


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

So that, from Product of Big-O Estimates:


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

For $x \ge 1$, we have:

So we have:


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

From Sum of Big-O Estimates, we finally have:


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