Sum Over Divisors of von Mangoldt is Logarithm

Theorem

Let $\Lambda$ be von Mangoldt's function.

Then for $n \ge 1$:

$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$

Proof

Let $n \ge 1$, and by the Fundamental Theorem of Arithmetic write $n = p_1^{e_1} \cdots p_k^{e_k}$ with $p_1, \ldots, p_k$ distinct primes and $e_1, \ldots, e_k > 0$.

Now $d \divides n$ if any only if $d = p_1^{f_1} \cdots p_k^{f_k}$ with $0 \le f_i \le e_i$ for $i = 1, \ldots, k$.

By the definition of $\Lambda$, for such $d$ we have $\map \Lambda d \ne 0$ if and only if there is exactly one $i \in \set {1, \ldots, k}$ such that $f_i > 0$.

If this is the case, let $d = p_i^{f_i}$.

Then:

$\map \Lambda d = \ln p_i$

Therefore:

$\ds \sum_{d \mathop \divides n} \map \Lambda d = \sum_{i \mathop = 1}^k e_i \ln p_i$

Also, we have:

\(\ds \ln n\)

\(=\)

\(\ds \ln (p_1^{e_1} \cdots p_k^{e_k})\)

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 1}^k e_i \ln p_i\)

Sum of Logarithms

Thus we indeed have:

$\ds \sum_{d \mathop \divides n} \map \Lambda d = \ln n$

$\blacksquare$