Sum Over Divisors of von Mangoldt is Logarithm

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

Then for $n \geq 1$


 * $\displaystyle \sum_{d | n} \Lambda(d) = \log n$

Proof
Let $n \geq 1$, 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 $d | n$ if any only if $d = p_1^{f_1}\cdots p_k^{f_k}$ with $0 \leq f_i \leq e_i$ for $i = 1,\ldots, k$.

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

If this is the case, say $d = p_i^{f_i}$ then $\Lambda(d) = \log p_i$. Therefore,


 * $\displaystyle \sum_{d | n}\Lambda(d) = \sum_{i = 1}^k e_i \log p_i$

Also, we have

Thus we indeed have


 * $\displaystyle \displaystyle \sum_{d | n}\Lambda(d) = \log n$