Proof
(related to Proposition: Natural Logarithm Sum of von Mangoldt Function Over Divisors)
- Let $n\in\mathbb N,$ $n > 0.$
- If $n=1$, then the sum $( * )$ is empty and we have the correct equation $0=0.$
- If $n > 1,$ consider the canonical representation $$n=\prod_{i=1}^\infty p_i^{e_i}=\prod_{p\mid n}p^e\quad ( * * )$$
where the second product is built for all primes dividing $n$ and the exponent $e$ depends on both, $n$ and the particular prime number $p$ in the product.
- Taking the natural logarithm on both sides of $( * * )$ we get
$$\log(n)=\sum_{p\mid n}e\cdot \log(p)=\sum_{p\mid a}(\Lambda(p)+\Lambda(p^2)+\ldots+\Lambda(p^e))=\sum_{d\mid n}\Lambda(d).$$
- This is the required sum of the von Mangoldt function over all divisors of $n.$
∎
Thank you to the contributors under CC BY-SA 4.0! 
- Github:
-
References
Bibliography
- Scheid Harald: "Zahlentheorie", Spektrum Akademischer Verlag, 2003, 3rd Edition
- Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927
