Proof of Von Mangoldt Dirichlet series

I am going to show a proof of \[\sum_{n\geq 1}\dfrac{\Lambda(n)}{n^s}=-\dfrac{\zeta'(s)}{\zeta(s)}\] We start off by writing this as \[\sum_{p=prime}\sum_{k=1}^\infty \dfrac{\ln(p)}{p^{sk}}=\sum_{p=prime} \dfrac{\ln(p)}{p^s-1}\] Let this be; we will use this later.

Lemma: ∑n≥1F(n)ns=ζ(s)∑p=primeF(p)ps−1\sum_{n≥1} \dfrac{F(n)}{n^s}=\zeta(s)\sum_{p=prime}\dfrac{F(p)}{p^s-1} where F(n) is a completely additive function.

Proof: we can split the sum over primes. F(pk)=kF(p)F(p^k)=kF(p). we use this and get ∑p=prime∑k=0∞F(pk)psk(∑p∤n1ns)=∑p=prime∑k=0∞kF(p)psk(ζ(s)−ζ(s)p−s)=∑p=primeF(p)ζ(s)(1−p−s)∑k=0∞kpsk=ζ(s)∑p=primeF(p)ps−1\large\sum_{p=prime}\sum_{k=0}^\infty \dfrac{F(p^k)}{p^{sk}}\left(\sum_{p\not\mid n} \dfrac{1}{n^s}\right)=\sum_{p=prime}\sum_{k=0}^\infty \dfrac{kF(p)}{p^{sk}}\left(\zeta(s)-\zeta(s)p^{-s}\right)\\=\sum_{p=prime}F(p)\zeta(s)(1-p^{-s})\sum_{k=0}^\infty\dfrac{k}{p^{sk}}= \zeta(s)\sum_{p=prime}\dfrac{F(p)}{p^s-1}

Now ln⁡(n)\ln(n) is a completely additive function, so ∑n=1∞ln⁡(n)ns=ζ(s)∑p=primeln⁡(p)ps−1\sum_{n=1}^\infty \dfrac{\ln(n)}{n^s}=\zeta(s)\sum_{p=prime}\dfrac{\ln(p)}{p^s-1} We know that −ζ′(s)=∑n=1∞ln⁡(n)ns-\zeta'(s)=\sum_{n=1}^\infty \dfrac{\ln(n)}{n^s} and putting in the RHS's summation in terms of Von Mangoldt: ∑n=1∞Λ(n)ns=−ζ′(s)ζ(s)\sum_{n=1}^\infty \dfrac{\Lambda(n)}{n^s}=-\dfrac{\zeta'(s)}{\zeta(s)}

#NumberTheory

Note by Aareyan Manzoor
5 years, 3 months ago

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Comments

I have changed a typo: From "completely multiplicative function" to "completely additive function" (regarding ln(n))

Other than that, great proof!

Extra: Alternative proof of lemma:

n=∏pj∣npjwjn=\prod _{p_j|n}^{ }p_j^{w_j}

F(n)=∑jwjF(pj)=∑pk∣nF(p)=1∗f(n)P(n)F(n)=\sum _j^{ }w_jF\left(p_j\right)=\sum _{p^k|n}^{ }F\left(p\right)=1*f(n)P(n)

Where f(n) satisfies f(p^k)=F(p) and P(n) is 1 if n=p^k and is 0 otherwise.

Julian Poon - 5 years, 3 months ago
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