I derived the relationship illustrated in (1) below which I believe converges for $s>0\lor\Re(s)>\frac{1}{2}$ assuming the Riemann hypothesis. The function $rad(n)$ is the radical or square-free kernel of $n$ (i.e. the greatest square-free divisor of $n$).

(1) $\quad\frac{\zeta(s+1)}{\zeta(s)}=\sum\limits_{n}\frac{b(n)}{n^s}\,,\quad b(n)=\frac{\mu(rad(n))\,\phi(n)\,rad(n)}{n^2}$

I noticed the question and answer at the following link define a similar relationship which is illustrated in (2) below where $d$ and $n$ are positive integers and $p$ is a prime.

Answer to Question on $\frac{\zeta(s+1)}{\zeta(s)}$

(2) $\quad\frac{\zeta(s+1)}{\zeta(s)}=\sum\limits_{n}\frac{c(n)}{n^s}\,,\quad c(n)=\sum\limits_{d\mid n}\frac{d}{n}\mu(d)=\frac{1}{n}\prod\limits_{p\mid n}(1-p)$

The $c(n)$ function defined in (2) above seems to evaluate exactly the same as the $b(n)$ function defined in (1) above, but the proof of the equivalence of the two is not totally obvious to me.

Question: What is the proof that $b(n)$=$c(n)$?


Both functions are multiplicative, so it suffices to check the equality at prime powers. For a prime $p$ and $k \geqslant 1$ we have

$$b(p^k) = \frac{\mu(\operatorname{rad}(p^k))\phi(p^k)\operatorname{rad}(p^k)}{p^{2k}} = \frac{\mu(p)(p-1)p^{k-1}p}{p^{2k}} = \frac{1-p}{p^k} = c(p^k),$$

thus $b = c$.

Concerning the multiplicativity, note that the pointwise product of multiplicative functions is multiplicative, and each of $n \mapsto \mu(\operatorname{rad}(n))$, $n \mapsto \phi(n)$, $n \mapsto \operatorname{rad}(n)$, $n \mapsto n^{-2}$ is multiplicative. Also the Dirichlet convolution of multiplicative functions is multiplicative, and $c = \mu \ast r$, where $r(n) = n^{-1}$.

  • $\begingroup$ @DanielFisher Thanks for taking the time to respond to my question. I was fairly sure of my derivation of $b(n)$, but your proof pretty much cements the correctness of it in my mind. $\endgroup$ – Steven Clark Mar 24 '18 at 17:29
  • $\begingroup$ @DanielFisher I did notice an anomaly at $n=1$. My implementations of $b(n)$ and the Dirichlet convolution $c(n)=\left(\mu(n)*\frac{1}{n}\right)(n)=\sum\limits_{d\mid n}\mu(d)\frac{d}{n}$ both evaluate to $1$ at $n=1$ which is the desired result, but my initial implementation of $c(n)=\frac{1}{n}\prod\limits_{p\mid n}(1-p)$ evaluated to $0$ at $n=1$. I adjusted my implementation to compensate for this anomaly, but I'm not sure I fully understand the justification for $c(1)=1$. $\endgroup$ – Steven Clark Mar 24 '18 at 17:30
  • $\begingroup$ Looks like your implementation includes a factor $(1 - 1)$ when $n = 1$. That can easily happen with factorisation code, one tends to think of the normal case $n > 1$ when there are prime factors. Depending on the used language, for the factorisation of $1$ one should return an empty list or so. $\endgroup$ – Daniel Fischer Mar 24 '18 at 18:17
  • $\begingroup$ @DanielFisher Your explanation is exactly what was happening. I was using the Mathematica $FactorInteger(n)$ function which includes $1$ in the list of factors only when $n=1$. I realized what was happening and adjusted my implementation, but my confusion is about $c(1)=\frac{1}{1}\prod\limits_{p|1}(1-p)=1$ which implies $\prod\limits_{p|1}(1-p)=1$ which doesn't make much sense to me. $\endgroup$ – Steven Clark Mar 24 '18 at 19:01
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    $\begingroup$ It's an empty product, and an empty product is defined to be $1$, just like an empty sum is defined to be $0$. $\endgroup$ – Daniel Fischer Mar 24 '18 at 19:06

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