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 $r(n) = n^{-1}$.