We can be sure, that $$\frac{\zeta(k)}{\zeta(k+1)}=\sum\limits_{n=1}^{\infty}\frac{\varphi(n)}{n}\cdot\frac{|\mu(n)|}{J_{k}(n)}= \sum\limits_{n=1}^{\infty}\frac{\varphi(n)}{n}\cdot\frac{1}{n^k}$$ Here $J_{k}(n)$ - Jordan totient function. First equation is pretty simple: $$\frac{\zeta(k)}{\zeta(k+1)}=\prod\limits_{p}^{\infty}\frac{p^k}{p^k-1}\cdot\frac{p^{k+1}-1}{p^{k+1}}=\prod\limits_{p}^{\infty}\frac{p^{k+1}-1}{p(p^k-1)}=\prod\limits_{p}^{\infty}1+\frac{p-1}{p(p^k-1)}$$ $$\frac{\zeta(k)}{\zeta(k+1)}=\sum\limits_{n=1}^{\infty}|\mu(n)|\cdot\prod\limits_{p|n}\frac{p-1}{p(p^k-1)}= \sum\limits_{n=1}^{\infty}\frac{\varphi(n)}{n}\cdot\frac{|\mu(n)|}{J_{k}(n)}$$ That's because for $|\mu(n)|$ $$\prod\limits_{p|n}p=n, \prod\limits_{p|n}p-1=\varphi(n), \prod\limits_{p|n}p^k-1=J_{k}(n)$$ How can I prove second?
If I made some mistakes, sorry for my English.
Note that coefficient of Dirichlet series of $fg$ is just the convolution of $f$ and $g$.
The convolution of $\varphi(n)/n$ and $1/n$ is: $$\sum_{d\mid n} \frac{\varphi(d)}{d} \frac{1}{n/d} = \frac{1}{n} \sum_{d\mid n} \varphi(d) = 1$$ Hence $$\left(\sum_{n=1}^\infty \frac{\varphi(n)}{n}\frac{1}{n^k} \right) \left( \sum_{n=1}^\infty \frac{1}{n}\frac{1}{n^k} \right) = \sum_{n=1}^\infty \frac{1}{n^k} = \zeta(k)$$