This question is about the evaluation of $a(n)$ defined in (1) below which is related to the Riemann zeta function $\zeta(s)$ as illustrated in (2) below.
(1) $\quad a(n)=\sum\limits_{d|n}\frac{\mu\left(\frac{n}{d}\right)}{d^2}=\frac{A046970(n)}{n^2}$
(2) $\quad \frac{\zeta(s+2)}{\zeta(s)}=\sum\limits_{n=1}^\infty a(n)\ n^{-s}\,,\quad\Re(s)>\frac{1}{2}\quad\text{(assuming Riemann hypothesis)}$
A046970 in formula (1) above is the Dirichlet inverse of the Jordan totient function $J_2(n)$ (A007434) where $J_k(n)$ is a generalization of the Euler totient function $\phi(n)$ which are both defined below.
(3) $\quad\phi(n)=n\prod\limits_{p|n}\left(1-\frac{1}{p}\right)$
(4) $\quad J_k(n)=n^k\prod\limits_{p|n}\left(1-\frac{1}{p^k}\right)$
The following figure illustrates $a(n)$ always seems to evaluate such that $-1<a(n)<1$ except at $n=1$ where $a(1)=1$, and there is also a noticeable banding in the values of $a(n)$.
Figure (1): Discrete Plot of $a(n)$
Question (1): Can it be proven that $-1<a(n)<1$ for all $n>1$? I've verified $-1<a(n)<1$ for $1<n\le 10^6$.
Question (2): Is there an explanation for the noticeable banding in the values of $a(n)$ in Figure (1), and does this banding pattern continue as $n\to\infty$?
If $f(n)$ is multiplicative, then $$ \sum_{d|n}\mu(d)f(d)=\sum_{p|n}[1-f(p)] $$
As a result, we have
$$ a(n)={1\over n^2}\sum_{d|n}\mu(d)d^2={1\over n^2}\sum_{p|n}(1-p^2) $$
Let $n$ be factorized as $\prod_{p|n}p^{e_p}$ then $$ |a(n)|=\left|\sum_{p|n}p^{-2e_p}(1-p^2)\right|=\prod_{p|n}p^{-2e_p}|p^{-2}-1|=\prod_{p|n}p^{2-2e_p}|1-p^{-2}| $$ Since $p^{2-2e_p}\le1$ and $|1-p^{-2}|<1$, we have $|a(n)|<1$. Since $a(n)\in\mathbb R$, this can be re-expressed as $$ -1<a(n )<1 $$