It is well known that
$$\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$$
which for $s=2$ gives a convergence to $6/\pi^2$.
If we consider the partial sum, Apostol gives
$$\sum_{n\leq x} \frac{\mu(n)}{n^2}=\frac{6}{\pi^2}+O\left(\frac{1}{x}\right)$$
This bound is obtained by simply noting that
$$ \sum_{n>x} \frac{\mu(n)}{n^2}< \sum_{n>x} \frac{1}{n^2}=O\left(\frac{1}{x}\right)$$
However, experimental calculations on this convergence suggests a smaller error size. This is also somewhat intuitive, as substituting $\mu(n)$ with $1$ in the estimation of the $O$ term significantly enlarges the true error.
Is there any stronger bound for this convergence? I did not find any useful reference on this issue in the literature.
If $M(x) = \sum_{n\le x} \mu(n)$ is the Mertens function, then we can write $$ \sum_{n>x} \frac{\mu(n)}{n^2} = \int_x^\infty \frac1{t^2}\,dM(t) = \frac{M(t)}{t^2}\bigg|_x^\infty - \int_x^\infty M(t)\,d\frac1{t^2} = -\frac{M(x)}{x^2} + \int_x^\infty \frac{2M(t)}{t^3}\,dt. $$ (Here the intermediate steps use Riemann–Stieltjes integrals, but the final expression can be verified by hand to equal the tail sum on the left.)
Note that the trivial upper bound $|M(x)| \le x$ recovers the $O(1/x)$ bound from above. Any nontrivial upper bound on $|M(x)|$ will yield a better upper bound here. (Indeed, through a more complicated argument, these two types of upper bounds are equivalent.)
It is known that $M(x) \ll x\exp(-c\sqrt{\log x})$ for some constant $c>0$ (this follows from the proof used for the prime number theorem); such an upper bound leads to the improved estimate $O(1/x\exp(c\sqrt{\log x})$ for your tail sum. It is also known that the Riemann hypothesis is equivalent to $M(x) \ll_\varepsilon x^{1/2+\varepsilon}$ for every $\varepsilon>0$ (more precise upper bounds are also known); this upper bound would lead to the improved estimate $O_\varepsilon(1/x^{3/2-\varepsilon})$ for your tail sum. Related arguments can show that one can never get a better estimate than $O(1/x^{3/2})$ for the tail sum.