This question is related to the following formula for Euler's constant $\gamma$ where $A$ is Glaisher's constant.
(1) $\quad\gamma=12\,\log(A)-\frac{\pi^2}{6}\sum\limits_{n=1}^N\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right),\quad N\to\infty$
The discrete plot in the following figure illustrates the error in formula (1) above as a function of $N$. The red evaluation points illustrate the error in formula (1) above where the Mertens function $M(N)=\sum\limits_{n=1}^N\mu(n)$ evaluates to zero.
Figure (1): Error in Formula (1) as a function of $N$
Question: Do the Prime Number Theorem and/or Riemann Hypothesis predict a limit on the accuracy of formula (1) for $\gamma$ as a function of $N$?
3/30/2019 Update:
Since $\sum_{n=1}^\infty\frac{\mu(n)}{n^2}=\frac{6}{\pi^2}$, formula (1) above can be simplified as follows.
(2) $\quad\gamma=12\,\log(A)-\log(2\,\pi)+\frac{\pi^2}{6}\sum\limits_{n=1}^N\frac{\mu(n)}{n^2}\,\log(n),\quad N\to\infty$
Formulas (1) and (2) above can be simplified further as follows.
(3) $\quad\gamma =12\,\log(A)-\log(2\,\pi)+\frac{6}{\pi^2}\,\zeta'(2)$
Yes and this is supposedly obvious. If $$\sum_{n=1}^N \mu(n) n^{-2} \log(2\pi/n) = C+O(N^a)$$ then $$\log(2\pi)/\zeta(s+2) + \zeta'(s+2)/\zeta(s+2)^2= s \int_1^\infty (\sum_{2 \le n \le x} \mu(n) n^{-2} \log(2\pi/n)) x^{-s-1}dx$$ is holomorphic for $\Re(s) > ...$
The converse is a matter of summation by parts to make $\sum_{n=1}^N \mu(n)$ appear as well as the converse theorems about its growth assuming the RH