# Do the Prime Number Theorem and/or Riemann Hypothesis predict a limit on the accuracy of this formula for $\gamma$?

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
• Before asking this question I briefly explored $f(x)=\sum\limits_{n=1}^x\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right)$ and $F(s)=s\int\limits_0^{\infty }f(x)x^{-s-1}\,dx=\sum\limits_{n=1}^N\frac{\mu(n)}{n^{s+2}}\,\log\left(\frac{2\,\pi}{n}\right),\, N\to\infty$. I'm not sure the Dirichlet transform of $\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right)$ is $\frac{\log(2\,\pi)}{\zeta(s+2)}+\frac{\zeta'(s+2)}{\zeta(s+2)^2}$ because if it were as I'd expect $F(s)$ to converge to this function for $\Re(s)>-1$as $N\to\infty$ which it doesn't seem to do. Mar 19, 2019 at 15:55
• ??? ${}{}{}{}{}{}{}$ Mar 19, 2019 at 16:09
• Shouldn't "$\log(2\pi)/\zeta(s+2) + \zeta'(s+2)/\zeta(s+2)^2= s \int_1^\infty (\sum_{2 \le n \le x} \mu(n) n^{-2} \log(2\pi/n)) x^{-s-1}dx$ is holomorphic for $\Re(s) > ...$" be "$\log(2\pi)/\zeta(s+2) - \zeta'(s+2)/\zeta(s+2)^2= s \int_0^\infty (\sum_{1 \le n \le x} \mu(n) n^{-2} \log(2\pi/n)) x^{-s-1}dx$ is holomorphic for $\Re(s) > ...$" (where ... is $-3/2$ assuming the Riemann Hypothesis)? Mar 21, 2019 at 20:11
• I don't understand how using summation by parts to separate out $M(x)=\sum\limits_{n=1}^x\mu(n)$ helps derive a bound for $f(x)=\sum\limits_{n=1}^x\frac{\mu(n)}{n^2}\,\log\left(\frac{2\,\pi}{n}\right)$ as the bound of $M(x)$ increases as $x$ increases whereas the bound of $f(x)$ decreases as $x$ increases. The second Chebyshev function $\psi(x)=\sum\limits_{n=1}^x\Lambda(n)$ is similar to $M(x)$ in that its bound increases as $x$ increases. Are there well-known arithmetic functions that have bounds predicted by the RH that decrease as $x$ increases similar to $f(x)$? Mar 22, 2019 at 16:26
• @StevenClark See the end of the other question. Then it reduces to $M(x) = O(x^a)$ implies $\sum_n M(x) (n^{-s}-(n+1)^{-s}) = \sum_n M(x) O(sn^{-s-1})$ converges for $\Re(s) > a$ (for the details look at en.wikipedia.org/wiki/Dirichlet_series#Abscissa_of_convergence ) Mar 23, 2019 at 2:50