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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.


Error in Formula (1) as a function of N

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) $\quad\gamma =12\,\log(A)-\log(2\,\pi)+\frac{6}{\pi^2}\,\zeta'(2)$

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1 Answer 1

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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

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  • $\begingroup$ 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. $\endgroup$ Mar 19, 2019 at 15:55
  • $\begingroup$ ??? ${}{}{}{}{}{}{}$ $\endgroup$
    – reuns
    Mar 19, 2019 at 16:09
  • $\begingroup$ 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)? $\endgroup$ Mar 21, 2019 at 20:11
  • $\begingroup$ 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 $f(x)$? $\endgroup$ Mar 22, 2019 at 16:26
  • $\begingroup$ @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 ) $\endgroup$
    – reuns
    Mar 23, 2019 at 2:50

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