Mertens function ($M(x)$)

The Mertens function is the summatory Möbius function

Let $\Delta (N)={\sum }_{n\le N}(\tau (n)-\log n-2\gamma )$. The Möbius divisor-sum identity and the Von Mangoldt divisor identity give $\mu \ast \tau =1$ and $\mu \ast \log =\Lambda$, soHere $\ast$ is Dirichlet convolution. The divisor summatory asymptotic bounds the terms with $d\le N/K$ in absolute value by $O(N/\sqrt{K})$. For $d>N/K$, group by $j=\lfloor N/d\rfloor <K$: the remaining sum is ${\sum }_{j<K}\Delta (j)(M(N/j)-M(N/(j+1)))=o_K(N)$. First let $N$ tend to infinity, then $K$ tend to infinity. This proves the Second Chebyshev function asymptotic from cancellation of the Mertens function.

Suppose $\Psi (x)=x+O(xE(x))$ and ${\sum }_{j\ge 0}{\sup }_{2^j\le u<2^{j+1}}|E(u)|<\infty$. The Möbius divisor-sum identity gives ${\sum }_{d\le x}\mu (d)/d=O(1)$. The exact Dirichlet convolution identity $\mu \log =-\mu \ast \Lambda$ then gives ${\sum }_{n\le x}\mu (n)\log n=O(x)$, because its error is a convergent dyadic harmonic sum. Since ${\sum }_{n\le x}\log (x/n)=O(x)$, the displayed Mertens function bound follows. The classical Prime number theorem error $E(x)=e^{-c\sqrt{\log x}}$ meets the summability condition. This is a deduction from the Chebyshev error itself, without a separate estimate for the reciprocal zeta function.

The classical zero-free region of the Riemann zeta function and a truncated Perron formula givefor some absolute constant $c>0$. Take contour height $T=\exp (\sqrt{\log x})$ and move the Perron contour to $\Re s=1-c_1/\log T$.