Möbius function

The Möbius function is $\mu (1)=1$, is zero on integers divisible by a prime square, and is $(-1{)}^r$ on products of $r$ distinct primes.

For $n>1$, expand ${\prod }_{p\mid n}(1-1)=0$ over the subsets of its distinct prime divisors. The coefficient of a squarefree divisor is its Möbius function value; nonsquarefree divisors have value zero. At $n=1$ the empty product is one. This identity is the multiplicative cancellation behind Möbius inversion and reciprocal Dirichlet series.

Put $A_j(x)={\sum }_{d\le x}\mu (d){\log }^j(x/d)/d$. The elementary estimates areThe identities ${\sum }_d\mu (d)\lfloor x/d\rfloor =1$ and ${\sum }_d\mu (d)H_{\lfloor x/d\rfloor }/d=1$ give the first two bounds. Convolve the harmonic divisor sum with $\mu (d)/d$: because $\mu \ast \tau =1$, its left side is $H_{\lfloor x\rfloor }$. Thus $\frac{1}{2}{A}_2+2\gamma A_1+c{A}_0=\log x+O(1)$. The summed error is $O(x^{-1/2}{\sum }_{d\le x}{d}^{-1/2}\log (2x/d))=O(1)$.

This elementary smoothing argument is used in Selberg's original proof and Ramaré's sieve lectures, Lemma 4.2.

If arithmetic functions $f,g$ satisfythen

The Mertens function is the summatory Möbius 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 $\Re s=1-c_1/\log T$.