Möbius function ($\mu (n)$)

The Möbius function assigns a cancellation weight to each positive integer:Thus a squarefree integer contributes a sign determined by its number of distinct prime factors; a repeated prime factor makes the weight zero. This packages the alternating signs of inclusion-exclusion into an arithmetic function. ^[ref]

For example, $\mu (2)=\mu (3)=-1$, $\mu (6)=1$, and $\mu (30)=(-1{)}^3=-1$, since $30=2\cdot 3\cdot 5$. In contrast, $\mu (4)=\mu (12)=\mu (60)=0$: all three have the square of the prime number $2$ as a divisor. On prime powers, $\mu (p)=-1$ and $\mu (p^a)=0$ for $a\ge 2$.

The Möbius function is a multiplicative function: for coprime integers, their prime factors are disjoint, so either a squared prime factor forces both sides of $\mu (mn)=\mu (m)\mu (n)$ to vanish, or the counts of distinct prime factors add and their signs multiply. It is not a completely multiplicative function, since $\mu (4)=0\ne 1=\mu (2{)}^2$. Its Möbius divisor-sum identity makes it the inverse of the constant-one arithmetic function under Dirichlet convolution: $\mu \ast 1=\epsilon$, the Identity for Dirichlet convolution. This is why the small set of weights $-1,0,1$ can undo an entire divisor sum.

For every positive integer $n$, the divisor sum of the Möbius function isThe sum includes all positive integer divisors, but only squarefree divisors contribute. Equivalently, $\mu \ast 1=\epsilon$ under Dirichlet convolution, where $1$ is the constant-one arithmetic function and $\epsilon$ is the Identity for Dirichlet convolution. ^[ref]

Each squarefree divisor chooses a subset of the distinct prime factors of $n$, contributing $(-1{)}^r$ when that subset has $r$ elements. Expanding a product therefore givesFor $n>1$ at least one prime factor occurs, so the product is zero. At $n=1$ it is an empty product, equal to one. This is the same alternating cancellation that drives inclusion-exclusion.

For $30=2\cdot 3\cdot 5$, all eight positive integer divisors are squarefree integers, givingPair each subset not containing $2$ with the subset obtained by inserting $2$: $(1,2),(3,6),(5,10),(15,30)$. Each pair cancels. The same nonzero terms occur for $60$, because its squarefree divisors are still the integer divisors of $30$; its other integer divisors have Möbius function value zero.

The Möbius divisor-sum identity is the cancellation behind Möbius inversion and reciprocal Dirichlet series: aggregation over integer divisors has an exact inverse.

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.

Möbius inversion recovers an arithmetic function from its divisor sum. For arithmetic functions $f,g$, the following statements are equivalent, with each equality required for every positive integer $n$:Here $d$ ranges over positive integer divisors of $n$ and $\mu$ is the Möbius function. No multiplicativity assumption is needed, and both sums are finite. The right side can also be written ${\sum }_{d\mid n}\mu (n/d)g(d)$ by exchanging complementary integer divisors. ^[ref]

The divisor sum $g(n)$ bundles $f(n)$ together with values at smaller integer divisors. Möbius inversion subtracts aggregates indexed by one distinct prime factor, adds those indexed by two, and continues with the alternating signs of inclusion-exclusion. A divisor containing a squared prime factor receives weight zero. The Möbius divisor-sum identity ensures that every unwanted contribution cancels exactly.

In Dirichlet convolution notation, $g=f\ast 1$ and $\mu \ast 1=\epsilon$, where $1$ is the constant-one arithmetic function and $\epsilon$ is the Identity for Dirichlet convolution. Associativity and commutativity give the short proof:Convolving $f=\mu \ast g$ with $1$ proves the reverse implication. At $n=1$, the formula reduces to $f(1)=g(1)$ because $\mu (1)=1$.

For a worked example, take $f(n)=\phi (n)$, Euler's totient function. The totient divisor-sum identity says that $g(n)=n$, the identity arithmetic function:This identity comes from classifying the integers $1,\dots ,n$ by their greatest common divisor with $n$: the group with $\gcd (j,n)=h$ has $\phi (n/h)$ members, since $j/h$ and $n/h$ are coprime integers. Summing the group sizes gives $n$, as detailed under totient divisor-sum identity.

Now undo this aggregation with the Möbius function:The nonzero terms use $d=1,2,3,6$; $d=4,12$ vanish because they contain the square of the prime number $2$. The four surviving integers coprime to $12$ are $1,5,7,11$.

This example also reveals why Möbius inversion is useful: it turns an easy count of all objects into a count satisfying a coprimality restriction. Expanding over subsets of distinct prime factors givesThe product is inclusion-exclusion over multiples of the prime factors. For $12$, it gives $12(1-1/2)(1-1/3)=4$. More generally, whenever a counting problem produces a divisor sum, Möbius inversion lets that aggregate be reversed without solving a new system of equations.

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 $\Re s=1-c_1/\log T$.