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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.

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 Dirichlet convolution of two multiplicative functions is a multiplicative function. For positive coprime integers $m,n$, every positive integer divisor of $mn$ has a unique form $de$ with $d\mid m$ and $e\mid n$, by prime factorization. The pairs $d,e$ and $m/d,n/e$ are each coprime integers, so the defining rules for the two multiplicative functions separate the sum:Also $(f\ast g)(1)=f(1)g(1)=1$. This short argument proves multiplicativity of both the divisor function $1\ast 1$ and the sum-of-divisors function $\mathrm{id}\ast 1$. The result need not be a completely multiplicative function, even when both inputs are: $1\ast 1=\tau$ fails at $2\cdot 2$. ^[ref]

Use the integer norm lower bound for rational approximation to a quadratic irrational. The integer $2q^2-a^2$ is nonzero, since $\sqrt{2}$ is irrational. If $a/q<2$, thenbecause $\sqrt{2}+a/q<\sqrt{2}+2<4$. If $a/q\ge 2$, the difference is at least $2-\sqrt{2}>1/4\ge 1/(4q^2)$. ThusIn particular, $\sqrt{2}$ is a badly approximable number.

To estimate the prime-weighted exponential sum, let $Q=\lfloor N^{1/2}\rfloor$. The Dirichlet approximation theorem gives a reduced fraction $a/q$ withCombining this with the lower bound gives $q>Q/4$. Thus $q\asymp N^{1/2}$; the approximation cannot have a small denominator.

The relevant analytic input is the prime exponential-sum estimate from Vaughan identity:Here is the structure behind that estimate. With Dirichlet convolution and cutoffs $U,V$, the Vaughan identity isIt follows from $\Lambda =\mu \ast \log$, $\log =\Lambda \ast 1$, and $\mu \ast 1={\delta }_1$. The first two Dirichlet convolution terms become type I sums with one short factor; the last becomes a bilinear type II sum with both factors bounded below. Split these sums into dyadic blocks and use $U=V=N^{2/5}$. For the type I sums, the inner finite geometric series has bound $\min (L,(2\Vert \alpha m\Vert {)}^{-1})$. For the type II sums, the Cauchy-Schwarz inequality followed by expansion introduces analogous sums in differences of indices. Rational approximation separates the phases into blocks of length comparable to $q$, bounding these reciprocal-distance sums. The Dirichlet convolution coefficients are divisor-bounded, producing the logarithmic factors. These steps give the displayed estimate; no unproved cancellation of arbitrary weighted sums is being assumed.

With $\alpha =\sqrt{2}$ and $q\asymp N^{1/2}$, the two denominator-dependent terms are $O(N^{3/4})$. ConsequentlyThe final passage absorbs the logarithm using the positive exponent margin $9/10-4/5=1/10$.

Extend $f$ by zero outside $[1,N]\cap \mathbb{Z}$. Here is a dyadic endpoint convention for small Type I and Type II sums. Write $D_M=[M,2M)\cap \mathbb{Z}$, with $M$ a power of two. Type I sums are $\delta$-small if, for every $M\le N^{1/100}$, dyadic $K$ with $MK\le N$, and arbitrary integer intervals $I_m\subset D_K$,Type II sums are $\delta$-small if, for all dyadic $M,K\in [N^{1/100},N^{99/100}]$ and sequences $|a_m|,|b_n|\le 1$,Zero extension handles boxes crossing the endpoint $mn=N$. Fixed choices of open or closed dyadic endpoints do not matter, but interval uniformity and uniformity in both bounded coefficient sequences do matter. These are the estimates used in the Vinogradov Type I–II method.

An interesting technical step is Fourier separation of interval cutoffs. For an integer interval $I_m$, let $\widehat{1_{I_m}}(\theta )={\sum }_{r\in I_m}e(-r\theta )$. Discrete Fourier inversion givesThe exponential geometric sum bound gives a common envelope $H_N(\theta )\ll \min (N,\Vert \theta {\Vert }^{-1})$, with ${\int }_0^1{H}_N\ll \log (2N)$. Divide the interval transform by this envelope: at each frequency it is a bounded coefficient depending on $m$, and $e(n\theta )$ is a bounded coefficient depending on $n$. The Type II sum assumption therefore bounds the interval-cutoff sum by $O(\delta N\log N)$. Choosing the outer unit phases to turn each inner sum into its modulus extends the Type I sum bound to boxes where both variables are in the Type II range.

Take $U=V=\lfloor N^{1/3}\rfloor$. With subscripts denoting truncation of arithmetic functions, the Vaughan identity iswhere $\ast$ now means Dirichlet convolution. For completeness it follows from $\mu \ast 1=ϵ$ and $\Lambda \ast 1=\log$: expand ${\mu }_{>U}\ast {\Lambda }_{>V}\ast 1$, cancel the short terms, and retain $\Lambda$. The short first term contributes $O(N^{1/3}\log N)$ to the sum against $f$.

The next two terms give Type I sums. The logarithmic term has outer coefficient $\mu (m)$, $m\le U$, and inner factor $\log n$, removed by partial summation at a cost $O(\log N)$. In the other term the outer variable is $m=cd\le UV$ with coefficientThis logarithmic coefficient bound in Vaughan identity avoids a loss from a divisor weight. On boxes with short outer variable use the hypothesis directly; on boxes where both variables are at least $N^{1/100}$ use the extended bound proved above. Boxes with inner variable less than $N^{1/100}$ contain at most $O(N^{2/3+1/100})$ pairs in total and are treated trivially. All remaining inner scales are at most $N^{99/100}$ whenever the outer scale exceeds $N^{1/100}$.

For the final convolution, group the variables asBoth variables lie between constant multiples of $N^{1/3}$ and $N^{2/3}$, well inside the Type II range. Divide $b(n)$ by $\log N$, and apply the Type II sum estimate. If the product cutoff is kept explicitly, Fourier separation of interval cutoffs costs only one extra logarithm; zero extension also permits its omission. There are $O({\log }^{2}N)$ boxes in the dyadic decomposition. The coefficient bound and, where needed, partial summation cost at most one more logarithm. These estimates give the convenient boundWith $\delta =(\log N{)}^{-20}$, each term is $o(N)$, proving the assertion. The grouping of the long term uses the nonnegativity of the Von Mangoldt function; more usual groupings instead control a divisor coefficient by a second-moment truncation. Ben Green's notes, Sections 2.3–2.4 describe that standard alternative.

Now take the sign version of the Thue–Morse sequence, $f(n)=(-1{)}^{s_2(n)}$. Concatenation of binary digits gives $f(a{2}^j+r)=f(a)f(r)$ for $0\le r<2^j$. Expanding over all choices of digits proves the Thue–Morse Fourier productPair adjacent factors. If $u=|\cos (\pi t)|$, their modulus issince the maximum on $[0,1]$ occurs at $u=1/\sqrt{3}$. Thus $|S_j(\theta )|\ll 2^{\sigma j}$ uniformly, where $\sigma =\frac{1}{2}{\log }_{2}B=2-\frac{3}{4}{\log }_{2}3<1$.

Every integer interval $J\subset [0,N]$ is a disjoint union of aligned binary blocks with at most two blocks of each length. On such a block the digit concatenation identity and the Thue–Morse Fourier product give the same bound, up to a unit phase. Summing the geometric series proves the uniform interval cancellation for binary digit parityFor a fixed $m$, the sum ${\sum }_{n\in I_m}f(mn)$ is a sum over multiples of $m$ in an integer interval $J\subset [1,N]$. The additive character orthogonality filter givesIts modulus is $O(N^{\sigma })$, independently of $m$ and the interval. Summing at most $O(N^{1/100})$ outer values proves the small Type I sums for binary digit parity:This is smaller than $(\log N{)}^{-20}$ for sufficiently large $N$. It proves the required Type I cancellation; the substantially harder Type II cancellation is not needed for this final clause.

Write $1(n)=1$ and let $\epsilon$ be the Identity for Dirichlet convolution. The Möbius function uses the empty-product convention $\mu (1)=1$; the Von Mangoldt function has $\Lambda (1)=0$. We first establish all the arithmetic function identities needed below. Every integer divisor of $n$ contributes one to $\tau (n)$, so $\tau =1\ast 1$. If $n>1$ has $r$ distinct prime factors, the squarefree divisors correspond to their subsets andAt $n=1$ the sum is one. Thus $\mu \ast 1=\epsilon$. Writing $n={\prod }_p{p}^{v_p(n)}$, the only divisors with nonzero Von Mangoldt function are $p,p^2,\dots ,p^{v_p(n)}$, and henceThe Dirichlet convolution is commutative by replacing $d$ with $n/d$, and associative because either bracketing sums $f(a)g(b)h(c)$ over $abc=n$. ConsequentlyThese are consequences of the definitions, rather than unproved invocations of Möbius inversion.

To identify the constant, integrate on each interval $[j,j+1)$:The tail is $O(1/J)$, since the integrand is at most $2/x^2$ for $x\ge 2$. Summation therefore gives $\gamma ={\lim }_{J\to \infty }(H_J-\log (J+1))$, where $H_J$ is the harmonic number, and $H_J=\log J+\gamma +O(1/J)$. Thus this is Euler's constant.

Put $r=\lfloor \sqrt{N}\rfloor$. Counting pairs of positive integers whose product is at most $N$, and subtracting the square counted twice, gives the Dirichlet hyperbola method identitySince $r=\sqrt{N}+O(1)$, the harmonic number estimate yieldsThe increasing natural logarithm also satisfies ${\sum }_{n\le N}\log n=N\log N-N+O(\log (2N))$ by comparison with its integral. Subtracting proves

For the second implication, denote the Mertens function by $M(x)$ and the Second Chebyshev function by $\psi (x)$. Apply the proved Dirichlet convolution identities to $b(n)=\tau (n)-\log n-2\gamma 1(n)$:Summing this identity over integers up to $N$ gives the exact relationFix an integer $K\ge 2$. The contribution to $E(N)$ from $d\le N/K$ has absolute value at mostFor $d>N/K$, group by $j=\lfloor N/d\rfloor$. This gives exactlyunder the assumed cancellation $M(x)=o(x)$; extending that assumption from integers to real $x$ uses $M(x)=M(\lfloor x\rfloor )$. Therefore ${\mathrm{limsup}}_{N\to \infty }|E(N)|/N\ll K^{-1/2}$. Letting $K\to \infty$ proves the prime number asymptotic from Mertens cancellation:

Let $\mu$ be the Möbius function. Its Möbius divisor-sum identity isIndeed, for $n>1$ the sum is the expansion of ${\prod }_{p\mid n}(1-1)$; nonsquarefree divisors contribute zero. Thus, for arithmetic functions $a,b$,To prove Möbius inversion, substitute the first formula into the second and collect the coefficient of $a(e)$. It is ${\sum }_{d\mid n/e}\mu (d)$, which is one for $e=n$ and zero otherwise. Conversely the same divisor interchange recovers $b$ from the displayed formula for $a$. In Dirichlet convolution notation this is simply $\mu \ast 1=\epsilon$, where $\epsilon$ is supported at one.

The Prime number theorem with classical zero-free-region error states that some absolute $c>0$ satisfiesHere $\Lambda$ is the Von Mangoldt function. An equivalent prime-counting form, after decreasing $c$ if needed, is $\pi (x)=\mathrm{Li}(x)+O(x{e}^{-c\sqrt{\log x}})$. The contributions of proper prime powers are $O(\sqrt{x}{\log }^{2}x)$ and can be absorbed into this error. We use $\Psi$ to keep it distinct from the Fourier transform appearing in Question 5.

Here is a direct deduction of the requested Mertens bound from a log-integrable Chebyshev error from the stated Prime number theorem. First, the divisor identity givesSecond, the exact Dirichlet convolution identityfollows from $\log =1\ast \Lambda$: multiplication of a Dirichlet convolution by $\log n$ differentiates its two factors, so applying it to $\mu \ast 1=\epsilon$ and convolving again with $\mu$ gives this formula. Summing it yieldsThe last harmonic sum is $O(1)$. To see this without an endpoint approximation, split $x/d$ into dyadic ranges $[2^j,2^{j+1})$. In each range $\sum 1/d=O(1)$, and its exponential factor is at most $e^{-c\sqrt{j\log 2}}$. Their sum over $j\ge 0$ converges. The finitely many terms with $1\le x/d<2$ obey the same estimate after adjusting the constant in the prime-number-theorem error. ThereforeFor the Mertens function $M(x)={\sum }_{n\le x}\mu (n)$,The second sum has absolute value at most ${\sum }_{n\le x}\log (x/n)=O(x)$, by an integral comparison or the Stirling formula. ConsequentlyThe absolute value in this conclusion is present in the original PDF.

A precise useful form of the principle is the type I–type II inverse principle for Möbius correlation. Normalize $|f(n)|\le 1$, and supposeFor any positive integer cutoffs $U,V$ with $U+V\le \delta X/2$, defineThen $|c_d|,|a_d|\le \tau (d)$, where $\tau$ is the divisor function, and at least one of the following two sums has modulus at least $\delta X/4$:The first is correlation against a controlled linear combination of indicators of multiples of small moduli, each a periodic function: this is the “somewhat periodic” branch. The second is a bilinear sum with independently weighted factors, both larger than the cutoffs: this is the “somewhat multiplicative” branch. The statement concerns these precise correlations, not an assertion that $f$ must itself be periodic or multiplicative.

For clarity, the Vaughan identity for the Möbius function proves this version immediately. Split $\mu ={\mu }_{\le U}+{\mu }_{>U}$ and likewise at $V$. Since $\mu \ast \mu \ast 1=\mu$,Multiply by $f(n)$ and sum. The first two terms contribute at most $U+V$ in modulus, and the remaining terms are $-T_{\mathrm{I}}+T_{\mathrm{II}}$. The triangle inequality gives the stated dichotomy. It is valid at finite $X$ and for any chosen cutoffs in the indicated range.

We now prove the required orthogonality without appealing to a stronger uniform exponential-sum theorem. Write $\alpha =\sqrt{2}$ and $e(u)=e^{2\pi iu}$. The Diophantine bound for the square root of two iswhere $\Vert u\Vert$ is distance to the nearest integer. If $a$ is that nearest integer, then $|2h^2-a^2|\ge 1$, while $h\sqrt{2}+a\le 4h$; dividing proves the bound.

Use the preceding Dirichlet convolution identity with $U=V=P=\lfloor X^{1/10}\rfloor$. The two short terms are $O(P)$. For the type I sum, the exponential geometric sum bound givesConsequentlyusing ${\sum }_{d\le Y}\tau (d)\ll Y\log (2Y)$ from counting factor pairs.

For the type II sum, perform a dyadic decomposition into $d\in (D,2D]$, $w\in (W,2W]$, keeping $dw\le X$. Nonempty blocks have $DW<X$, while $D,W\gg P$. A block $B$ satisfies, by the Cauchy-Schwarz inequality and the divisor-square summatory bound,To justify the cutoff, after expanding the square the permissible $d$ for a pair $w,w^{\prime }$ still form an interval, with upper endpoint $\min (2D,X/\max (w,w^{\prime }))$. Its exponential sum has the same geometric bound. Diagonal pairs supply $DW$; for each nonzero difference $h=w-w^{\prime }$ there are $O(W)$ pairs.

The points $0,\alpha ,2\alpha ,\dots ,\lfloor W\rfloor \alpha$ modulo one are separated by at least $1/(4W)$. Counting points in consecutive distance bands around zero therefore givesSubstitution yields the bilinear cancellation for badly approximable phases estimateThere are $O({\log }^{2}X)$ blocks, so $T_{\mathrm{II}}\ll X^{19/20}{\log }^{4}X$. The divisor-square bound itself can be obtained elementarily from $\tau (n{)}^2\le {\tau }_4(n)$ prime-power by prime-power and ${\sum }_{n\le Y}{\tau }_4(n)\le Y(1+\log Y{)}^3$; no uncontrolled coefficient bound is being suppressed.

Combining the short, type I and type II terms,Hence

Write $f(n)=3^{\omega (n)}$, where $\omega$ is the prime omega function. If $p^a\Vert n$, its contribution to the Dirichlet convolution $(f\ast \Lambda )(n)$ isSince $3(a-1+1/3)\ge a$, summing over the prime factors proves the log-weighted convolution bound for three to the prime omega, $f(n)\log n\le 3(f\ast \Lambda )(n)$. On $\sqrt{D}\le n\le D$, $\log n\ge \frac{1}{2}\log D$, henceThis is the required inequality. The standard Chebyshev estimate $\psi (t)\ll t$ bounds its right side by $D(\log D{)}^{-1}{\sum }_{m\le D}f(m)/m$. To bound the latter sum, use multiplicativity and extend to all numbers with prime factors at most $D$:The Mertens second theorem states ${\sum }_{p\le D}1/p=\log \log D+O(1)$. Taking logarithms of the product, with a summable $O(p^{-2})$ remainder, gives $3\log \log D+O(1)$. Therefore the product is $O({\log }^{3}D)$, proving the summatory bound for three to the prime omega in the required range:

Let $1(n)=1$ and $ϵ(n)=1_{n=1}$, the identity for Dirichlet convolution. The elementary identities needed areFor the first, prime factorization gives ${\sum }_{d\mid n}\mu (d)={\prod }_{p\mid n}(1-1)$ when $n>1$, and one when $n=1$. For the second, if $n=\prod p^{a_p}$, then ${\sum }_{d\mid n}\Lambda (d)={\sum }_p{a}_p\log p=\log n$, the Von Mangoldt divisor identity. Convolving the second identity with $\mu$ proves the third.

For $U,V\ge 1$, let the subscripts $\le U$, $>U$, $\le V$, $>V$ denote truncations. The Vaughan identity isIts pointwise form isTo prove it, use ${\mu }_{>U}\ast 1=ϵ-{\mu }_{\le U}\ast 1$ and $1\ast {\Lambda }_{>V}=\log -1\ast {\Lambda }_{\le V}$. This givesand rearrangement proves the formula, as in the Vaughan identity proof.

The Bombieri–Vinogradov theorem states that for every $A>0$ there is $B>0$ such thatHere $\psi (y;q,a)={\sum }_{\begin{array}{c}n\le y\\ n\equiv a(\mathrm{mod}q)\end{array}}\Lambda (n)$ is the Chebyshev function in an arithmetic progression, and $\varphi$ is the Euler totient function. In particular, partial summation givesafter choosing the logarithmic saving in the weighted theorem sufficiently large and absorbing the prime power terms. This is the offset logarithmic integral function.

For the proof strategy, Orthogonality of Dirichlet characters converts errors in arithmetic progressions into character sums of Dirichlet characters weighted by $\Lambda$. Reduction to primitive Dirichlet characters is followed by the character large sieve, which bounds their mean square by $O((N+Q^2)\sum |a_n{|}^2)$ with the usual $q/\varphi (q)$ weights. The Vaughan identity separates the weighted sums into short terms, Type I sums and Type II sums. Direct inner-sum estimates handle the Type I sums; the Cauchy-Schwarz inequality and the large sieve control the Type II sums. The Siegel–Walfisz theorem supplies arbitrarily strong logarithmic savings for small moduli, where an average estimate alone would not suffice. A dyadic decomposition, suitable $U,V$ and a sufficiently large $B$ absorb the divisor and logarithmic losses. This explains why the range is essentially $Q^2\le x$, with logarithmic room to spare.

The Euler totient function counts integers $1\le a\le n$ with $\gcd (a,n)=1$. The Möbius function has $\mu (1)=1$, is zero when a prime square divides $n$, and equals $(-1{)}^r$ when $n$ is a product of $r$ distinct primes. Consequentlybecause for $n>1$ the sum over squarefree divisors is $(1-1{)}^r$.

The Möbius inversion formula is $g(n)={\sum }_{d\mid n}\mu (d)f(n/d)$. Substituting the divisor sum defining $f$ and collecting the coefficient of $g(e)$ givesThis proof uses only finite sums, so no convergence hypothesis is needed.

Classifying $1,\dots ,n$ by their greatest common divisor with $n$ gives ${\sum }_{d\mid n}\phi (d)=n$: the class with greatest common divisor $n/d$ contains exactly $\phi (d)$ integers. Applying the Möbius inversion formula yieldsThe Riemann zeta function is $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\mathrm{Re}s>1$. Absolute convergence justifies multiplication of Dirichlet series. The divisor identity for the Möbius function gives $\sum \mu (n)n^{-s}=1/\zeta (s)$ there. For $\mathrm{Re}s>2$ both factors below converge absolutely, and their Dirichlet convolution is the Euler totient function:

The Möbius function has $\mu (1)=1$, $\mu (n)=0$ when $n$ is divisible by a prime square, and $\mu (n)=(-1{)}^r$ when $n$ is a product of $r$ distinct primes. The Riemann zeta function is initially the Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\sigma =\Re s>1$. Both this Dirichlet series and $\sum \mu (n)n^{-s}$ have absolute convergence, because $|\mu (n)|\le 1$ and $\sum n^{-\sigma }<\infty$.

For a finite set of primes $p\le y$, multiplication of absolutely convergent geometric series and unique prime factorization giveEvery omitted integer has a prime factor exceeding $y$, so it exceeds $y$. The difference from $\zeta (s)$ has absolute value at most ${\sum }_{n>y}{n}^{-\sigma }\to 0$. This proves the Euler product, with the product interpreted as the limit of its finite products:

For the reciprocal identity, absolute convergence justifies multiplying and regrouping the two Dirichlet series by their product index. Their Dirichlet convolution coefficient at $m$ isThe coefficient calculation uses only the squarefree integers dividing $m$. Consequently the reciprocal identity is an identity of absolutely convergent series:

An arithmetic function $f$ is multiplicative if $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime integers. If $g(n)={\sum }_{d\mid n}f(d)$ and $(m,n)=1$, every positive integer divisor of $mn$ is uniquely $d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Henceso $g$ is multiplicative.

The Möbius function is defined by $\mu (1)=1$, $\mu (n)=0$ if the square of a prime number divides $n$, and $\mu (n)=(-1{)}^r$ if $n$ is a product of $r$ distinct prime numbers. Coprime integers have disjoint sets of prime factors, so this definition immediately gives $\mu (mn)=\mu (m)\mu (n)$ when $(m,n)=1$.

For $n=p_1^{a_1}\cdots p_r^{a_r}>1$, only the square-free divisors contribute, and thereforeFor $n=1$ the sum is $1$. Thus $1\ast \mu =\epsilon$ in the language of Dirichlet convolution. Convolving $g=1\ast f$ with $\mu$ gives the Möbius inversion formula

If $f(n)=n$, its divisor sum is the sum-of-divisors function. Conversely, if $g(n)=n$, Möbius inversion givesusing the standard divisor identity for the Euler totient function. Therefore the two requested answers are