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The pointwise product of arithmetic functions multiplies two arithmetic functions at the same input: $(f\cdot g)(n)=f(n)g(n)$. Thus $(1\cdot 1)(12)=1$ and $(\mathrm{id}\cdot 1)(12)=12$, using the constant-one arithmetic function and the identity arithmetic function. These differ from the corresponding Dirichlet convolution values $6$ and $28$, because Dirichlet convolution sums over every positive integer divisor.

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.

Put $g(n)=\mu (n{)}^2/\phi (n)$. As a Dirichlet convolution, $g=h\ast (n\mapsto 1/n)$, where $h$ is a multiplicative arithmetic function with $h(p)=1/(p(p-1))$, $h(p^2)=-1/(p(p-1))$, and $h(p^j)=0$ for $j\ge 3$. The local sums are one, so ${\sum }_{n}h(n)=1$. The series ${\sum }_n|h(n)|\log (2n)$ converges, since the prime-local errors are $O((\log p)/p^2)$. Consequently $G(z)={\sum }_{d\le z}h(d)H_{\lfloor z/d\rfloor }=\log z+O(1)$ by the harmonic number estimate $H_m=\log m+O(1)$. This identifies the leading constant in the one-dimensional Selberg upper-bound sieve.

The sum-of-divisors function adds all positive integer divisors of a positive integer, including the number itself:It is the divisor sum of the identity arithmetic function $\mathrm{id}(n)=n$, so $\sigma =\mathrm{id}\ast 1$ under Dirichlet convolution. Both factors are multiplicative functions, hence so is the sum-of-divisors function. The notation ${\sigma }_1$ distinguishes it from other weighted divisor sums. ^[ref]

For a prime number $p$ and a nonnegative exponent $a$, the positive integer divisors of the prime power $p^a$ are precisely $1,p,\dots ,p^a$. Their finite geometric series givesFor example, $\sigma (8)=1+2+4+8=15=(2^4-1)/(2-1)$. For a prime factorization $n={\prod }_{k=1}^r{p}_k^{a_k}$ with distinct prime numbers $p_k$ and positive exponents $a_k$, multiplicativity givesFor $n=1$, this is an empty product, equal to $1$. ^[ref]

The distributive property makes the formula especially beautiful: choosing one term from each factor produces each positive integer divisor exactly once, by the fundamental theorem of arithmetic. For $12=4\cdot 3=2^2\cdot 3$, the choices form a three-by-two grid with entries $2^i{3}^j$, where $0\le i\le 2$ and $0\le j\le 1$. Adding its rows givesThe divisor function counts six cells; the sum-of-divisors function adds their six values. Once the prime factorization is known, the product formula avoids testing every candidate integer divisor. Shared prime factors require care: $\sigma (4)=7\ne 9=\sigma (2{)}^2$, so the sum-of-divisors function is not a completely multiplicative function.

Subtracting the number itself gives the proper-divisor sum $s(n)=\sigma (n)-n$. Comparing $s(n)$ with $n$ distinguishes perfect numbers, deficient numbers, and abundant numbers. For $6$, $s(6)=1+2+3=6$; for $28$, $s(28)=1+2+4+7+14=28$. Both are perfect numbers. The calculation $s(8)=1+2+4=7<8$ makes $8$ a deficient number, while $s(12)=1+2+3+4+6=16>12$ makes $12$ an abundant number. These comparisons turn the sizes of integer divisors into a classification of positive integers. ^[ref]

This connection is ancient: Euclid's Elements, Book IX, Proposition 36, constructs a perfect number by multiplying the last term of $1,2,\dots ,2^{k-1}$ by their sum, when that sum is a prime number. In modern notation, put $q=2^k-1$ and $N=2^{k-1}q$, with $k\ge 2$ and $q$ a prime number. The sum-of-divisors function proves the construction immediately:Consequently the proper-divisor sum is $s(N)=N$. Taking $k=2,3$ produces the perfect numbers $6,28$. ^[ref]

The divisor sum of Euler's totient function is the identity arithmetic function:Equivalently, $\phi \ast 1=\mathrm{id}$ under Dirichlet convolution, with $1$ the constant-one arithmetic function. ^[ref]

To see why, group the integers $j=1,\dots ,n$ by $h=\gcd (j,n)$. Each $h$ is a positive integer divisor of $n$. Within its group, $j=ha$ with $1\le a\le n/h$ and $\gcd (a,n/h)=1$, by the scaling identity for greatest common divisors. Dividing by $h$ gives a bijection with the integers counted by $\phi (n/h)$. The groups form a partition of a set, soFor $n=12$, the groups with greatest common divisor $h=1,2,3,4,6,12$ are respectively $\{1,5,7,11\},\{2,10\},\{3,9\},\{4,8\},\{6\},\{12\}$. Their sizes are $4,2,2,2,1,1$, totaling $12$. Möbius inversion reverses this divisor sum to recover Euler's totient function from $g(n)=n$.

For Dirichlet convolution, let $1(n)=1$ and $ϵ(n)=1_{n=1}$. Prime factorization gives $\mu \ast 1=ϵ$ and $\Lambda \ast 1=\log$. With subscripts denoting truncation at $U,V$,Rearranging proves the Vaughan identity.

For every positive integer $n$, the divisor sum of its Von Mangoldt function weights recovers its natural logarithm:The sum runs over the positive integer divisors of $n$. For $n=1$, its only term is $\Lambda (1)=0=\log 1$. Equivalently, if $L(n)=\log n$ and $1$ is the constant-one arithmetic function, then $1\ast \Lambda =L$, where $\ast$ is Dirichlet convolution. ^[ref]

For $12=2^2\cdot 3$, the positive integer divisors are $1,2,3,4,6,12$. Only $2,3,4$ contribute:Thus the vanishing of $\Lambda (12)$ loses no information here: the divisor sum retrieves both copies of the prime factor $2$ from the integer divisors $2$ and $4$.

Proof outline: by the fundamental theorem of arithmetic, write $n={\prod }_{i=1}^r{p}_i^{a_i}$ with distinct prime numbers $p_i$ and positive exponents $a_i$. Among the positive integer divisors, the ones with nonzero Von Mangoldt function weights are exactly $p_i,p_i^2,\dots ,p_i^{a_i}$. ConsequentlyThe natural logarithm turns the prime factorization product into a sum, and the prime powers greater than one supply exactly the required multiplicities.