= Möbius divisor-sum identity
{c}
{title2=$\sum_{d\mid n}\mu(d)=1_{n=1}$}
For every <positive integer> $n$, the <divisor sum> of the <Möbius function> is
$$
\sum_{d\mid n}\mu(d)=\begin{cases}1&n=1,\\0&n>1.\end{cases}
$$
The sum includes all positive <integer divisors>, but only <squarefree divisors> contribute. Equivalently, $\mu*\mathbf1=\varepsilon$ under <Dirichlet convolution>, where $\mathbf1$ is the <constant-one arithmetic function> and $\varepsilon$ is the <Identity for Dirichlet convolution>. https://darijgrinberg.gitlab.io/floor.pdf[Darij Grinberg, MIT 18.781 lecture notes, Propositions 2.4.6 and 2.4.7]{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 gives
$$
\sum_{d\mid n}\mu(d)=\prod_{p\mid n}(1-1).
$$
For $n>1$ 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 $30=2\cdot3\cdot5$, all eight positive <integer divisors> are <squarefree integers>, giving
$$
\mu(1)+\mu(2)+\mu(3)+\mu(5)+\mu(6)+\mu(10)+\mu(15)+\mu(30)
=1-1-1-1+1+1+1-1=0.
$$
Pair 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.
\Image[/mobius-divisor-cancellation.png]
{title=The eight <integer divisors> of 30 form a <cube> of <subsets> of its <prime factors>; adding one <prime factor> reverses the <Möbius function> sign, and all contributions cancel}
The <Möbius divisor-sum identity> is the cancellation behind <Möbius inversion> and reciprocal <Dirichlet series>: aggregation over <integer divisors> has an exact inverse.
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