メビウスの反転公式
∑ 1 ≤ n ≤ x μ ( n ) g ( x n ) = ∑ 1 ≤ n ≤ x μ ( n ) ∑ 1 ≤ m ≤ x / n f ( x m n ) = ∑ 1 ≤ n ≤ x μ ( n ) ∑ 1 ≤ m ≤ x / n ∑ 1 ≤ r ≤ x [ r = m n ] f ( x r ) = ∑ 1 ≤ r ≤ x f ( x r ) ∑ 1 ≤ n ≤ x μ ( n ) ∑ 1 ≤ m ≤ x / n [ m = r / n ] rearranging the summation order = ∑ 1 ≤ r ≤ x f ( x r ) ∑ n | r μ ( n ) = ∑ 1 ≤ r ≤ x f ( x r ) ε ( r ) = f ( x ) since ε ( r ) = 0 except when r = 1 {\displaystyle {\begin{aligned}\sum _{1\leq n\leq x}\mu (n)g\left({\frac {x}{n}}\right)&=\sum _{1\leq n\leq x}\mu (n)\sum _{1\leq m\leq x/n}f\left({\frac {x}{mn}}\right)\\&=\sum _{1\leq n\leq x}\mu (n)\sum _{1\leq m\leq x/n}\sum _{1\leq r\leq x}[r=mn]f\left({\frac {x}{r}}\right)\\&=\sum _{1\leq r\leq x}f\left({\frac {x}{r}}\right)\sum _{1\leq n\leq x}\mu (n)\sum _{1\leq m\leq x/n}[m=r/n]\qquad {\text{rearranging the summation order}}\\&=\sum _{1\leq r\leq x}f\left({\frac {x}{r}}\right)\sum _{n|r}\mu (n)\\&=\sum _{1\leq r\leq x}f\left({\frac {x}{r}}\right)\varepsilon (r)\\&=f(x)\qquad {\text{since }}\varepsilon (r)=0{\text{ except when }}r=1\end{aligned}}}
二つ目の一般化では α(n) が 1 に取って代わるが、証明は本質的に同一である。