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Da φ {\displaystyle \varphi } multiplikativ ist, ist ∑ n = 1 ∞ φ ( n ) n s = ∏ p ∑ k = 0 ∞ φ ( p k ) p k s {\displaystyle \sum _{n=1}^{\infty }{\frac {\varphi (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\varphi (p^{k})}{p^{ks}}}} Wegen φ ( p k ) = { 1 k = 0 p k − p k − 1 k > 0 {\displaystyle \varphi (p^{k})=\left\{{\begin{matrix}1&k=0\\p^{k}-p^{k-1}&k>0\end{matrix}}\right.} ist dies ∏ p ( 1 + ∑ k = 1 ∞ p k − p k − 1 p k s ) = ∏ p 1 − p − s 1 − p 1 − s = ζ ( 1 − s ) ζ ( s ) {\displaystyle \prod _{p}\left(1+\sum _{k=1}^{\infty }{\frac {p^{k}-p^{k-1}}{p^{ks}}}\right)=\prod _{p}{\frac {1-p^{-s}}{1-p^{1-s}}}={\frac {\zeta (1-s)}{\zeta (s)}}} .
Da ψ {\displaystyle \psi \,} multiplikativ ist, ist ∑ n = 1 ∞ ψ ( n ) n s = ∏ p ∑ k = 0 ∞ ψ ( p k ) p k s {\displaystyle \sum _{n=1}^{\infty }{\frac {\psi (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\psi (p^{k})}{p^{ks}}}} Wegen ψ ( p k ) = { 1 k = 0 p k + p k − 1 k > 0 {\displaystyle \psi (p^{k})=\left\{{\begin{matrix}1&k=0\\p^{k}+p^{k-1}&k>0\end{matrix}}\right.} ist dies ∏ p ( 1 + ∑ k = 1 ∞ p k + p k − 1 p k s ) = ∏ p 1 + p − s 1 − p 1 − s = ∏ p 1 − p − 2 s ( 1 − p − s ) ( 1 − p 1 − s ) = ζ ( s ) ζ ( s − 1 ) ζ ( 2 s ) {\displaystyle \prod _{p}\left(1+\sum _{k=1}^{\infty }{\frac {p^{k}+p^{k-1}}{p^{ks}}}\right)=\prod _{p}{\frac {1+p^{-s}}{1-p^{1-s}}}=\prod _{p}{\frac {1-p^{-2s}}{(1-p^{-s})(1-p^{1-s})}}={\frac {\zeta (s)\,\zeta (s-1)}{\zeta (2s)}}} .
Da μ {\displaystyle \mu \,} multiplikativ ist, ist ∑ n = 1 ∞ μ ( n ) n s = ∏ p ∑ k = 0 ∞ μ ( p k ) p k s = ∏ p ( 1 − 1 p s ) = 1 ζ ( s ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\mu (p^{k})}{p^{ks}}}=\prod _{p}\left(1-{\frac {1}{p^{s}}}\right)={\frac {1}{\zeta (s)}}} .
∑ n = 1 ∞ 1 n s ∑ n = 1 ∞ Λ ( n ) n s = ∑ n = 1 ∞ ( 1 ⋆ Λ ) ( n ) n s {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{s}}}\,\sum _{n=1}^{\infty }{\frac {\Lambda (n)}{n^{s}}}=\sum _{n=1}^{\infty }{\frac {(1\star \Lambda )(n)}{n^{s}}}} , wobei ( 1 ⋆ Λ ) ( n ) = ∑ d | n Λ ( d ) = log n {\displaystyle (1\star \Lambda )(n)=\sum _{d|n}\Lambda (d)=\log n} ist. Also ist ζ ( s ) ∑ n = 1 ∞ Λ ( n ) n s = ∑ n = 1 ∞ log n n s = − ζ ′ ( s ) {\displaystyle \zeta (s)\sum _{n=1}^{\infty }{\frac {\Lambda (n)}{n^{s}}}=\sum _{n=1}^{\infty }{\frac {\log n}{n^{s}}}=-\zeta '(s)} .
Da τ {\displaystyle \tau \,} multiplikativ ist, ist ∑ n = 1 ∞ τ 2 ( n ) n s = ∏ p ∑ k = 0 ∞ τ 2 ( p k ) p k s = ∏ p ∑ k = 0 ∞ ( k + 1 ) 2 p k s = ∏ p 1 − p − 2 s ( 1 − p − s ) 4 = ζ 4 ( s ) ζ ( 2 s ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\tau ^{2}(n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\tau ^{2}(p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {(k+1)^{2}}{p^{ks}}}=\prod _{p}{\frac {1-p^{-2s}}{\left(1-p^{-s}\right)^{4}}}={\frac {\zeta ^{4}(s)}{\zeta (2s)}}} .
Da τ {\displaystyle \tau \,} multiplikativ ist, ist ∑ n = 1 ∞ τ ( n 2 ) n s = ∏ p ∑ k = 0 ∞ τ ( p 2 k ) p k s = ∏ p ∑ k = 0 ∞ 2 k + 1 p k s = ∏ p 1 − p − 2 s ( 1 − p − s ) 3 = ζ 3 ( s ) ζ ( 2 s ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\tau (n^{2})}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\tau (p^{2k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {2k+1}{p^{ks}}}=\prod _{p}{\frac {1-p^{-2s}}{\left(1-p^{-s}\right)^{3}}}={\frac {\zeta ^{3}(s)}{\zeta (2s)}}} .
Da μ {\displaystyle \mu \,} multiplikativ ist, ist ∑ n = 1 ∞ μ 2 ( n ) n s = ∏ p ∑ k = 0 ∞ μ 2 ( p k ) p k s = ∏ p ( 1 + 1 p s ) = ∏ p 1 − p − 2 s 1 − p − s = ζ ( s ) ζ ( 2 s ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\mu ^{2}(n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\mu ^{2}(p^{k})}{p^{ks}}}=\prod _{p}\left(1+{\frac {1}{p^{s}}}\right)=\prod _{p}{\frac {1-p^{-2s}}{1-p^{-s}}}={\frac {\zeta (s)}{\zeta (2s)}}} .
Da 2 ω {\displaystyle 2^{\omega }\,} multiplikativ ist, ist ∑ n = 1 ∞ 2 ω ( n ) n s = ∏ p ∑ k = 0 ∞ 2 ω ( p k ) p k s = ∏ p ( 1 + ∑ k = 1 ∞ 2 p k s ) {\displaystyle \sum _{n=1}^{\infty }{\frac {2^{\omega (n)}}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {2^{\omega (p^{k})}}{p^{ks}}}=\prod _{p}\left(1+\sum _{k=1}^{\infty }{\frac {2}{p^{ks}}}\right)} = ∏ p ( 1 + 2 p − s 1 − p − s ) = ∏ p 1 + p − s 1 − p − s = ∏ p 1 − p − 2 s ( 1 − p − s ) 2 = ζ 2 ( s ) ζ ( 2 s ) {\displaystyle =\prod _{p}\left(1+{\frac {2\,p^{-s}}{1-p^{-s}}}\right)=\prod _{p}{\frac {1+p^{-s}}{1-p^{-s}}}=\prod _{p}{\frac {1-p^{-2s}}{(1-p^{-s})^{2}}}={\frac {\zeta ^{2}(s)}{\zeta (2s)}}} .
Da λ {\displaystyle \lambda \,} multiplikativ ist, ist ∑ n = 1 ∞ λ ( n ) n s = ∏ p ∑ k = 0 ∞ λ ( p k ) p k s = ∏ p ∑ k = 0 ∞ ( − 1 ) k p k s = ∏ p 1 1 + p − s = ∏ p 1 − p − s 1 − p − 2 s = ζ ( 2 s ) ζ ( s ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\lambda (p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{p^{ks}}}=\prod _{p}{\frac {1}{1+p^{-s}}}=\prod _{p}{\frac {1-p^{-s}}{1-p^{-2s}}}={\frac {\zeta (2s)}{\zeta (s)}}} .
∑ n = 1 ∞ τ ( n ) n s = ∏ p ∑ k = 0 ∞ τ ( p k ) p k s = ∏ p ∑ k = 0 ∞ k + 1 p k s = ∏ p 1 ( 1 − p − s ) 2 = ζ 2 ( s ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\tau (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\tau (p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {k+1}{p^{ks}}}=\prod _{p}{\frac {1}{(1-p^{-s})^{2}}}=\zeta ^{2}(s)} .
∑ n = 1 ∞ τ ( n ) n s = ∑ n = 1 ∞ ( ∑ i ⋅ j = n 1 ) 1 n s = ∑ i , j = 1 ∞ 1 ( i j ) s = ( ∑ i = 1 ∞ 1 i s ) ( ∑ j = 1 ∞ 1 j s ) = ζ 2 ( s ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\tau (n)}{n^{s}}}=\sum _{n=1}^{\infty }\left(\sum _{i\cdot j=n}1\right){\frac {1}{n^{s}}}=\sum _{i,j=1}^{\infty }{\frac {1}{(ij)^{s}}}=\left(\sum _{i=1}^{\infty }{\frac {1}{i^{s}}}\right)\left(\sum _{j=1}^{\infty }{\frac {1}{j^{s}}}\right)=\zeta ^{2}(s)}
∑ n = 1 ∞ σ ( n ) n s = ∏ p ∑ k = 0 ∞ σ ( p k ) p k s = ∏ p ∑ k = 0 ∞ 1 − p k + 1 1 − p p k s = ∏ p 1 ( 1 − p − s ) ( 1 − p 1 − s ) = ζ ( s ) ζ ( s − 1 ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\sigma (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\sigma (p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\frac {1-p^{k+1}}{1-p}}{p^{ks}}}=\prod _{p}{\frac {1}{(1-p^{-s})(1-p^{1-s})}}=\zeta (s)\,\zeta (s-1)} .
∑ n = 1 ∞ σ ( n ) n s = ∑ n = 1 ∞ ( ∑ i ⋅ j = n j ) 1 n s = ∑ i , j = 1 ∞ j ( i j ) s = ( ∑ i = 1 ∞ 1 i s ) ( ∑ j = 1 ∞ 1 j s − 1 ) = ζ ( s ) ζ ( s − 1 ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\sigma (n)}{n^{s}}}=\sum _{n=1}^{\infty }\left(\sum _{i\cdot j=n}j\right){\frac {1}{n^{s}}}=\sum _{i,j=1}^{\infty }{\frac {j}{(ij)^{s}}}=\left(\sum _{i=1}^{\infty }{\frac {1}{i^{s}}}\right)\left(\sum _{j=1}^{\infty }{\frac {1}{j^{s-1}}}\right)=\zeta (s)\,\zeta (s-1)}
∑ n = 1 ∞ σ a ( n ) n s = ∏ p ∑ k = 0 ∞ σ a ( p k ) p k s = ∏ p ∑ k = 0 ∞ 1 − p a ( k + 1 ) 1 − p a p k s = ∏ p 1 ( 1 − p − s ) ( 1 − p a − s ) = ζ ( s ) ζ ( s − a ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\sigma _{a}(p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\frac {1-p^{a(k+1)}}{1-p^{a}}}{p^{ks}}}=\prod _{p}{\frac {1}{(1-p^{-s})(1-p^{a-s})}}=\zeta (s)\,\zeta (s-a)} .
∑ n = 1 ∞ σ a ( n ) n s = ∑ n = 1 ∞ ( ∑ i ⋅ j = n j a ) 1 n s = ∑ i , j = 1 ∞ j a ( i j ) s = ( ∑ i = 1 ∞ 1 i s ) ( ∑ j = 1 ∞ 1 j s − a ) = ζ ( s ) ζ ( s − a ) {\displaystyle \sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)}{n^{s}}}=\sum _{n=1}^{\infty }\left(\sum _{i\cdot j=n}j^{a}\right){\frac {1}{n^{s}}}=\sum _{i,j=1}^{\infty }{\frac {j^{a}}{(ij)^{s}}}=\left(\sum _{i=1}^{\infty }{\frac {1}{i^{s}}}\right)\left(\sum _{j=1}^{\infty }{\frac {1}{j^{s-a}}}\right)=\zeta (s)\,\zeta (s-a)}
Da σ a σ b {\displaystyle \sigma _{a}\,\sigma _{b}} multiplikativ ist, ist ∑ n = 1 ∞ σ a ( n ) σ b ( n ) n s = ∏ p ∑ k = 0 ∞ σ a ( p k ) σ b ( p k ) p k s {\displaystyle \sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)\,\sigma _{b}(n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\sigma _{a}(p^{k})\,\sigma _{b}(p^{k})}{p^{ks}}}} = ∏ p ∑ k = 0 ∞ 1 − p a ( k + 1 ) 1 − p a 1 − p b ( k + 1 ) 1 − p b p k s = ∏ p ( 1 1 − p a 1 1 − p b ∑ k = 0 ∞ 1 − p a ( k + 1 ) − p b ( k + 1 ) + p ( a + b ) ( k + 1 ) p k s ) {\displaystyle =\prod _{p}\sum _{k=0}^{\infty }{\frac {{\frac {1-p^{a(k+1)}}{1-p^{a}}}\,{\frac {1-p^{b(k+1)}}{1-p^{b}}}}{p^{ks}}}=\prod _{p}\left({\frac {1}{1-p^{a}}}\,{\frac {1}{1-p^{b}}}\,\sum _{k=0}^{\infty }{\frac {1-p^{a(k+1)}-p^{b(k+1)}+p^{(a+b)(k+1)}}{p^{ks}}}\right)} = ∏ p ( 1 1 − p a 1 1 − p b ( 1 1 − p − s − p a 1 − p − ( s − a ) − p b 1 − p − ( s − b ) + p a + b 1 − p − ( s − a − b ) ) ) {\displaystyle =\prod _{p}\left({\frac {1}{1-p^{a}}}\,{\frac {1}{1-p^{b}}}\,\left({\frac {1}{1-p^{-s}}}-{\frac {p^{a}}{1-p^{-(s-a)}}}-{\frac {p^{b}}{1-p^{-(s-b)}}}+{\frac {p^{a+b}}{1-p^{-(s-a-b)}}}\right)\right)} = ∏ p 1 − p − ( 2 s − a − b ) ( 1 − p − s ) ( 1 − p − ( s − a ) ) ( 1 − p − ( s − b ) ) ( 1 − p − ( s − a − b ) ) = ζ ( s ) ζ ( s − a ) ζ ( s − b ) ζ ( s − a − b ) ζ ( 2 s − a − b ) {\displaystyle =\prod _{p}{\frac {1-p^{-(2s-a-b)}}{(1-p^{-s})(1-p^{-(s-a)})(1-p^{-(s-b)})(1-p^{-(s-a-b)})}}={\frac {\zeta (s)\,\zeta (s-a)\,\zeta (s-b)\,\zeta (s-a-b)}{\zeta (2s-a-b)}}} .
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