Formelsammlung Mathematik: Unendliche Reihen: Dirichletreihen

Zurück zu Unendliche Reihen

1

$\sum _{n=1}^{\infty }{\frac {\varphi (n)}{n^{s}}}={\frac {\zeta (s-1)}{\zeta (s)}}\qquad {\text{Re}}(s)>2$

Beweis

Da $\varphi$ multiplikativ ist, ist $\sum _{n=1}^{\infty }{\frac {\varphi (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\varphi (p^{k})}{p^{ks}}}$

Wegen $\varphi (p^{k})=\left\{{\begin{matrix}1&k=0\\p^{k}-p^{k-1}&k>0\end{matrix}}\right.$     ist dies

$\prod _{p}\left(1+\sum _{k=1}^{\infty }{\frac {p^{k}-p^{k-1}}{p^{ks}}}\right)=\prod _{p}{\frac {1-p^{-s}}{1-p^{1-s}}}={\frac {\zeta (1-s)}{\zeta (s)}}$.

 

2

$\sum _{n=1}^{\infty }{\frac {\psi (n)}{n^{s}}}={\frac {\zeta (s)\,\zeta (s-1)}{\zeta (2s)}}\qquad {\text{Re}}(s)>2$

Beweis

Da $\psi \,$ multiplikativ ist, ist $\sum _{n=1}^{\infty }{\frac {\psi (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\psi (p^{k})}{p^{ks}}}$

Wegen $\psi (p^{k})=\left\{{\begin{matrix}1&k=0\\p^{k}+p^{k-1}&k>0\end{matrix}}\right.$     ist dies

$\prod _{p}\left(1+\sum _{k=1}^{\infty }{\frac {p^{k}+p^{k-1}}{p^{ks}}}\right)=\prod _{p}{\frac {1+p^{-s}}{1-p^{1-s}}}=\prod _{p}{\frac {1-p^{-2s}}{(1-p^{-s})(1-p^{1-s})}}={\frac {\zeta (s)\,\zeta (s-1)}{\zeta (2s)}}$.

 

3

$\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}={\frac {1}{\zeta (s)}}\qquad {\text{Re}}(s)>1$

Beweis

Da $\mu \,$ multiplikativ ist, ist $\sum _{n=1}^{\infty }{\frac {\mu (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\mu (p^{k})}{p^{ks}}}=\prod _{p}\left(1-{\frac {1}{p^{s}}}\right)={\frac {1}{\zeta (s)}}$.

 

4

$\sum _{n=1}^{\infty }{\frac {\Lambda (n)}{n^{s}}}=-{\frac {\zeta '(s)}{\zeta (s)}}\qquad {\text{Re}}(s)>1$

Beweis

$\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}\,\sum _{n=1}^{\infty }{\frac {\Lambda (n)}{n^{s}}}=\sum _{n=1}^{\infty }{\frac {(1\star \Lambda )(n)}{n^{s}}}$, wobei $(1\star \Lambda )(n)=\sum _{d|n}\Lambda (d)=\log n$ ist.

Also ist $\zeta (s)\sum _{n=1}^{\infty }{\frac {\Lambda (n)}{n^{s}}}=\sum _{n=1}^{\infty }{\frac {\log n}{n^{s}}}=-\zeta '(s)$.

 

5

$\sum _{n=1}^{\infty }{\frac {\tau ^{2}(n)}{n^{s}}}={\frac {\zeta ^{4}(s)}{\zeta (2s)}}\qquad {\text{Re}}(s)>1$

Beweis

Da $\tau \,$ multiplikativ ist, ist $\sum _{n=1}^{\infty }{\frac {\tau ^{2}(n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\tau ^{2}(p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {(k+1)^{2}}{p^{ks}}}=\prod _{p}{\frac {1-p^{-2s}}{\left(1-p^{-s}\right)^{4}}}={\frac {\zeta ^{4}(s)}{\zeta (2s)}}$.

 

6

$\sum _{n=1}^{\infty }{\frac {\tau (n^{2})}{n^{s}}}={\frac {\zeta ^{3}(s)}{\zeta (2s)}}\qquad {\text{Re}}(s)>1$

Beweis

Da $\tau \,$ multiplikativ ist, ist $\sum _{n=1}^{\infty }{\frac {\tau (n^{2})}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\tau (p^{2k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {2k+1}{p^{ks}}}=\prod _{p}{\frac {1-p^{-2s}}{\left(1-p^{-s}\right)^{3}}}={\frac {\zeta ^{3}(s)}{\zeta (2s)}}$.

 

7

$\sum _{n=1}^{\infty }{\frac {\mu ^{2}(n)}{n^{s}}}={\frac {\zeta (s)}{\zeta (2s)}}\qquad {\text{Re}}(s)>1$

Beweis

Da $\mu \,$ multiplikativ ist, ist $\sum _{n=1}^{\infty }{\frac {\mu ^{2}(n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\mu ^{2}(p^{k})}{p^{ks}}}=\prod _{p}\left(1+{\frac {1}{p^{s}}}\right)=\prod _{p}{\frac {1-p^{-2s}}{1-p^{-s}}}={\frac {\zeta (s)}{\zeta (2s)}}$.

 

8

$\sum _{n=1}^{\infty }{\frac {2^{\omega (n)}}{n^{s}}}={\frac {\zeta ^{2}(s)}{\zeta (2s)}}\qquad {\text{Re}}(s)>1$

Beweis

Da $2^{\omega }\,$ multiplikativ ist, ist $\sum _{n=1}^{\infty }{\frac {2^{\omega (n)}}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {2^{\omega (p^{k})}}{p^{ks}}}=\prod _{p}\left(1+\sum _{k=1}^{\infty }{\frac {2}{p^{ks}}}\right)$

$=\prod _{p}\left(1+{\frac {2\,p^{-s}}{1-p^{-s}}}\right)=\prod _{p}{\frac {1+p^{-s}}{1-p^{-s}}}=\prod _{p}{\frac {1-p^{-2s}}{(1-p^{-s})^{2}}}={\frac {\zeta ^{2}(s)}{\zeta (2s)}}$.

 

9

$\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}={\frac {\zeta (2s)}{\zeta (s)}}\qquad {\text{Re}}(s)>1$

Beweis

Da $\lambda \,$ multiplikativ ist, ist $\sum _{n=1}^{\infty }{\frac {\lambda (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\lambda (p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{p^{ks}}}=\prod _{p}{\frac {1}{1+p^{-s}}}=\prod _{p}{\frac {1-p^{-s}}{1-p^{-2s}}}={\frac {\zeta (2s)}{\zeta (s)}}$.

 

10

$\sum _{n=1}^{\infty }{\frac {\tau (n)}{n^{s}}}=\zeta ^{2}(s)\qquad {\text{Re}}(s)>1$

1. Beweis

$\sum _{n=1}^{\infty }{\frac {\tau (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\tau (p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {k+1}{p^{ks}}}=\prod _{p}{\frac {1}{(1-p^{-s})^{2}}}=\zeta ^{2}(s)$.

 

2. Beweis

$\sum _{n=1}^{\infty }{\frac {\tau (n)}{n^{s}}}=\sum _{n=1}^{\infty }\left(\sum _{i\cdot j=n}1\right){\frac {1}{n^{s}}}=\sum _{i,j=1}^{\infty }{\frac {1}{(ij)^{s}}}=\left(\sum _{i=1}^{\infty }{\frac {1}{i^{s}}}\right)\left(\sum _{j=1}^{\infty }{\frac {1}{j^{s}}}\right)=\zeta ^{2}(s)$

 

11

$\sum _{n=1}^{\infty }{\frac {\sigma (n)}{n^{s}}}=\zeta (s)\,\zeta (s-1)\qquad {\text{Re}}(s)>2$

1. Beweis

$\sum _{n=1}^{\infty }{\frac {\sigma (n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\sigma (p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\frac {1-p^{k+1}}{1-p}}{p^{ks}}}=\prod _{p}{\frac {1}{(1-p^{-s})(1-p^{1-s})}}=\zeta (s)\,\zeta (s-1)$.

 

2. Beweis

$\sum _{n=1}^{\infty }{\frac {\sigma (n)}{n^{s}}}=\sum _{n=1}^{\infty }\left(\sum _{i\cdot j=n}j\right){\frac {1}{n^{s}}}=\sum _{i,j=1}^{\infty }{\frac {j}{(ij)^{s}}}=\left(\sum _{i=1}^{\infty }{\frac {1}{i^{s}}}\right)\left(\sum _{j=1}^{\infty }{\frac {1}{j^{s-1}}}\right)=\zeta (s)\,\zeta (s-1)$

 

12

$\sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)}{n^{s}}}=\zeta (s)\,\zeta (s-a)\qquad {\text{Re}}(s)>a+1$

1. Beweis

$\sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\sigma _{a}(p^{k})}{p^{ks}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\frac {1-p^{a(k+1)}}{1-p^{a}}}{p^{ks}}}=\prod _{p}{\frac {1}{(1-p^{-s})(1-p^{a-s})}}=\zeta (s)\,\zeta (s-a)$.

 

2. Beweis

$\sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)}{n^{s}}}=\sum _{n=1}^{\infty }\left(\sum _{i\cdot j=n}j^{a}\right){\frac {1}{n^{s}}}=\sum _{i,j=1}^{\infty }{\frac {j^{a}}{(ij)^{s}}}=\left(\sum _{i=1}^{\infty }{\frac {1}{i^{s}}}\right)\left(\sum _{j=1}^{\infty }{\frac {1}{j^{s-a}}}\right)=\zeta (s)\,\zeta (s-a)$

 

13

$\sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)\,\sigma _{b}(n)}{n^{s}}}={\frac {\zeta (s)\,\zeta (s-a)\,\zeta (s-b)\,\zeta (s-a-b)}{\zeta (2s-a-b)}}$

Beweis

Da $\sigma _{a}\,\sigma _{b}$ multiplikativ ist, ist $\sum _{n=1}^{\infty }{\frac {\sigma _{a}(n)\,\sigma _{b}(n)}{n^{s}}}=\prod _{p}\sum _{k=0}^{\infty }{\frac {\sigma _{a}(p^{k})\,\sigma _{b}(p^{k})}{p^{ks}}}$

$=\prod _{p}\sum _{k=0}^{\infty }{\frac {{\frac {1-p^{a(k+1)}}{1-p^{a}}}\,{\frac {1-p^{b(k+1)}}{1-p^{b}}}}{p^{ks}}}=\prod _{p}\left({\frac {1}{1-p^{a}}}\,{\frac {1}{1-p^{b}}}\,\sum _{k=0}^{\infty }{\frac {1-p^{a(k+1)}-p^{b(k+1)}+p^{(a+b)(k+1)}}{p^{ks}}}\right)$

$=\prod _{p}\left({\frac {1}{1-p^{a}}}\,{\frac {1}{1-p^{b}}}\,\left({\frac {1}{1-p^{-s}}}-{\frac {p^{a}}{1-p^{-(s-a)}}}-{\frac {p^{b}}{1-p^{-(s-b)}}}+{\frac {p^{a+b}}{1-p^{-(s-a-b)}}}\right)\right)$

$=\prod _{p}{\frac {1-p^{-(2s-a-b)}}{(1-p^{-s})(1-p^{-(s-a)})(1-p^{-(s-b)})(1-p^{-(s-a-b)})}}={\frac {\zeta (s)\,\zeta (s-a)\,\zeta (s-b)\,\zeta (s-a-b)}{\zeta (2s-a-b)}}$.