Formelsammlung Mathematik: Unendliche Reihen: Reihen gewisser Produkte

Zurück zu Unendliche Reihen

$\sum _{k=n}^{\infty }{\frac {1}{k\,(k+1)}}={\frac {1}{n}}$

$\sum _{k=1}^{\infty }{\frac {(1-z)^{k+n}\,(-1)^{n}}{k\,(k+1)\cdots (k+n)}}={\frac {z^{n}}{n!}}\,(H_{n}-\log z)+\sum _{k=1}^{n}{\frac {(-1)^{k}}{k!\,k}}\,{\frac {z^{n-k}}{(n-k)!}}$

$\sum _{k=1}^{\infty }{\frac {1}{k\,(k+1)\cdots (k+n)}}={\frac {1}{n\cdot n!}}$

1. Beweis

$\sum _{k=1}^{\infty }{\frac {n!}{k\,(k+1)\cdots (k+n)}}=\sum _{k=1}^{\infty }{\frac {(k-1)!\,n!}{(k+n)!}}=\sum _{k=1}^{\infty }B(k,n+1)=B(1,n)={\frac {1}{n}}$

 

2. Beweis

Definiert man $S_{n}=\sum _{k=1}^{n}{\frac {1}{k\,(k+1)\cdots (k+n)}}$, so ist

$S_{n-1}=\sum _{k=1}^{n}{\frac {k+n}{k\,(k+1)\cdots (k+n)}}=\sum _{k=1}^{n}{\frac {1}{(k+1)\cdots (k+n)}}+n\,S_{n}$.

Also ist $S_{n-1}-n\,S_{n}=\sum _{k=0}^{n}{\frac {1}{(k+1)\cdots (k+n)}}-{\frac {1}{n!}}=S_{n-1}-{\frac {1}{n!}}$. Daraus folgt $S_{n}={\frac {1}{n\cdot n!}}$.

 

$\sum _{k=1}^{\infty }{\frac {1}{k\,(k+2)\,(k+4)\cdots (k+2n)}}={\frac {1}{2n}}\left({\frac {2^{n}\,n!}{(2n)!}}+{\frac {1}{2^{n}\,n!}}\right)$

$\sum _{k=1}^{\infty }{\frac {1}{k^{2}\,(k+1)^{2}\cdots (k+n)^{2}}}={\frac {1}{n!^{2}}}{2n \choose n}\left({\frac {\pi ^{2}}{6}}-3\sum _{k=1}^{n}{\frac {1}{k^{2}{2k \choose k}}}\right)$

$\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}\,(2k+3)^{2}\cdots (2k+2n+1)^{2}}}={\frac {1}{n!^{2}}}{\frac {2n \choose n}{2^{2n}}}\left({\frac {\pi ^{2}}{8}}-{\frac {1}{4}}\sum _{k=1}^{n}{\frac {(3k-1)\,2^{4k}}{k^{3}{2k \choose k}^{3}}}\right)$

$\sum _{k=1}^{\infty }{\frac {1}{k^{3}\,(k+1)^{3}\cdots (k+2n)^{3}}}={\frac {(-1)^{n}\,(3n)!}{(2n)!^{3}\,n!^{3}}}\left(\zeta (3)+{\frac {1}{4}}\sum _{k=1}^{n}(-1)^{k}\,{\frac {(56k^{2}-32k+5)\,(k-1)!^{3}}{(3k)!\,(2k-1)^{2}}}\right)$