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# Formelsammlung Mathematik: Unendliche Reihen: Reihen mit harmonischen Zahlen

Zurück zu Unendliche Reihen

${\displaystyle \sum _{n=1}^{\infty }H_{n}\,x^{n}=-{\frac {\log(1-x)}{1-x}}\qquad |x|<1}$

${\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}}{k^{2}}}=2\,\zeta (3)}$

${\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}}{k^{3}}}={\frac {\pi ^{4}}{72}}}$

${\displaystyle 2\,\sum _{k=1}^{\infty }{\frac {H_{k}}{k^{n}}}=(n+2)\,\zeta (n+1)-\sum _{k=1}^{n-2}\zeta (n-k)\,\zeta (k+1)\qquad n\in \mathbb {Z} ^{\geq 2}}$

${\displaystyle 2\sum _{k=1}^{\infty }(-1)^{k-1}\,{\frac {H_{k}}{k}}={\frac {\pi ^{2}}{6}}-(\log 2)^{2}}$

${\displaystyle 2\sum _{k=1}^{\infty }(-1)^{k-1}\,{\frac {H_{k}}{k^{2n}}}=(2n+1)\eta (2n+1)-2\sum _{k=0}^{n-1}\eta (2k)\,\zeta (2n+1-2k)\qquad n\in \mathbb {Z} ^{\geq 1}}$

${\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}}{k\,2^{k}}}={\frac {\pi ^{2}}{12}}}$

${\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}}{k^{2}\,2^{k}}}=\zeta (3)-{\frac {\pi ^{2}}{12}}\log 2}$

${\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}^{2}}{k^{2}}}={\frac {17\pi ^{4}}{360}}}$

${\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}^{2}}{(k+1)^{2}}}={\frac {11\pi ^{4}}{360}}}$

${\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k,r}}{k^{n}}}={\begin{Bmatrix}\zeta (m)\left({\frac {1}{2}}-{\frac {(-1)^{r}}{2}}{m-1 \choose r}-{\frac {(-1)^{r}}{2}}{m-1 \choose n}\right)\\\\+{\frac {1-(-1)^{r}}{2}}\,\zeta (n)\,\zeta (r)\\\\+(-1)^{r}\sum \limits _{k=1}^{\lfloor r/2\rfloor }{m-2k-1 \choose n-1}\,\zeta (2k)\,\zeta (m-2k)\\\\+(-1)^{r}\sum \limits _{k=1}^{\lfloor n/2\rfloor }{m-2k-1 \choose r-1}\,\zeta (2k)\,\zeta (m-2k)\end{Bmatrix}}\qquad n,r\in \mathbb {Z} ^{\geq 2}}$ und ${\displaystyle m=n+r\,}$ ungerade