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# Formelsammlung Mathematik: Unendliche Reihen: Reihen nach Ramanujan

Zurück zu Unendliche Reihen

##### 1
${\displaystyle \sum _{k=1}^{\infty }{\frac {\coth(k\pi )}{(k\pi )^{4n-1}}}=\sum _{k=0}^{2n}(-1)^{k-1}\,{\frac {\zeta (2k)}{\pi ^{2k}}}\,{\frac {\zeta (4n-2k)}{\pi ^{4n-2k}}}\qquad n\in \mathbb {Z} ^{>0}}$

##### 2
${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k}\,{\text{csch}}(k\pi )}{(k\pi )^{4n-1}}}=\sum _{k=0}^{2n}(-1)^{k-1}\,{\frac {\eta (2k)}{\pi ^{2k}}}\,{\frac {\eta (4n-2k)}{\pi ^{4n-2k}}}\qquad n\in \mathbb {Z} }$

##### 3
${\displaystyle \sum _{k=0}^{\infty }{\frac {\tanh \left((2k+1){\frac {\pi }{2}}\right)}{\left((2k+1){\frac {\pi }{2}}\right)^{4n-1}}}=\sum _{k=0}^{2n}(-1)^{k-1}\,{\frac {\lambda (2k)}{\left({\frac {\pi }{2}}\right)^{2k}}}\,{\frac {\lambda (4n-2k)}{\left({\frac {\pi }{2}}\right)^{4n-2k}}}\qquad n\in \mathbb {Z} ^{>0}}$

##### 4
${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}\,{\text{sech}}\left((2k+1){\frac {\pi }{2}}\right)}{\left((2k+1){\frac {\pi }{2}}\right)^{4n+1}}}=\sum _{k=0}^{2n}(-1)^{k}\,{\frac {\beta (2k+1)}{\left({\frac {\pi }{2}}\right)^{2k+1}}}\,{\frac {\beta (4n-2k+1)}{\left({\frac {\pi }{2}}\right)^{4n-2k+1}}}\qquad n\in \mathbb {Z} }$

##### 5
${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}\,{\text{sech}}\left({\sqrt {3}}\,(2k+1){\frac {\pi }{2}}\right)}{(2k+1){\frac {\pi }{2}}}}={\frac {1}{12}}}$

##### 6
${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}\,{\text{sech}}\left({\frac {1}{\sqrt {3}}}\,(2k+1){\frac {\pi }{2}}\right)}{(2k+1){\frac {\pi }{2}}}}={\frac {5}{12}}}$

##### 7
${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}\,{\text{sech}}\left({\sqrt {3}}\,(2k+1){\frac {\pi }{2}}\right)}{\left((2k+1){\frac {\pi }{2}}\right)^{7}}}={\frac {1}{180}}}$

##### 8
${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}\,{\text{sech}}\left({\frac {1}{\sqrt {3}}}\,(2k+1){\frac {\pi }{2}}\right)}{\left((2k+1){\frac {\pi }{2}}\right)^{7}}}={\frac {143}{27\cdot 180}}}$

##### 9
${\displaystyle \sum _{k=0}^{\infty }{\frac {(2k+1)^{4n+1}}{e^{(2k+1)\pi }+1}}=\left(2^{4n+1}-1\right)\,{\frac {B_{4n+2}}{4\,(2n+1)}}\qquad n\in \mathbb {Z} ^{\geq 0}}$

##### 10
${\displaystyle \sum _{k=1}^{\infty }{\frac {k^{-4n+1}}{e^{2k\pi }-1}}=-{\frac {(2\pi )^{4n-1}}{4}}\,\sum _{k=0}^{2n}(-1)^{k}\,{\frac {B_{2k}}{(2k)!}}\,{\frac {B_{4n-2k}}{(4n-2k)!}}-{\frac {1}{2}}\,\zeta (4n-1)\qquad n\in \mathbb {Z} ^{\geq 0}}$

##### 11
${\displaystyle \sum _{k=1}^{\infty }{\frac {k^{4n+1}}{e^{2k\pi }-1}}={\frac {B_{4n+2}}{4\,(2n+1)}}\qquad n\in \mathbb {Z} ^{>0}}$

##### 12
${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k\,(e^{2\pi k}-1)}}={\frac {1}{4}}\log \left({\frac {4}{\pi }}\right)-{\frac {\pi }{12}}+\log \Gamma \left({\frac {3}{4}}\right)}$

##### 13
${\displaystyle \sum _{k=1}^{\infty }{\frac {k}{(-1)^{k}\,e^{{\sqrt {3}}\pi k}-1}}={\frac {1}{24}}-{\frac {1}{4{\sqrt {3}}\pi }}}$

##### 14
${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {1}{\cosh(k\pi )}}={\frac {1}{\sqrt {2\pi }}}\,{\frac {\Gamma \left({\frac {1}{4}}\right)}{\Gamma \left({\frac {3}{4}}\right)}}}$

##### 15
${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {\pi }{\cosh ^{2}\left((2k+1){\frac {\pi }{2}}\right)}}=1}$

##### 16
${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k\,\sinh(k\pi )}}={\frac {\pi }{12}}-{\frac {1}{2}}\log 2}$

##### 17
${\displaystyle \sum _{k\in \mathbb {Z} }{\frac {k\pi }{\sinh k\pi }}={\frac {1}{2}}+{\frac {1}{8}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{\Gamma ^{2}\left({\frac {3}{4}}\right)}}}$

##### 18
${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{\sinh ^{2}(k\pi )}}={\frac {1}{6}}-{\frac {1}{2\pi }}}$

##### 19
${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{\sinh(2^{k}\pi )}}=\coth \pi -1}$

##### 20
${\displaystyle \sum _{k=0}^{\infty }{-{\frac {1}{2}} \choose k}^{3}=\left({\frac {\Gamma \left({\frac {9}{8}}\right)}{\Gamma \left({\frac {5}{4}}\right)\,\Gamma \left({\frac {7}{8}}\right)}}\right)^{2}}$

##### 21
${\displaystyle \sum _{k=0}^{\infty }(4k+1){-{\frac {1}{2}} \choose k}^{3}={\frac {2}{\pi }}}$

##### 22
${\displaystyle \sum _{k=0}^{\infty }(8k+1){-{\frac {1}{4}} \choose k}^{4}={\frac {2{\sqrt {2}}}{\Gamma \left({\frac {1}{2}}\right)\,\Gamma ^{2}\left({\frac {3}{4}}\right)}}}$

##### 23
${\displaystyle \sum _{k=0}^{\infty }(4k+1){-{\frac {1}{2}} \choose k}^{5}={\frac {2}{\Gamma ^{4}\left({\frac {3}{4}}\right)}}}$