Formelsammlung Mathematik: Unendliche Reihen: Reihen nach Ramanujan

Zurück zu Unendliche Reihen

1

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

1. Beweis

Es sei $f(z)={\frac {\cot z\,\coth z}{z^{4n-1}}}$ und $\gamma _{N}\,$ das Quadrat mit den Ecken $(\pm 1\pm i)\,\left(N+{\frac {1}{2}}\right)\,\pi$.

Wegen $\cot z=-{\frac {2}{z}}\,\sum _{k=0}^{\infty }{\frac {\zeta (2k)}{\pi ^{2k}}}\,z^{2k}$ und $\coth z=-{\frac {2}{z}}\,\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {\zeta (2k)}{\pi ^{2k}}}\,z^{2k}$

ist nach der Cauchy-Produktformel $f(z)={\frac {\cot z\,\coth z}{z^{4n-1}}}={\frac {4}{z^{4n+1}}}\,\sum _{m=0}^{\infty }\sum _{k=0}^{m}(-1)^{k}\,{\frac {\zeta (2k)}{\pi ^{2k}}}\,{\frac {\zeta (2m-2k)}{\pi ^{2m-2k}}}\,z^{2m}$.

Das Reihenglied der Form ${\text{Koeffizient}}\cdot {\frac {1}{z}}$ erhält man für $m=2n\,$.

Daher ist ${\text{res}}f{\big |}_{z=0}=4\,\sum _{k=0}^{2n}(-1)^{k}\,{\frac {\zeta (2k)}{\pi ^{2k}}}\,{\frac {\zeta (4n-2k)}{\pi ^{4n-2k}}}$.

Des Weiteren ist ${\text{res}}f{\big |}_{z=k\pi }=\lim _{z\to k\pi }{\frac {z-k\pi }{\sin z}}\,{\frac {\cos z\,\coth z}{z^{4n-1}}}={\frac {\coth k\pi }{(k\pi )^{4n-1}}}$

und ${\text{res}}f{\big |}_{z=ik\pi }=\lim _{z\to ik\pi }{\frac {z-ik\pi }{\sinh z}}\,{\frac {\cot z\,\cosh z}{z^{4n-1}}}={\frac {\coth k\pi }{(k\pi )^{4n-1}}}$.

Wegen $\sin z=\sin(x+iy)=\sin x\,\cosh y+i\cos x\,\sinh y$ ist

$|\sin z|^{2}=\sin ^{2}x\,\underbrace {\cosh ^{2}y} _{1+\sinh ^{2}y}+\cos ^{2}x\,\sinh ^{2}y=\sin ^{2}x+\sinh ^{2}y$.

Das ist mindestens $1\,$ wenn $x\,$ oder $y\,$ gleich $\pm \left(N+{\frac {1}{2}}\right)\,\pi$ ist.

Wegen $\cot ^{2}z={\frac {1}{\sin ^{2}z}}-1$ ist daher für alle $z\in \gamma _{N}\qquad |\cot z|^{2}\leq 1+1$.

Und da aus $z\in \gamma _{N}\qquad iz\in \gamma _{N}$ folgt, ist $|\coth z|^{2}=|\cot iz|^{2}\,$ ebenfalls $\leq 2$.

Somit ist $|f(z)|\leq {\frac {2}{|z|^{4n-1}}}$ für alle $z\in \gamma _{N}$.

Daraus folgt $\left|\oint _{\gamma _{N}}f\,dz\right|\leq 8\left(N+{\frac {1}{2}}\right)\,{\frac {2}{\left(N+{\frac {1}{2}}\right)^{4n-1}}}\to 0$ für $N\to \infty \,$.

Die Summe aller Residuen muss also $0\,$ ergeben:

$4\,\sum _{k=1}^{\infty }{\frac {\coth k\pi }{(k\pi )^{4n-1}}}+4\sum _{k=0}^{2n}(-1)^{k}\,{\frac {\zeta (2k)}{\pi ^{2k}}}\,{\frac {\zeta (4n-2k)}{\pi ^{4n-2k}}}=0$

 

2. Beweis

Es ist ${\frac {\pi }{k}}\,\coth(k\pi )=\sum _{\ell \in \mathbb {Z} }{\frac {1}{k^{2}+\ell ^{2}}}={\frac {1}{k^{2}}}+2\sum _{\ell =1}^{\infty }{\frac {1}{k^{2}+\ell ^{2}}}$

Teile durch $(k^{2})^{2n-1}\,:\quad {\frac {\pi }{k^{4n-1}}}\,\coth(k\pi )={\frac {1}{k^{4n}}}+2\sum _{\ell =1}^{\infty }{\frac {1}{(k^{2})^{2n-1}\,(k^{2}+\ell ^{2})}}$

und summiere nach $k\,$ von $1\,$ bis $\infty \,:\quad \pi \sum _{k=1}^{\infty }{\frac {\coth(k\pi )}{k^{4n-1}}}=\zeta (4n)+2\sum _{k,\ell =1}^{\infty }{\frac {1}{(k^{2})^{2n-1}\,(k^{2}+\ell ^{2})}}$

Aus Symmetriegründen kann $\sum _{k,\ell =1}^{\infty }{\frac {1}{(k^{2})^{2n-1}\,(k^{2}+\ell ^{2})}}$ auch als $\sum _{k,\ell =1}^{\infty }{\frac {1}{(\ell ^{2})^{2n-1}\,(k^{2}+\ell ^{2})}}$ geschrieben werden.

Daher ist $2\sum _{k,\ell =1}^{\infty }{\frac {1}{(k^{2})^{2n-1}\,(k^{2}+\ell ^{2})}}=\sum _{k,\ell =1}^{\infty }{\frac {1}{(\ell ^{2})^{2n-1}\,(k^{2}+\ell ^{2})}}+\sum _{k,\ell =1}^{\infty }{\frac {1}{(k^{2})^{2n-1}\,(k^{2}+\ell ^{2})}}$

$=\sum _{k,\ell =1}^{\infty }{\frac {1}{(k^{2})^{2n-1}\,(\ell ^{2})^{2n-1}}}\,{\frac {(k^{2})^{2n-1}+(\ell ^{2})^{2n-1}}{k^{2}+\ell ^{2}}}$,

wobei ${\frac {(k^{2})^{2n-1}+(\ell ^{2})^{2n-1}}{k^{2}+\ell ^{2}}}=\sum _{r=1}^{2n-1}(-1)^{r-1}\,(k^{2})^{2n-1-r}\,(\ell ^{2})^{r-1}$ ist.

$\pi \sum _{k=1}^{\infty }{\frac {\coth(k\pi )}{k^{4n-1}}}=\zeta (4n)+\sum _{k,\ell =1}^{\infty }\sum _{r=1}^{2n-1}(-1)^{r-1}{\frac {1}{(k^{2})^{r}\,(\ell ^{2})^{2n-r}}}$

$=\zeta (4n)+\sum _{r=1}^{2n-1}(-1)^{r-1}\,\zeta (2r)\,\zeta (4n-2r)=\sum _{r=0}^{2n}(-1)^{r-1}\,\zeta (2r)\,\zeta (4n-2r)$

 

2

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

ohne Beweis

 

3

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

ohne Beweis

 

4

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

ohne Beweis

 

5

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

ohne Beweis

 

6

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

ohne Beweis

 

7

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

ohne Beweis

 

8

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

ohne Beweis

 

9

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

ohne Beweis

 

10

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

ohne Beweis

 

11

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

ohne Beweis

 

12

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

ohne Beweis

 

13

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

Beweis

Verwende folgende Reihenidentität von Ramanujan:

Ist $f(z)=\sum _{n=1}^{\infty }{\frac {2\pi nz}{e^{2\pi nz}-1}}$, so gilt $f(z)+f\left({\frac {1}{z}}\right)={\frac {\pi }{12}}\left(z+{\frac {1}{z}}\right)-{\frac {1}{2}}$ für ${\text{Re}}(z)>0\,$.

Setzt man $z=e^{\frac {i\pi }{6}}={\frac {{\sqrt {3}}+i}{2}}$, so ist

$f(z)+f\left({\frac {1}{z}}\right)=\sum _{n=1}^{\infty }{\frac {n\pi ({\sqrt {3}}+i)}{(-1)^{n}\,e^{{\sqrt {3}}n\pi }-1}}+\sum _{n=1}^{\infty }{\frac {n\pi ({\sqrt {3}}-i)}{(-1)^{n}\,e^{{\sqrt {3}}n\pi }-1}}=2\pi {\sqrt {3}}\sum _{n=1}^{\infty }{\frac {n}{(-1)^{n}\,e^{{\sqrt {3}}n\pi }-1}}$

und ${\frac {\pi }{12}}\left(z+{\frac {1}{z}}\right)-{\frac {1}{2}}={\frac {\pi }{12}}{\sqrt {3}}-{\frac {1}{2}}$.

Daraus folgt dann $\sum _{k=1}^{\infty }{\frac {k}{(-1)^{k}\,e^{{\sqrt {3}}\pi k}-1}}={\frac {1}{24}}-{\frac {1}{4{\sqrt {3}}\pi }}$ .

 

14

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

ohne Beweis

 

15

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

ohne Beweis

 

16

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

ohne Beweis

 

17

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

ohne Beweis

 

18

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

Beweis

Verwende folgende Reihenidentität von Ramanujan:

Ist $f(z)=\sum _{n=1}^{\infty }{\frac {1}{n\,(e^{2\pi nz}-1)}}$, so gilt $f(z)-f\left({\frac {1}{z}}\right)={\frac {1}{2}}\log z-{\frac {\pi }{12}}\left(z-{\frac {1}{z}}\right)$ für ${\text{Re}}(z)>0\,$.

Differenziert man beide Seiten, $f'(z)+f'\left({\frac {1}{z}}\right)\,{\frac {1}{z^{2}}}={\frac {1}{2z}}-{\frac {\pi }{12}}\left(1+{\frac {1}{z^{2}}}\right)$, und setzt $z=1\,$, so ist $2\,f'(1)={\frac {1}{2}}-{\frac {\pi }{6}}$.

Auf der anderen Seite ist $f'(z)=-{\frac {\pi }{2}}\sum _{n=1}^{\infty }{\frac {1}{\sinh ^{2}n\pi z}}$, und somit $2f'(1)=-\pi \sum _{n=1}^{\infty }{\frac {1}{\sinh ^{2}n\pi }}$.

 

19

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

Beweis

Aus der Identität

$2\arctan {\frac {\tanh x}{\tan x}}=\arctan {\frac {\tanh 2x}{\tan 2x}}+\arctan {\frac {\sinh 2x}{\sin 2x}}$

folgt unmittelbar

${\frac {1}{2^{k}}}\arctan {\frac {\sinh 2^{k}x}{\sin 2^{k}x}}={\frac {1}{2^{k-1}}}\arctan {\frac {\tanh 2^{k-1}x}{\tan 2^{k-1}x}}-{\frac {1}{2^{k}}}\arctan {\frac {\tanh 2^{k}x}{\tan 2^{k}x}}$.

Daraus ergibt sich die Teleskopsumme

$\sum _{k=1}^{n}{\frac {1}{2^{k}}}\arctan {\frac {\sinh 2^{k}x}{\sin 2^{k}x}}=\arctan {\frac {\tanh x}{\tan x}}-{\frac {1}{2^{n}}}\arctan {\frac {\tanh 2^{n}x}{\tan 2^{n}x}}$.

Differenziert man, so ist

$\sum _{k=1}^{n}{\frac {\sin(2^{k}x)\,\cosh(2^{k}x)-\sinh(2^{k}x)\,\cos(2^{k}x)}{\sinh ^{2}(2^{k}x)+\sin ^{2}(2^{k}x)}}$

$={\frac {\sin(x)\,\cos(x)-\sinh(x)\,\cosh(x)}{\sinh ^{2}(x)+\sin ^{2}(x)}}-{\frac {\sin(2^{n}x)\,\cos(2^{n}x)-\sinh(2^{n}x)\,\cosh(2^{n}x)}{\sinh ^{2}(2^{n}x)+\sin ^{2}(2^{n}x)}}$.

Für $n\to \infty \,$ geht letzter Bruch gegen $-\operatorname {sgn}(x)\,$.

Setzt man $x=\pi \,$, so erhält man

$\sum _{k=1}^{\infty }{\frac {-\sinh(2^{k}\pi )}{\sinh ^{2}(2^{k}\pi )}}={\frac {-\sinh(\pi )\,\cosh(\pi )}{\sinh ^{2}(\pi )}}+1$, gleichbedeutend mit $\sum _{k=1}^{\infty }{\frac {1}{\sinh(2^{k}\pi )}}=\coth \pi -1$.

 

20

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

ohne Beweis

 

21

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

ohne Beweis

 

22

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

ohne Beweis

 

23

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

ohne Beweis