User:Van bui 314

Inspired by your website

These are the formulas that I derived by visiting your website for the last fews months.

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Definition of:

$\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=1+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+\cdots$

$\eta (s)=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}}=1-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+\cdots$

$\lambda (s)=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{s}}}=1+{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}+{\frac {1}{7^{s}}}+\cdots$

$\beta (s)=\sum _{n=0}^{\infty }{\frac {(-1)^{k}}{(2n+1)^{s}}}=1-{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}-{\frac {1}{7^{s}}}+\cdots$

Where p is prime numbers

$\phi ={\frac {{\sqrt {5}}+1}{2}}$

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Jacobi theta function

$J_{2}(0,e^{-\pi })=J_{4}(0,e^{-\pi })=\left({\frac {\pi }{2}}\right)^{\frac {1}{4}}\cdot {\frac {1}{\Gamma \left({\frac {3}{4}}\right)}}\cdots (1)$

${\frac {J_{1}^{'}(0,e^{-\pi })}{J_{2}(0,e^{-\pi })}}=\left({\frac {\pi ^{2}}{2}}\right)^{\frac {1}{4}}\cdot {\frac {1}{\Gamma ^{2}\left({\frac {3}{4}}\right)}}\cdots (2)$

$J_{1}^{'}(0,e^{-\pi })={\frac {(\pi )^{\frac {3}{4}}}{2^{\frac {1}{2}}}}\cdot {\frac {1}{\Gamma ^{3}\left({\frac {3}{4}}\right)}}$

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Infinite product

$\pi ={\frac {\prod _{n=1}^{\infty }\left[1+{\frac {1}{4(3n)^{2}-1}}\right]}{\sum _{n=1}^{\infty }{\frac {1}{4^{n}}}}}\cdots (1)$

${\frac {\Gamma ^{2}({\frac {1}{4}})}{\pi ^{\frac {3}{2}}}}={\frac {\prod _{n=1}^{\infty }\left[1+{\frac {1}{(4n-1)^{2}-1}}\right]}{\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}}}\cdots (2)$

${\frac {\Gamma ^{2}({\frac {1}{4}})}{\pi ^{\frac {5}{2}}}}={\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}}\cdot {{\prod _{n=1}^{\infty }\left[1+{\frac {1}{(4n-1)^{2}-1}}\right]^{2}}{\left[1+{\frac {1}{(4n+1)^{2}-1}}\right]}}\cdots (3)$

$\left[{\frac {\Gamma ^{(}{\frac {1}{4}})}{2}}\right]^{2}\cdot {\frac {1}{\sqrt {2\pi }}}={\frac {3}{2}}\cdot {\frac {4}{5}}\cdot {\frac {7}{6}}\cdot {\frac {8}{9}}\cdot {\frac {11}{10}}\cdot {\frac {12}{13}}=\prod _{k=2,4,6,...}^{\infty }\left({\frac {k+1}{k}}\right)^{(-1)^{{\frac {k}{2}}+1}}\cdots (4)$

${\frac {4}{\pi }}=\lim _{n\to \infty }(4n+1)\cdot \left({\frac {1}{2}}\cdot {\frac {3}{4}}\cdot {\frac {5}{6}}\cdots {\frac {2n-1}{2n}}\right)^{2}\cdots (5)$

$2^{\frac {1}{4}}=\prod _{n=1}^{\infty }{\frac {e^{(2n-1)\pi }+1}{e^{(3n-1)\pi }}}\cdot {\frac {(e^{n\pi }+1)^{3}}{e^{2n\pi }+1}}\cdots (6)$

$\pi =\lim _{n\to \infty }\left[\prod _{k=1}^{n}\left(1+{\frac {1}{2k-1}}\right)^{2}-\prod _{k=1}^{n-1}\left(1+{\frac {1}{2k-1}}\right)^{2}\right]\cdots (7)$

${\frac {e^{\pi }}{2^{3}}}=\prod _{n=1,3,5,...}^{\infty }\left({\frac {e^{n\pi }}{e^{n\pi }-1}}\right)^{24}\cdots (8)$

${\frac {e^{\pi }}{2^{6}}}=\prod _{n=1,3,5,...}^{\infty }\left({\frac {e^{n\pi }}{e^{n\pi }+1}}\right)^{24}\cdots (9)$

${\frac {e^{2\pi }}{2^{9}}}=\prod _{n=1,3,5,...}^{\infty }\left({\frac {e^{2n\pi }}{e^{2n\pi }-1}}\right)^{24}\cdots (10)$

$2=\prod _{n=1,3,5,...}^{\infty }\left({\frac {e^{n\pi }+1}{e^{n\pi }-1}}\right)^{8}\cdots (11)$

$\left({\frac {2}{\pi }}\right)^{\frac {1}{4}}\cdot {\Gamma \left({\frac {3}{4}}\right)}=\prod _{n=1}^{\infty }\left({\frac {e^{n\pi }+1}{e^{n\pi }-1}}\right)\cdots (12)$

$2^{\frac {1}{8}}=\prod _{n=1,3,5,...}^{\infty }\left({\frac {e^{n\pi }+1}{e^{n\pi }-1}}\right)\cdots (13)$

$\left({\frac {2}{\pi ^{2}}}\right)^{\frac {1}{8}}\cdot \Gamma \left({\frac {3}{4}}\right)=\prod _{n=1}^{\infty }{\frac {e^{2n\pi }+1}{e^{2n\pi }-1}}\cdots (14)$

$2=\prod _{n=1}^{\infty }\left[1+{\frac {1}{e^{(2n-1)\pi }-1}}\right]^{20}\left[\left(1+{\frac {1}{e^{n\pi }}}\right)\left(1-{\frac {1}{e^{n\pi }}}\right)^{2}\right]^{4}\cdots (15)$

$2=\prod _{n=1}^{\infty }\left(1+{\frac {1}{e^{n\pi }}}\right)^{20}\left(1-{\frac {1}{e^{2n\pi }+1}}\right)^{8}\left(1-{\frac {1}{e^{(2n-1)\pi }}}\right)^{4}\cdots (16)$

$2=\prod _{n=1}^{\infty }\left(1+{\frac {1}{e^{n\pi }}}\right)^{14}\left(1-{\frac {1}{e^{n\pi }}}\right)^{8}\left(1+{\frac {1}{e^{(2n-1)\pi }-1}}\right)^{10}\cdots (17)$

$e^{\pi }=\prod _{n=1}^{\infty }\left(1+{\frac {1}{e^{n\pi }-1}}\right)^{30}\left(1+{\frac {1}{e^{(2n-1)\pi }-1}}\right)^{42}\left(1-{\frac {1}{e^{2n\pi }}}\right)^{54}\cdots (18)$

${\frac {e^{\frac {\pi }{24}}}{2^{\frac {1}{8}}}}=\prod _{n=1}^{\infty }\left(1+{\frac {1}{e^{n\pi }}}\right)\cdots (19)$

${\frac {2^{\frac {1}{8}}}{e^{\frac {\pi }{24}}}}=\prod _{n=1}^{\infty }\left(1-{\frac {1}{e^{(2n-1)\pi }}}\right)\cdots (20)$

${\frac {2^{\frac {1}{4}}}{e^{\frac {\pi }{24}}}}=\prod _{n=1}^{\infty }\left(1+{\frac {1}{e^{(2n-1)\pi }}}\right)\cdots (21)$

${\frac {2^{\frac {3}{8}}}{e^{\frac {\pi }{12}}}}=\prod _{n=1}^{\infty }\left(1-{\frac {1}{e^{2(2n-1)\pi }}}\right)\cdots (22)$

${\frac {e^{\frac {\pi }{12}}\cdot \pi ^{\frac {1}{4}}}{2^{\frac {1}{2}}\cdot \Gamma \left({\frac {3}{4}}\right)}}=\prod _{n=1}^{\infty }\left(1-{\frac {1}{e^{2n\pi }}}\right)\cdots (23)$

${\frac {e^{\frac {\pi }{24}}\cdot \pi ^{\frac {1}{4}}}{2^{\frac {3}{8}}\cdot \Gamma \left({\frac {3}{4}}\right)}}=\prod _{n=1}^{\infty }\left(1-{\frac {1}{e^{n\pi }}}\right)\cdots (24)$

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Infinite sum

$\pi =72\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(e^{n\pi }-1)}}+24\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(e^{2n\pi }-1)}}\cdots (1)$

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Beta function

$ln\left[{\frac {\Gamma ^{2}({\frac {1}{4}})}{4{\sqrt {2\pi }}}}\right]=\sum _{n=1}^{\infty }{\frac {1-\beta (n)}{n}}\cdots (1)$

$ln\left[{\frac {\Gamma ^{4}({\frac {1}{4}})}{16\pi ^{2}}}\right]=\sum _{k=1}^{\infty }{\frac {1-\beta (2k)}{k}}\cdots (2)$

$ln{\sqrt {\frac {\pi }{2}}}=\sum _{k=0}^{\infty }{\frac {1-\beta (2k+1)}{2k+1}}\cdots (3)$

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prime number

$ln{\sqrt {\frac {7}{6}}}=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\sum _{p=2}^{\infty }{\frac {1}{p^{8k+4}}}\cdots (4)$

$ln{\sqrt {\frac {5}{2}}}=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\sum _{p=2}^{\infty }{\frac {1}{p^{4k+2}}}\cdots (5)$

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Zeta function

$ln\left({\frac {\pi }{sinh\pi }}\right)=\sum _{n=1}^{\infty }{\frac {(-1)^{k}\zeta (2n)}{n}}\cdots (6)$

$ln{\sqrt {\frac {3}{2}}}=\sum _{k=0}^{\infty }{\frac {\zeta (6k+3)-1}{2k+1}}\cdots (7)$

$ln{\sqrt {\frac {sinh\pi }{\pi }}}=\sum _{k=0}^{\infty }{\frac {\zeta (4k+2)-1}{2k+1}}\cdots (8)$

$ln\pi -\gamma =\sum _{k=2}^{\infty }{\frac {\zeta (k)}{k\cdot 2^{k-1}}}\cdots (9)$

$ln2-\gamma =\sum _{k=1}^{\infty }{\frac {\zeta (2k+1)}{(2k+1)2^{2k}}}\cdots (10)$

$ln(2\pi )=\sum _{n=1}^{\infty }{\frac {\left(2^{2n+1}+1\right)\zeta (2n)-2^{2n+1}}{2^{2n}\cdot {n}}}\cdots (11)$

$ln\left({\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{2^{2n}n}}\cdots (12)$

$ln\left({\frac {\pi }{3}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{6^{2n}n}}\cdots (13)$

$ln\left({\frac {\pi \phi }{5}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{10^{2n}n}}\cdots (14)$

$ln\left({\frac {2\pi \phi }{5{\sqrt {\phi +2}}}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{5^{2n}n}}\cdots (15)$

$ln(2)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{n}}\cdots (16)$

$ln\left({\frac {4\pi }{sinh\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (4n)-1}{n}}\cdots (17)$

${\gamma }=\lim _{k\to \infty }\left[\sum _{n=1}^{k}{{\frac {\zeta (2n)}{n}}-ln2k}\right]\cdots (18)$

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Eta function

$ln\left({\frac {2}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {(-1)^{n}\eta (n)}{n}}\cdots (19)$

$ln{\sqrt {2}}=\sum _{k=0}^{\infty }{\frac {1-\eta (2k+1)}{2k+1}}\cdots (20)$

$ln\left({\frac {8}{\pi ^{2}}}\right)=\sum _{n=1}^{\infty }{\frac {\eta (2n)-1}{n}}\cdots (21)$

$ln\left({\frac {2}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\eta (n)-1}{n}}\cdots (22)$

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Lambda function

$ln\left({\frac {4}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\lambda (2n)-1}{n}}\cdots (23)$

$ln\left({\frac {1}{\pi }}\right)+2-\gamma =2\sum _{n=2}^{\infty }{\frac {\lambda (n)-1}{n}}\cdots (24)$

${\gamma }=\lim _{k\to \infty }\left[\sum _{n=1}^{k}{{\frac {\lambda (2n)}{m}}-ln\left({\frac {4k}{\pi }}\right)}\right]\cdots (25)$

$\gamma +ln\left({\frac {4}{\pi }}\right)=2\sum _{n=2}^{\infty }{\frac {(-1)^{n}\lambda (n)}{n}}\cdots (26)$

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$\pi +ln\left({\frac {1}{2^{3}}}\right)=24\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(e^{n\pi }-1)}}\cdots (27)$

$2\pi +ln\left({\frac {1}{2^{9}}}\right)=24\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(e^{2n\pi }-1)}}\cdots (28)$

$ln2=24\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(e^{n\pi }-1)}}-8\sum _{n=1}^{\infty }{\frac {1}{n(e^{n\pi }-1)}}\cdots (29)$

${4\pi }+ln\left({\frac {1}{2^{18}}}\right)=24\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {1}{e^{2k\pi }+1}}+{\frac {1}{e^{2k\pi }-1}}\right)\cdots (30)$

${\frac {1}{8}}ln2=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\left({\frac {1}{e^{(2k+1)\pi }+1}}+{\frac {1}{e^{(2k+1)\pi }-1}}\right)\cdots (31)$

${\pi }+ln\left({\frac {1}{2^{6}}}\right)=12\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}\left({\frac {1}{e^{n\pi }+1}}+{\frac {1}{e^{n\pi }-1}}\right)\cdots (32)$

$ln{\frac {\pi ^{\frac {1}{4}}}{\Gamma ({\frac {3}{4}})}}=2\sum _{k=0}^{\infty }{\frac {1}{(2k+1)(e^{\pi (2k+1)}+1)}}\cdots (33)$

$ln\left[\left({\frac {2}{\pi }}\right)^{\frac {1}{4}}\cdot {\Gamma \left({\frac {3}{4}}\right)}\right]=2\sum _{k=0}^{\infty }{\frac {1}{(2k+1)(e^{\pi (2k+1)}-1)}}\cdots (34)$

$ln\left[\Gamma \left({\frac {3}{4}}\right)\cdot \left({\frac {2}{\pi ^{2}}}\right)^{\frac {1}{8}}\right]=\sum _{n=1}^{\infty }{\frac {1}{2k+1}}\left({\frac {1}{e^{(2n-1)\pi }-1}}-{\frac {1}{e^{(2n-1)\pi }+1}}\right)\cdots (35)$

x > 1

$ln2=\sum _{n=1}^{\infty }{\frac {1}{(2^{x}-x-1)n}}\sum _{r=1}^{2^{x}-2}{\frac {2^{x}-r-1}{(r+1)^{2n}}}\cdots (36)$

$ln3=\sum _{k=0}^{\infty }{\frac {1}{2^{2k}\cdot (2k+1)}}\cdots (37)$

$ln2=\sum _{n=1}^{\infty }{\frac {1}{n}}\left({\frac {1}{2^{2n}}}+{\frac {2}{3^{2n}}}+{\frac {2}{4^{2n}}}+{\frac {1}{5^{2n}}}\right)\cdots (38)$

$ln\left({\frac {2{\sqrt {2}}}{e}}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\left[{\frac {1}{4}}\left({\frac {1}{3^{2n}}}\right)+{\frac {1}{8}}\left({\frac {1}{5^{2n}}}+{\frac {1}{7^{2n}}}\right)+{\frac {1}{16}}\left({\frac {1}{9^{2n}}}+{\frac {1}{11^{2n}}}+{\frac {1}{13^{2n}}}+{\frac {1}{15^{2n}}}\right)+\cdots \right]\cdots (39)$

$\sum _{k=1}^{\infty }{\frac {\zeta (2k)\left[(2^{2k}-1)\zeta (2k)-2^{2k}\right]}{k}}+ln\left({\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{r=1}^{\infty }\sum _{a=3}^{\infty }\left[{\frac {1}{(2ar-a-1)^{2n}}}+{\frac {2}{(2ar-a)^{2n}}}+{\frac {1}{(2ar-a+1)^{2n}}}\right]\cdots (40)$

Where K = 8.700... is the constant from polygon circumscribing.

$ln\left({\frac {2K}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{r=1}^{\infty }\sum _{k=1}^{\infty }\left[{\frac {1}{(2rk+r-1)^{2n}}}+{\frac {2}{(2rk+r)^{2n}}}+{\frac {1}{(2rk+r+1)^{2n}}}\right]\cdots (41)$

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Inspired by your website

These are the formulas that I derived by visiting your website for the last fews months.

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Definition of:

$\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=1+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+\cdots$

$\eta (s)=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}}=1-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+\cdots$

$\lambda (s)=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{s}}}=1+{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}+{\frac {1}{7^{s}}}+\cdots$

$\beta (s)=\sum _{n=0}^{\infty }{\frac {(-1)^{k}}{(2n+1)^{s}}}=1-{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}-{\frac {1}{7^{s}}}+\cdots$

Where p is prime numbers

$\phi ={\frac {{\sqrt {5}}+1}{2}}$

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Jacobi theta function

$J_{4}(0,e^{-1})=\left({\frac {\pi }{2}}\right)^{\frac {1}{4}}\cdot {\frac {1}{\Gamma \left({\frac {3}{4}}\right)}}\cdots (1)$

${\frac {J_{2}(0,e^{-1})}{J_{1}^{'}(0,e^{-1})}}=\left({\frac {\pi ^{2}}{2}}\right)^{\frac {1}{4}}\cdot {\frac {1}{\Gamma ^{2}\left({\frac {3}{4}}\right)}}\cdots (2)$

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Infinite product

$\pi ={\frac {\prod _{n=1}^{\infty }\left[1+{\frac {1}{4(3n)^{2}-1}}\right]}{\sum _{n=1}^{\infty }{\frac {1}{4^{n}}}}}\cdots (1)$

${\frac {\Gamma ^{2}({\frac {1}{4}})}{\pi ^{\frac {3}{2}}}}={\frac {\prod _{n=1}^{\infty }\left[1+{\frac {1}{(4n-1)^{2}-1}}\right]}{\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}}}\cdots (2)$

${\frac {\Gamma ^{2}({\frac {1}{4}})}{\pi ^{\frac {5}{2}}}}={\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}}\cdot {{\prod _{n=1}^{\infty }\left[1+{\frac {1}{(4n-1)^{2}-1}}\right]^{2}}{\left[1+{\frac {1}{(4n+1)^{2}-1}}\right]}}\cdots (3)$

$\left[{\frac {\Gamma ^{(}{\frac {1}{4}})}{2}}\right]^{2}\cdot {\frac {1}{\sqrt {2\pi }}}={\frac {3}{2}}\cdot {\frac {4}{5}}\cdot {\frac {7}{6}}\cdot {\frac {8}{9}}\cdot {\frac {11}{10}}\cdot {\frac {12}{13}}=\prod _{k=2,4,6,...}^{\infty }\left({\frac {k+1}{k}}\right)^{(-1)^{{\frac {k}{2}}+1}}\cdots (4)$

${\frac {4}{\pi }}=\lim _{n\to \infty }(4n+1)\cdot \left({\frac {1}{2}}\cdot {\frac {3}{4}}\cdot {\frac {5}{6}}\cdots {\frac {2n-1}{2n}}\right)^{2}\cdots (5)$

$2^{\frac {1}{4}}=\prod _{n=1}^{\infty }{\frac {e^{(2n-1)\pi }+1}{e^{(3n-1)\pi }}}\cdot {\frac {(e^{n\pi }+1)^{3}}{e^{2n\pi }+1}}\cdots (6)$

$\pi =\lim _{n\to \infty }\left[\prod _{k=1}^{n}\left(1+{\frac {1}{2k-1}}\right)^{2}-\prod _{k=1}^{n-1}\left(1+{\frac {1}{2k-1}}\right)^{2}\right]\cdots (7)$

${\frac {e^{\pi }}{2^{3}}}=\prod _{n=1,3,5,...}^{\infty }\left({\frac {e^{n\pi }}{e^{n\pi }-1}}\right)^{24}\cdots (8)$

${\frac {e^{2\pi }}{2^{9}}}=\prod _{n=1,3,5,...}^{\infty }\left({\frac {e^{2n\pi }}{e^{2n\pi }-1}}\right)^{24}\cdots (9)$

$2=\prod _{n=1,3,5,...}^{\infty }\left({\frac {e^{n\pi }+1}{e^{n\pi }-1}}\right)^{8}\cdots (10)$

$\left({\frac {2}{\pi }}\right)^{\frac {1}{4}}\cdot {\Gamma \left({\frac {3}{4}}\right)}=\prod _{n=1}^{\infty }\left({\frac {e^{n\pi }+1}{e^{n\pi }-1}}\right)\cdots (11)$

$2=\prod _{n=1}^{\infty }\left[1+{\frac {1}{e^{(2n-1)\pi }-1}}\right]^{20}\left[\left(1+{\frac {1}{e^{n\pi }}}\right)\left(1-{\frac {1}{e^{n\pi }}}\right)^{2}\right]^{4}\cdots (12)$

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$ln\left[{\frac {\Gamma ^{2}({\frac {1}{4}})}{4{\sqrt {2\pi }}}}\right]=\sum _{n=1}^{\infty }{\frac {1-\beta (n)}{n}}\cdots (1)$

$ln\left[{\frac {\Gamma ^{4}({\frac {1}{4}})}{16\pi ^{2}}}\right]=\sum _{k=1}^{\infty }{\frac {1-\beta (2k)}{k}}\cdots (2)$

$ln{\sqrt {\frac {\pi }{2}}}=\sum _{k=0}^{\infty }{\frac {1-\beta (2k+1)}{2k+1}}\cdots (3)$

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$ln{\sqrt {\frac {7}{6}}}=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\sum _{p=2}^{\infty }{\frac {1}{p^{8k+4}}}\cdots (4)$

$ln{\sqrt {\frac {5}{2}}}=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\sum _{p=2}^{\infty }{\frac {1}{p^{4k+2}}}\cdots (5)$

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$ln\left({\frac {\pi }{sinh\pi }}\right)=\sum _{n=1}^{\infty }{\frac {(-1)^{k}\zeta (2n)}{n}}\cdots (6)$

$ln{\sqrt {\frac {3}{2}}}=\sum _{k=0}^{\infty }{\frac {\zeta (6k+3)-1}{2k+1}}\cdots (7)$

$ln{\sqrt {\frac {sinh\pi }{\pi }}}=\sum _{k=0}^{\infty }{\frac {\zeta (4k+2)-1}{2k+1}}\cdots (8)$

$ln\pi -\gamma =\sum _{k=2}^{\infty }{\frac {\zeta (k)}{k\cdot 2^{k-1}}}\cdots (9)$

$ln2-\gamma =\sum _{k=1}^{\infty }{\frac {\zeta (2k+1)}{(2k+1)2^{2k}}}\cdots (10)$

$ln(2\pi )=\sum _{n=1}^{\infty }{\frac {\left(2^{2n+1}+1\right)\zeta (2n)-2^{2n+1}}{2^{2n}\cdot {n}}}\cdots (11)$

$ln\left({\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{2^{2n}n}}\cdots (12)$

$ln\left({\frac {\pi }{3}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{6^{2n}n}}\cdots (13)$

$ln\left({\frac {\pi \phi }{5}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{10^{2n}n}}\cdots (14)$

$ln\left({\frac {2\pi \phi }{5{\sqrt {\phi +2}}}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{5^{2n}n}}\cdots (15)$

$ln(2)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{n}}\cdots (16)$

$ln\left({\frac {4\pi }{sinh\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (4n)-1}{n}}\cdots (17)$

${\gamma }=\lim _{k\to \infty }\left[\sum _{n=1}^{k}{{\frac {\zeta (2n)}{n}}-ln2k}\right]\cdots (18)$

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$ln\left({\frac {2}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {(-1)^{n}\eta (n)}{n}}\cdots (19)$

$ln{\sqrt {2}}=\sum _{k=0}^{\infty }{\frac {1-\eta (2k+1)}{2k+1}}\cdots (20)$

$ln\left({\frac {8}{\pi ^{2}}}\right)=\sum _{n=1}^{\infty }{\frac {\eta (2n)-1}{n}}\cdots (21)$

$ln\left({\frac {2}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\eta (n)-1}{n}}\cdots (22)$

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$ln\left({\frac {4}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\lambda (2n)-1}{n}}\cdots (23)$

$ln\left({\frac {1}{\pi }}\right)+2-\gamma =2\sum _{n=2}^{\infty }{\frac {\lambda (n)-1}{n}}\cdots (24)$

${\gamma }=\lim _{k\to \infty }\left[\sum _{n=1}^{k}{{\frac {\lambda (2n)}{m}}-ln\left({\frac {4k}{\pi }}\right)}\right]\cdots (25)$

$\gamma +ln\left({\frac {4}{\pi }}\right)=2\sum _{n=2}^{\infty }{\frac {(-1)^{n}\lambda (n)}{n}}\cdots (26)$

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${\frac {\pi }{6}}-{\frac {3}{4}}ln2=\sum _{k=1}^{\infty }{\frac {1}{k}}\left({\frac {1}{e^{2k\pi }+1}}+{\frac {1}{e^{2k\pi }-1}}\right)\cdots (27)$

${\frac {1}{8}}ln2=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\left({\frac {1}{e^{(2k+1)\pi }+1}}+{\frac {1}{e^{(2k+1)\pi }-1}}\right)\cdots (28)$

${\frac {\pi }{12}}-{\frac {1}{2}}ln2=\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}\left({\frac {1}{e^{n\pi }+1}}+{\frac {1}{e^{n\pi }-1}}\right)\cdots (29)$

$ln{\frac {\pi ^{\frac {1}{4}}}{\Gamma ({\frac {3}{4}})}}=2\sum _{k=0}^{\infty }{\frac {1}{(2k+1)(e^{\pi (2k+1)}+1)}}\cdots (30)$

$ln\left[\left({\frac {2}{\pi }}\right)^{\frac {1}{4}}\cdot {\Gamma \left({\frac {3}{4}}\right)}\right]=2\sum _{k=0}^{\infty }{\frac {1}{(2k+1)(e^{\pi (2k+1)}-1)}}\cdots (31)$

x > 1

$ln2=\sum _{n=1}^{\infty }{\frac {1}{(2^{x}-x-1)n}}\sum _{r=1}^{2^{x}-2}{\frac {2^{x}-r-1}{(r+1)^{2n}}}\cdots (32)$

$ln3=\sum _{k=0}^{\infty }{\frac {1}{2^{2k}\cdot (2k+1)}}\cdots (33)$

$ln2=\sum _{n=1}^{\infty }{\frac {1}{n}}\left({\frac {1}{2^{2n}}}+{\frac {2}{3^{2n}}}+{\frac {2}{4^{2n}}}+{\frac {1}{5^{2n}}}\right)\cdots (34)$

$ln\left({\frac {2{\sqrt {2}}}{e}}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\left[{\frac {1}{4}}\left({\frac {1}{3^{2n}}}\right)+{\frac {1}{8}}\left({\frac {1}{5^{2n}}}+{\frac {1}{7^{2n}}}\right)+{\frac {1}{16}}\left({\frac {1}{9^{2n}}}+{\frac {1}{11^{2n}}}+{\frac {1}{13^{2n}}}+{\frac {1}{15^{2n}}}\right)+\cdots \right]\cdots (35)$

$\sum _{k=1}^{\infty }{\frac {\zeta (2k)\left[(2^{2k}-1)\zeta (2k)-2^{2k}\right]}{k}}+ln\left({\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{r=1}^{\infty }\sum _{a=3}^{\infty }\left[{\frac {1}{(2ar-a-1)^{2n}}}+{\frac {2}{(2ar-a)^{2n}}}+{\frac {1}{(2ar-a+1)^{2n}}}\right]\cdots (36)$

Where K = 8.700... is the constant from polygon circumscribing.

$ln\left({\frac {2K}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{r=1}^{\infty }\sum _{k=1}^{\infty }\left[{\frac {1}{(2rk+r-1)^{2n}}}+{\frac {2}{(2rk+r)^{2n}}}+{\frac {1}{(2rk+r+1)^{2n}}}\right]\cdots (37)$

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Inspired by your website

These are the formulas that I derived by visiting your website for the last fews months.

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Definition of:

$\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=1+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+\cdots$

$\eta (s)=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}}=1-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+\cdots$

$\lambda (s)=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{s}}}=1+{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}+{\frac {1}{7^{s}}}+\cdots$

$\beta (s)=\sum _{n=0}^{\infty }{\frac {(-1)^{k}}{(2n+1)^{s}}}=1-{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}-{\frac {1}{7^{s}}}+\cdots$

Where p is prime numbers

$\phi ={\frac {{\sqrt {5}}+1}{2}}$

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I like your symmetric formula. I found one, not quite symmetric.

$\pi ={\frac {\prod _{n=1}^{\infty }\left[1+{\frac {1}{4(3n)^{2}-1}}\right]}{\sum _{n=1}^{\infty }{\frac {1}{4^{n}}}}}\cdots (1)$

${\frac {\Gamma ^{2}({\frac {1}{4}})}{\pi ^{\frac {3}{2}}}}={\frac {\prod _{n=1}^{\infty }\left[1+{\frac {1}{(4n-1)^{2}-1}}\right]}{\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}}}\cdots (2)$

${\frac {\Gamma ^{2}({\frac {1}{4}})}{\pi ^{\frac {5}{2}}}}={\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}}\cdot {{\prod _{n=1}^{\infty }\left[1+{\frac {1}{(4n-1)^{2}-1}}\right]^{2}}{\left[1+{\frac {1}{(4n+1)^{2}-1}}\right]}}\cdots (3)$

$\left[{\frac {\Gamma ^{(}{\frac {1}{4}})}{2}}\right]^{2}\cdot {\frac {1}{\sqrt {2\pi }}}={\frac {3}{2}}\cdot {\frac {4}{5}}\cdot {\frac {7}{6}}\cdot {\frac {8}{9}}\cdot {\frac {11}{10}}\cdot {\frac {12}{13}}=\prod _{k=2,4,6,...}^{\infty }\left({\frac {k+1}{k}}\right)^{(-1)^{{\frac {k}{2}}+1}}\cdots (4)$

${\frac {4}{\pi }}=\lim _{n\to \infty }(4n+1)\cdot \left({\frac {1}{2}}\cdot {\frac {3}{4}}\cdot {\frac {5}{6}}\cdots {\frac {2n-1}{2n}}\right)^{2}\cdots (5)$

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$ln\left[{\frac {\Gamma ^{2}({\frac {1}{4}})}{4{\sqrt {2\pi }}}}\right]=\sum _{n=1}^{\infty }{\frac {1-\beta (n)}{n}}\cdots (1)$

$ln\left[{\frac {\Gamma ^{4}({\frac {1}{4}})}{16\pi ^{2}}}\right]=\sum _{k=1}^{\infty }{\frac {1-\beta (2k)}{k}}\cdots (2)$

$ln{\sqrt {\frac {\pi }{2}}}=\sum _{k=0}^{\infty }{\frac {1-\beta (2k+1)}{2k+1}}\cdots (3)$

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$ln{\sqrt {\frac {7}{6}}}=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\sum _{p=2}^{\infty }{\frac {1}{p^{8k+4}}}\cdots (4)$

$ln{\sqrt {\frac {5}{2}}}=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\sum _{p=2}^{\infty }{\frac {1}{p^{4k+2}}}\cdots (5)$

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$ln\left({\frac {\pi }{sinh\pi }}\right)=\sum _{n=1}^{\infty }{\frac {(-1)^{k}\zeta (2n)}{n}}\cdots (6)$

$ln{\sqrt {\frac {3}{2}}}=\sum _{k=0}^{\infty }{\frac {\zeta (6k+3)-1}{2k+1}}\cdots (7)$

$ln{\sqrt {\frac {sinh\pi }{\pi }}}=\sum _{k=0}^{\infty }{\frac {\zeta (4k+2)-1}{2k+1}}\cdots (8)$

$ln\pi -\gamma =\sum _{k=2}^{\infty }{\frac {\zeta (k)}{k\cdot 2^{k-1}}}\cdots (9)$

$ln2-\gamma =\sum _{k=1}^{\infty }{\frac {\zeta (2k+1)}{(2k+1)2^{2k}}}\cdots (10)$

$ln(2\pi )=\sum _{n=1}^{\infty }{\frac {\left(2^{2n+1}+1\right)\zeta (2n)-2^{2n+1}}{2^{2n}\cdot {n}}}\cdots (11)$

$ln\left({\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{2^{2n}n}}\cdots (12)$

$ln\left({\frac {\pi }{3}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{6^{2n}n}}\cdots (13)$

$ln\left({\frac {\pi \phi }{5}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{10^{2n}n}}\cdots (14)$

$ln\left({\frac {2\pi \phi }{5{\sqrt {\phi +2}}}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{5^{2n}n}}\cdots (15)$

$ln(2)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{n}}\cdots (16)$

$ln\left({\frac {4\pi }{sinh\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (4n)-1}{n}}\cdots (17)$

${\gamma }=\lim _{k\to \infty }\left[\sum _{n=1}^{k}{{\frac {\zeta (2n)}{n}}-ln2k}\right]\cdots (18)$

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$ln\left({\frac {2}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {(-1)^{n}\eta (n)}{n}}\cdots (19)$

$ln{\sqrt {2}}=\sum _{k=0}^{\infty }{\frac {1-\eta (2k+1)}{2k+1}}\cdots (20)$

$ln\left({\frac {8}{\pi ^{2}}}\right)=\sum _{n=1}^{\infty }{\frac {\eta (2n)-1}{n}}\cdots (21)$

$ln\left({\frac {2}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\eta (n)-1}{n}}\cdots (22)$

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$ln\left({\frac {4}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\lambda (2n)-1}{n}}\cdots (23)$

${\gamma }=\lim _{k\to \infty }\left[\sum _{n=1}^{k}{{\frac {\lambda (2n)}{m}}-ln\left({\frac {4k}{\pi }}\right)}\right]\cdots (24)$

$\gamma +ln\left({\frac {4}{\pi }}\right)=2\sum _{n=2}^{\infty }{\frac {(-1)^{n}\lambda (n)}{n}}\cdots (25)$

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${\frac {\pi }{12}}-{\frac {1}{2}}ln2=\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}\left({\frac {1}{e^{n\pi }+1}}+{\frac {1}{e^{n\pi }-1}}\right)\cdots (26)$

$ln{\frac {\pi ^{\frac {1}{4}}}{\Gamma ({\frac {3}{4}})}}=2\sum _{k=0}^{\infty }{\frac {1}{(2k+1)(e^{\pi (2k+1)}+1)}}\cdots (27)$

x > 1

$ln2=\sum _{n=1}^{\infty }{\frac {1}{(2^{x}-x-1)n}}\sum _{r=1}^{2^{x}-2}{\frac {2^{x}-r-1}{(r+1)^{2n}}}\cdots (28)$

$ln3=\sum _{k=0}^{\infty }{\frac {1}{2^{2k}\cdot (2k+1)}}\cdots (29)$

$ln2=\sum _{n=1}^{\infty }{\frac {1}{n}}\left({\frac {1}{2^{2n}}}+{\frac {2}{3^{2n}}}+{\frac {2}{4^{2n}}}+{\frac {1}{5^{2n}}}\right)\cdots (30)$

$ln\left({\frac {2{\sqrt {2}}}{e}}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\left[{\frac {1}{4}}\left({\frac {1}{3^{2n}}}\right)+{\frac {1}{8}}\left({\frac {1}{5^{2n}}}+{\frac {1}{7^{2n}}}\right)+{\frac {1}{16}}\left({\frac {1}{9^{2n}}}+{\frac {1}{11^{2n}}}+{\frac {1}{13^{2n}}}+{\frac {1}{15^{2n}}}\right)+\cdots \right]\cdots (31)$

$\sum _{k=1}^{\infty }{\frac {\zeta (2k)\left[(2^{2k}-1)\zeta (2k)-2^{2k}\right]}{k}}+ln\left({\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{r=1}^{\infty }\sum _{a=3}^{\infty }\left[{\frac {1}{(2ar-a-1)^{2n}}}+{\frac {2}{(2ar-a)^{2n}}}+{\frac {1}{(2ar-a+1)^{2n}}}\right]\cdots (32)$

Where K = 8.700... is the constant from polygon circumscribing.

$ln\left({\frac {2K}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{r=1}^{\infty }\sum _{k=1}^{\infty }\left[{\frac {1}{(2rk+r-1)^{2n}}}+{\frac {2}{(2rk+r)^{2n}}}+{\frac {1}{(2rk+r+1)^{2n}}}\right]\cdots (33)$

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Kawarism

Definition of:

$\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=1+{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}+{\frac {1}{4^{s}}}+\cdots$

$\eta (s)=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n^{s}}}=1-{\frac {1}{2^{s}}}+{\frac {1}{3^{s}}}-{\frac {1}{4^{s}}}+\cdots$

$\lambda (s)=\sum _{n=0}^{\infty }{\frac {1}{(2n+1)^{s}}}=1+{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}+{\frac {1}{7^{s}}}+\cdots$

$\beta (s)=\sum _{n=0}^{\infty }{\frac {(-1)^{k}}{(2n+1)^{s}}}=1-{\frac {1}{3^{s}}}+{\frac {1}{5^{s}}}-{\frac {1}{7^{s}}}+\cdots$

$B(s)={\frac {1}{3^{s}}}+{\frac {1}{7^{s}}}+{\frac {1}{11^{s}}}+\cdots$

Where p is prime numbers

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I like your symmetric formula. I found one, not quite symmetric.

$\pi ={\frac {\prod _{n=1}^{\infty }\left(1+{\frac {1}{4(3n)^{2}-1}}\right)}{\sum _{n=1}^{\infty }{\frac {1}{4^{n}}}}}\cdots (1)$

${\frac {\Gamma ^{2}({\frac {1}{4}})}{\pi ^{\frac {3}{2}}}}=\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}\cdot {\prod _{n=1}^{\infty }\left(1+{\frac {1}{(4n-1)^{2}-1}}\right)}\cdots (2)$

${\frac {\Gamma ^{2}({\frac {1}{4}})}{\pi ^{\frac {5}{2}}}}={\frac {\prod _{n=1}^{\infty }\left[1+{\frac {1}{(4k-1)^{2}-1}}\right]^{2}\cdot {\prod _{n=1}^{\infty }\left[1+{\frac {1}{(4n+1)^{2}-1}}\right]}}{\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-1}}}}\cdots (3)$

$ln\left[\left({\frac {4}{\pi }}\right)^{2}\cdot {\frac {1}{e}}\right]=\sum _{n=2}^{\infty }(-1)^{n-1}\left[{\frac {1}{n}}-ln\left({\frac {n-1}{n}}\cdot {\frac {n+1}{n}}\right)\right]\cdots (4)$

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$ln\left[{\frac {\Gamma ^{2}({\frac {1}{4}})}{4\pi }}\right]=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{[8n(n+1)+1]^{2k+1}}}\cdots (-6)$

$ln\left[{\frac {\Gamma ^{4}({\frac {1}{4}})}{16\pi ^{2}}}\right]=\sum _{k=1}^{\infty }{\frac {1-\beta (2k)}{k}}\cdots (-5)$

$ln{\frac {\Gamma ^{2}({\frac {1}{4}})}{2\pi ^{\frac {3}{2}}}}=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{k=1}^{\infty }{\frac {1}{(4k-1)^{2n}}}\cdots (-4)$

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

$ln{\sqrt {\frac {7}{6}}}=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\sum _{p=2}^{\infty }{\frac {1}{p^{8k+4}}}\cdots (-2)$

$ln{\sqrt {\frac {5}{2}}}=\sum _{k=0}^{\infty }{\frac {1}{2k+1}}\sum _{p=2}^{\infty }{\frac {1}{p^{4k+2}}}\cdots (-1)$

$ln{\sqrt {\frac {3}{2}}}=\sum _{k=0}^{\infty }{\frac {\zeta (6k+3)-1}{2k+1}}\cdots (0)$

$ln{\sqrt {\frac {sinh\pi }{\pi }}}=\sum _{k=0}^{\infty }{\frac {\zeta (4k+2)-1}{2k+1}}\cdots (1)$

$ln{\frac {2}{\sqrt {\pi }}}=\sum _{k=0}^{\infty }{\frac {\beta (2k+1)-\eta (2k+1)}{2k+1}}\cdots (2)$

$ln{\sqrt {\frac {\pi }{2}}}=\sum _{k=0}^{\infty }{\frac {1-\beta (2k+1)}{2k+1}}\cdots (3)$

$ln{\sqrt {2}}=\sum _{k=0}^{\infty }{\frac {1-\eta (2k+1)}{2k+1}}\cdots (4)$

$ln\pi -\gamma =\sum _{k=2}^{\infty }{\frac {\zeta (k)}{k\cdot 2^{k-1}}}\cdots (5)$

$ln2-\gamma =\sum _{k=1}^{\infty }{\frac {\zeta (2k+1)}{(2k+1)2^{2k}}}\cdots (6)$

$ln(\pi )=\sum _{n=1}^{\infty }{\frac {1}{n}}\left[{\zeta (2n)}\left(1+{\frac {1}{2^{2n}}}\right)-1\right]\cdots (7)$

$ln(2\pi )=\sum _{n=1}^{\infty }{\frac {\left(2^{2n+1}+1\right)\zeta (2n)-2^{2n+1}}{2^{2n}\cdot {n}}}\cdots (8)$

$ln\left({\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{2^{2n}n}}\cdots (9)$

$ln\left({\frac {\pi }{3}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{6^{2n}n}}\cdots (10)$

$ln\left({\frac {4}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\lambda (2n)-1}{n}}\cdots (11)$

$ln\left({\frac {8}{\pi ^{2}}}\right)=\sum _{n=1}^{\infty }{\frac {\eta (2n)-1}{n}}\cdots (12)$

$ln\left({\frac {2}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\eta (n)-1}{n}}\cdots (13)$

$ln(2)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{n}}\cdots (14)$

$ln\left({\frac {4}{\pi }}\right)-\sum _{n=1}^{\infty }{\frac {2B(2n)}{n}}=\sum _{n=1}^{\infty }{\frac {\beta (2n)-1}{n}}\cdots (15)$

x > 1

$ln2=\sum _{n=1}^{\infty }{\frac {1}{(2^{x}-x-1)n}}\sum _{r=1}^{2^{x}-2}{\frac {2^{x}-r-1}{(r+1)^{2n}}}\cdots (16)$

$ln3=\sum _{k=0}^{\infty }{\frac {1}{2^{2k}\cdot (2k+1)}}\cdots (17)$

$ln2=\sum _{n=1}^{\infty }{\frac {1}{n}}\left({\frac {1}{2^{2n}}}+{\frac {2}{3^{2n}}}+{\frac {2}{4^{2n}}}+{\frac {1}{5^{2n}}}\right)\cdots (18)$

$ln\left({\frac {\pi \phi }{5}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{10^{2n}n}}\cdots (19)$

$ln\left({\frac {2\pi \phi }{5{\sqrt {\phi +2}}}}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{5^{2n}n}}\cdots (20)$

$ln\left({\frac {4\pi }{sinh\pi }}\right)=\sum _{n=1}^{\infty }{\frac {\zeta (4n)-1}{n}}\cdots (21)$

${\gamma }=\lim _{k\to \infty }\left[\sum _{n=1}^{k}{{\frac {\zeta (2n)}{n}}-ln2k}\right]\cdots (22)$

${\gamma }=\lim _{k\to \infty }\left[\sum _{n=1}^{k}{{\frac {\lambda (2n)}{m}}-ln\left({\frac {4k}{\pi }}\right)}\right]\cdots (23)$

$ln\left({\frac {2{\sqrt {2}}}{e}}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\left[{\frac {1}{4}}\left({\frac {1}{3^{2n}}}\right)+{\frac {1}{8}}\left({\frac {1}{5^{2n}}}+{\frac {1}{7^{2n}}}\right)+{\frac {1}{16}}\left({\frac {1}{9^{2n}}}+{\frac {1}{11^{2n}}}+{\frac {1}{13^{2n}}}+{\frac {1}{15^{2n}}}\right)+\cdots \right]\cdots (24)$

$\sum _{k=1}^{\infty }{\frac {\zeta (2k)\left[(2^{2k}-1)\zeta (2k)-2^{2k}\right]}{k}}+ln\left({\frac {\pi }{2}}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{r=1}^{\infty }\sum _{a=3}^{\infty }\left[{\frac {1}{(2ar-a-1)^{2n}}}+{\frac {2}{(2ar-a)^{2n}}}+{\frac {1}{(2ar-a+1)^{2n}}}\right]\cdots (25)$

Where K = 8.700... is the constant from polygon circumscribing.

$ln\left({\frac {2K}{\pi }}\right)=\sum _{n=1}^{\infty }{\frac {1}{n}}\sum _{r=1}^{\infty }\sum _{k=1}^{\infty }\left[{\frac {1}{(2rk+r-1)^{2n}}}+{\frac {2}{(2rk+r)^{2n}}}+{\frac {1}{(2rk+r+1)^{2n}}}\right]\cdots (26)$