User:Van bui 314

Inspired by your website

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

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

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

Jacobi theta function

$$J_2(0,e^{-\pi})=J_4(0,e^{-\pi})=\left(\frac{\pi}{2}\right)^{\frac{1}{4}}\cdot \frac{1}{\Gamma\left(\frac{3}{4}\right)}\cdots(1)$$

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

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

Infinite product

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

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

$$\frac{\Gamma^2(\frac{1}{4})}{\pi^{\frac{5}{2}}}={\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}\cdot \cdots(3)$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Infinite sum

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

Beta function

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

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

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

prime number

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

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

Zeta function

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

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

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

$$ln\pi-\gamma=\sum_{k=2}^{\infty}\frac{\zeta(k)}{k\cdot2^{k-1}}\cdots(9)$$

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

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

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

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

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

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

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

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

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

Eta function

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

$$ln\sqrt2=\sum_{k=0}^{\infty}\frac{1-\eta(2k+1)}{2k+1}\cdots(20)$$

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

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

Lambda function

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

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

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

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

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

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

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

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

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

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

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

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

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

x > 1

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

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

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

$$ln\left(\frac{2\sqrt2}{e}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\left[\frac{1}{4}\left(\frac{1}{3^{2n}}\right)+ \frac{1}{8}\left(\frac{1}{5^{2n}}+\frac{1}{7^{2n}}\right)+\frac{1}{16}\left(\frac{1}{9^{2n}}+\frac{1}{11^{2n}}+ \frac{1}{13^{2n}}+\frac{1}{15^{2n}}\right)+\cdots\right]\cdots(39)$$

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

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

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

Inspired by your website

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

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

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

Jacobi theta function

$$J_4(0,e^{-1})=\left(\frac{\pi}{2}\right)^{\frac{1}{4}}\cdot \frac{1}{\Gamma\left(\frac{3}{4}\right)}\cdots(1)$$

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

Infinite product

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

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

$$\frac{\Gamma^2(\frac{1}{4})}{\pi^{\frac{5}{2}}}={\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}\cdot \cdots(3)$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$ln\pi-\gamma=\sum_{k=2}^{\infty}\frac{\zeta(k)}{k\cdot2^{k-1}}\cdots(9)$$

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

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

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

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

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

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

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

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

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

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

$$ln\sqrt2=\sum_{k=0}^{\infty}\frac{1-\eta(2k+1)}{2k+1}\cdots(20)$$

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

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

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

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

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

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

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

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

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

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

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

x > 1

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

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

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

$$ln\left(\frac{2\sqrt2}{e}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\left[\frac{1}{4}\left(\frac{1}{3^{2n}}\right)+ \frac{1}{8}\left(\frac{1}{5^{2n}}+\frac{1}{7^{2n}}\right)+\frac{1}{16}\left(\frac{1}{9^{2n}}+\frac{1}{11^{2n}}+ \frac{1}{13^{2n}}+\frac{1}{15^{2n}}\right)+\cdots\right]\cdots(35)$$

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

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

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

Inspired by your website

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

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

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

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

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

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

$$\frac{\Gamma^2(\frac{1}{4})}{\pi^{\frac{5}{2}}}={\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}\cdot \cdots(3)$$

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

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

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

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

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

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

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

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

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

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

$$ln\pi-\gamma=\sum_{k=2}^{\infty}\frac{\zeta(k)}{k\cdot2^{k-1}}\cdots(9)$$

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

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

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

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

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

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

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

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

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

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

$$ln\sqrt2=\sum_{k=0}^{\infty}\frac{1-\eta(2k+1)}{2k+1}\cdots(20)$$

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

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

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

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

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

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

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

x > 1

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

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

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

$$ln\left(\frac{2\sqrt2}{e}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\left[\frac{1}{4}\left(\frac{1}{3^{2n}}\right)+ \frac{1}{8}\left(\frac{1}{5^{2n}}+\frac{1}{7^{2n}}\right)+\frac{1}{16}\left(\frac{1}{9^{2n}}+\frac{1}{11^{2n}}+ \frac{1}{13^{2n}}+\frac{1}{15^{2n}}\right)+\cdots\right]\cdots(31)$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$ln\sqrt2=\sum_{k=0}^{\infty}\frac{1-\eta(2k+1)}{2k+1}\cdots(4)$$

$$ln\pi-\gamma=\sum_{k=2}^{\infty}\frac{\zeta(k)}{k\cdot2^{k-1}}\cdots(5)$$

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

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

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

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

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

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

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

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

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

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

x > 1

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

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

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

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

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

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

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

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

$$ln\left(\frac{2\sqrt2}{e}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\left[\frac{1}{4}\left(\frac{1}{3^{2n}}\right)+ \frac{1}{8}\left(\frac{1}{5^{2n}}+\frac{1}{7^{2n}}\right)+\frac{1}{16}\left(\frac{1}{9^{2n}}+\frac{1}{11^{2n}}+ \frac{1}{13^{2n}}+\frac{1}{15^{2n}}\right)+\cdots\right]\cdots(24)$$

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

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

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