User:Adam314

$$ln\left(\prod_{r=1,3,5,...}^{\infty}\frac{n^r+1}{n^r-1}\right)=\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{n^{r(2k+1)}+1}+\frac{1}{n^{r(2k+1)}-1}\right)$$

$$ln\left(\prod_{r=1}^{\infty}\frac{n^r+1}{n^r-1}\right)=2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{n^{r(2k+1)}-1}\right)$$

$$ln\left(\prod_{r=1}^{\infty}\frac{n^r}{n^r+1}\right)=\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\left(\frac{1}{n^k-1}\right)$$

$$ln\left(\prod_{r=1,3,5,...}^{\infty}\frac{n^r}{n^r+1}\right)=\sum_{k=1}^{\infty}\frac{(-1)^k}{2k}\left(\frac{1}{n^k-1}+\frac{1}{n^k+1}\right)$$

$$ln\left(\prod_{r=2,4,6,...}^{\infty}\frac{n^r}{n^r+1}\right)=\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\left(\frac{1}{n^{2k}-1}\right)$$

$$ln\left(\prod_{r=1}^{\infty}\frac{n^r}{n^r-1}\right)=\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{n^k-1}\right)$$

$$ln\left(\prod_{r=1,3,5,...}^{\infty}\frac{n^r}{n^r-1}\right)=\sum_{k=1}^{\infty}\frac{1}{2k}\left(\frac{1}{n^k-1}+\frac{1}{n^k+1}\right)$$

$$ln\left(\prod_{r=2,4,6,...}^{\infty}\frac{n^r}{n^r-1}\right)=\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{n^{2k}-1}\right)$$

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

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

$$ln\left[\prod_{n=1,3,5,...}^{\infty}\left(\frac{n^r}{n^r+1}\right)^{(-1)^{\frac{n-1}{2}}}\right]=\sum_{k=1}^{\infty}\frac{(-1)^k\beta(rk)}{k}$$

$$ln\left(\prod_{n=1,3,5,...}^{\infty}\frac{n^r}{n^r+1}\right)=\sum_{k=1}^{\infty}\frac{(-1)^k\lambda(rk)}{k}$$

$$ln\left[\prod_{n=1}^{\infty}\left(\frac{n^r}{n^r+1}\right)^{(-1)^{n+1}}\right]=\sum_{k=1}^{\infty}\frac{(-1)^k\eta(rk)}{k}$$

$$ln\left(\prod_{n=1}^{\infty}\frac{n^r}{n^r+1}\right)=\sum_{k=1}^{\infty}\frac{(-1)^k\zeta(rk)}{k}$$

$$\frac{\Gamma^2(\frac{1}{4})}{4\sqrt{2\pi}}=\prod_{k=2,4,6,...}^{\infty}\left(\frac{k+1}{k}\right)^{(-1)^{\frac{k}{2}}}$$

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

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

$$\gamma+ln\left(\frac{4}{\pi}\right)=2\sum_{n=2}^{\infty}\frac{(-1)^n\lambda(n)}{n}\cdots(-6)$$

$$\frac{\Gamma^2(\frac{1}{4})}{2\pi^{\frac{3}{2}}}=\frac{3^2}{3^2-1}\cdot\frac{7^2}{7^2-1}\cdot\frac{11^2}{11^2-1}\cdots$$

$$\frac{\Gamma^2(\frac{1}{4})}{8\pi^{\frac{1}{2}}}=\frac{5^2-1}{5^2}\cdot\frac{9^2-1}{9^2}\cdot\frac{13^2-1}{13^2}\cdots$$

$$\frac{2\Gamma^2(\frac{1}{4})}{\pi^{\frac{5}{2}}}=\left(\frac{3^2}{3^2-1}\right)^2\cdot\frac{5^2}{5^2-1}\cdot\left(\frac{7^2}{7^2-1}\right)^2 \cdot\frac{9^2}{9^2-1}\cdots\left(\frac{11^2}{11^2-1}\right)^2\cdots$$

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

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

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

Jacobi theta function

$$ln\left[J_{3}(0,n^{-1})\right]=2\sum_{k=0}^{\infty}\frac{1}{(2k+1)(n^{2k+1}+1)}$$

$$ln\left[J_{4}(0,n^{-1})\right]=-2\sum_{k=0}^{\infty}\frac{1}{(2k+1)(n^{2k+1}-1)}$$

$$ln\frac{J_{2}(0,n^{-1})}{J_{1}^{'}(0,n^{-1})}=2\sum_{k=0}^{\infty}\frac{1}{2k+1}\left(\frac{1}{n^{2k+1}-1}-\frac{1}{n^{2k+1}+1}\right)$$

$$\frac{1}{J_{3}(0,n^{-1})\cdot{J_{4}(0,n^{-1})}}=\frac{J_{2}(0,n^{-1})}{J_{1}^{'}(0,n^{-1})}$$

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

$$-2\sum_{k=0}^{\infty}\frac{1}{(2k+1)(n^{2k+1}-1)}=ln\left[1+2\sum_{k=1}^{\infty}\frac{(-1)^k}{n^{k^2}}\right]$$

$$ln\sqrt{\prod_{r=1}^{\infty}\frac{n^r+1}{n^r-1}}=\sum_{k=0}^{\infty}\frac{1}{(2k+1)(n^{2k+1}-1)}$$

$$\prod_{p=2}^{\infty}\frac{p^4+1}{p^4-1}=\frac{7}{6}$$

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

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

$$ln\sqrt{\frac{\pi}{asin\left(\frac{\pi}{a}\right)}}=\sum_{k=0}^{\infty}\frac{1}{2k+1}\left[\sum_{r=1}^{\infty}\frac{1}{(2ar-1)^{2k+1}}-\sum_{r=1}^{\infty}\frac{1}{(2ar+1)^{2k+1}}\right]$$

$$ln\left(cos\sqrt{\frac{\pi}{a}}\right)=\sum_{k=0}^{\infty}\frac{1}{2k+1}\left[\sum_{r=0}^{\infty}\frac{1} {[\left(2r+1)a+1\right]^{2k+1}}-\sum_{r=0}^{\infty}\frac{1}{[\left(2r+1)-1\right]^{(2k+1)a}}\right] $$

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

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

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

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

$$ln\sqrt{\prod_{n=2}^{\infty}\frac{n^r+1}{n^r-1}}=\sum_{k=0}^{\infty}\frac{\zeta(2rk+r)-1}{2k+1}$$

Using identity (1) to derive other

$$\eta(k)+\zeta(k)=2\lambda(k)\cdots(1)$$

$$ln\left(\frac{n+2}{n}\right)=\frac{2}{n+1}\cdot\sum_{k=0}^{\infty}\frac{1}{(n+1)^{2k}\cdot(2k+1)}$$

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

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

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

$$\pi=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{4(3n)^2-1}\right)}{\sum_{n=1}^{\infty}\frac{1}{4^n} }=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{4(3n)^2-1}\right)}{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^n}  }=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{4(3n)^2-1}\right)}{\sum_{n=1}^{\infty}\frac{(ln3)^{2n}}{2(2n)!}  }$$

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

$$\sum_{k=1}^{\infty}\frac{(4^k-1)\zeta(2k)\left(4^k\left[\zeta(2k)-1\right]-1\right)}{4^kk}=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{r=1}^{\infty}\sum_{a=3}^{\infty}\left[\frac{1}{(2ar-a-1)^{2n}}+\frac{2}{(2ar-a)^{2n}}+\frac{1}{(2ar-a+1)^{2n}}\right]$$

Simplify version

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

$$lnK=\sum_{k=1}^{\infty}\frac{\zeta(2k)\left[(2^{2k}-1)\zeta(2k)-2^{2k}\right]}{k}+ln\left(\frac{\pi}{2}\right)= \sum_{n=1}^{\infty}\frac{1}{n}\sum_{r=1}^{\infty}\sum_{a=3}^{\infty}\left[\frac{1}{(2ar-a-1)^{2n}} +\frac{2}{(2ar-a)^{2n}}+\frac{1}{(2ar-a+1)^{2n}}\right]\cdots(17)$$

$$ln\left(\frac{2K}{\pi}\right)=\sum_{k=1}^{\infty}\frac{\zeta(2k)\left[(2^{2k}-1)\zeta(2k)-2^{2k}\right]}{k}$$

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

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

pi and gamma

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

pi and phi

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

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

$$ln\left(\frac{3\pi}{5\phi}\right)=\sum_{n=1}^{\infty}\frac{n}\left(\frac{3}{10}\right)^{2n}$$

$$ln\left(\frac{4\pi}{5\sqrt{\phi+2}}\right)=\sum_{n=1}^{\infty}\frac{n}\left(\frac{4}{10}\right)^{2n}$$

Logarithms

$$ln\left(\frac{16}{\pi^3}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\left[{\zeta(2n)}\left(1-\frac{3}{2^{2n}}\right)-1\right]$$

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

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

r > 0

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

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

r > 1

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

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

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

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

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

$$ln2=\sum_{n=1}^{\infty}\frac{\zeta(2n)-1}{n}$$

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

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

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

b > 1

$$ln\left(\frac{b^{b+1}a!}{a^{a+1}b!}\right)={b-a}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)}{\left[ \frac{1}{(b-1)^n}+\frac{1}{(b-2)^n}+\cdots+\frac{1}{a^n}\right]}$$

$$ln2=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{r=1}^{\infty}\left[\frac{1}{(6r-4)^{2n}}+\frac{2}{(6r-3)^{2n}}+\frac{1}{(6r-2)^{2n}}\right]$$

$$ln6=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{r=2}^{\infty}\left[\frac{1}{(r)^{2n}}+\frac{2}{(r+1)^{2n}}+\frac{1}{(r+2)^{2n}}\right]$$

a > 2

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

$$ln\left[\frac{120\pi}{sinh(2\pi)}\right]=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=3}^{\infty}\left[\frac{1}{(m^2-3)^{2n}}+\frac{2}{(m^2-2)^{2n}}+\frac{3}{(m^2-1)^{2n}}+\frac{4}{(m^2)^{2n}}+\frac{3}{(m^2+1)^{2n}}+\frac{2}{(m^2+2)^{2n}}+\frac{1}{(m^2+3)^{2n}}\right]$$

General pi

$$\frac{(2a+b){\pi}}{8}=a-\frac{a-b}{3}+\frac{a-2b}{5}-\frac{a-3b}{7}+\cdots= \sum_{r=0}^{\infty}{\frac{a-br}{2r+1}(-1)^{r}}$$

a = 1 and b = 0

$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots= \sum_{r=0}^{\infty}{\frac{1}{2r+1}(-1)^{r}}$$

General

$$ln\left(\frac{n+d}{n}\right)=\sum_{m=1}^{\infty}\frac{1}{m}\left[{d\zeta(2m)}-d\sum_{r=1}^{n}\frac{1}{r^{2m}}-\sum_{q=1}^{d-1}\frac{d-q}{(q+n)^{2m}}\right]$$

$$ln\left(\frac{n^s}{n^s-1}\right)=\sum_{r=1}^{\infty}\left(\frac{1}{n^{sr}r}\right)$$

$$ln\left(\frac{2^d}{1+d}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{q=1}^{d-1}\frac{d-q}{(q+1)^{2n}}$$

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

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

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

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

a > 1

$$ln\left[\frac{\pi}{asin\left(\frac{\pi}{a}\right)}\right]=\sum_{n=1}^{\infty}\frac{a^{2n}n}$$

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

$$ln\left(\frac{12\pi^{2}}{1+cosh(\pi\sqrt3)}\right)=\sum_{n=1}^{\infty}\frac{\zeta(6n)-1}{n}$$

Twin primes constant

$$ln\left(\frac{1}{C_{2}}\right)=\sum_{n=1}^{\infty}\frac{1}{2^{2n}n}\sum_{p=3}^{\infty}\left(\frac{2}{p-1}\right)^{2n}$$

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

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

Special conditions (k=0)

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

$$ln\left(\frac{4}{\pi}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{3^{2n}}+ \frac{1}{5^{2n}}+\frac{1}{7^{2n}}+\cdots\right)=\sum_{n=1}^{\infty}\frac{1}{n}\left[\sum_{k=1}^{\infty}\frac{1}{(2k+1)^{2n}}\right]= \sum_{n=1}^{\infty}\frac{\lambda(2n)-1}{n}$$

Arctan relating to Pi, Phi and Fibonacci

$$\frac{5\pi}{60}=arctan\left(\frac{1}{\sqrt{3}+4}\right)+arctan\left(\frac{1}{5\sqrt{3}+10}\right) +arctan\left(\frac{1}{7\sqrt{3}+16}\right)$$

$$\frac{\pi}{4}=arctan\left(\frac{1}{\sqrt{3}+1}\right)+arctan\left(\frac{1}{3\sqrt{3}+6}\right) +2arctan\left(\frac{1}{\sqrt{3}+4}\right)\cdots(1)$$

$$\frac{\pi}{4}=arctan\left(\frac{1}{nF_{2p-1}+1}\right) +\sum_{r=1}^narctan\left[{\frac{1}{r^2F_{2p-1}+r(2-F_{2p-1}) +\frac{2-F_{2p-1}}{F_{2p-1}}}}\right]$$

$$\frac{\pi}{4}=arctan\left(\frac{1}{2\cdot1^2}\right)+arctan\left(\frac{1}{2\cdot2^2}\right)+arctan\left(\frac{1}{2\cdot3^2}\right)+\cdots+\arctan\left(\frac{1}{2r^2}\right)+ arctan\left(\frac{1}{2r+1}\right)\cdots(1)$$

$$\frac{\pi}{4}=\sum_{r=1}^{\infty}arctan\left({\frac{1}{2r^2}}\right)\cdots(24)$$

$$\frac{\pi}{4}=\sum_{r=1}^{\infty}arctan\left({\frac{1}{r^2+r+1}}\right)\cdots(25)$$

Special conditions r = 1  yield,

$$\frac{\pi}{4}=arctan\left(\frac{1}{2}\right)+arctan\left(\frac{1}{3}\right)$$which is Euler formula r = 2 gives,

$$\frac{\pi}{4}=arctan\left(\frac{1}{2}\right)+arctan\left(\frac{1}{8}\right)+arctan\left(\frac{1}{5}\right)$$which is Daze formula

$$\frac{\pi}{2}= arctan\left(\frac{2}{1^2}\right)+arctan\left(\frac{2}{3^2}\right)+arctan\left(\frac{2}{5^2}\right) +arctan\left(\frac{2}{7^2}\right)+\cdots(2)$$

$$arctan\left(\frac{1}{xF_{2n-1}+F_{2n}}\right)= arctan\left(\frac{1}{{x}F_{2n-1}+F_{2n+1}}\right)+arctan\left(\frac{1}{{x^2}F_{2n-1}+{x}F_{2n+2}+F_{2n+2}}\right)\cdots(3) $$

$$arctan\left(\frac{1}{\phi}\right)=arctan\left(\frac{1}{2}\right)+arctan\left(\frac{1}{13}\right)+arctan\left(\frac{1}{89}\right)+arctan\left(\frac{1}{610}\right)+\cdots(4)$$

$$\frac{\pi}{4}=arctan\left(\frac{1}{2}\right)+arctan\left(\frac{1}{2}\right)-arctan\left(\frac{1}{5}\right)+arctan\left(\frac{1}{13}\right)-arctan\left(\frac{1}{34}\right)+\cdots(5)$$

$$\frac{\pi}{4}=narctan(\frac{1}{A})+arctan(\frac{{n\choose n}{A}^n-{n\choose n-1}{A^{n-1}}-{n\choose n-2}{A^{n-2}}+{n\choose n-3}{A^{n-3}}+ \cdots}{{n\choose n}{A^n}+{n\choose n-1}{A^{n-1}}-{n\choose n-2}{A^{n-2}}-{n\choose n-3}{A^{n-3}}+\cdots})\cdots(6)$$

$$\frac{\pi}{4}=arctan\left(\frac{1}{a}\right)+arctan\left(\frac{a-1}{a+1}\right)\cdots(7)$$

$$\frac{\pi}{4}=arctan\left(\frac{1}{a}\right)+arctan\left(\frac{1}{b}\right)+arctan\left(\frac{ab-(a+b)-1}{ab+(a+b)-1}\right)\cdots(8)$$

$$\frac{\pi}{4}=arctan\left(\frac{1}{a}\right)+arctan\left(\frac{1}{b}\right)+arctan\left(\frac{1}{c}\right)+ arctan\left(\frac{abc-(ab+ac+bc)-(a+b+c)+1}{abc+(ab+ac+bc)-(a+b+c)-1}\right)\cdots(9)$$

$$\frac{n\pi}{4}+arctan\left(\frac{{n\choose n-1}{\phi}^{4k(n-1)}-{n\choose n-3}{\phi^{4k(n-3)}}+{n\choose n-5}{\phi^{4k(n-5)}}-{n\choose n-7}{\phi^{4k(n-7)}}+ \cdots}{{n\choose n}{\phi^{4kn}}-{n\choose n-2}{\phi^{4k(n-2)}}+{n\choose n-4}{\phi^{4k(n-4)}}-{n\choose n-6}{\phi^{4k(n-6)}}+\cdots}\right)=narctan\left(\frac{L_{2k}}{F_{2k}\sqrt{5}}\right)\cdots(10)$$

$$\frac{n\pi}{4}+arctan\left(\frac{{n\choose n-1}{\phi}^{(4k-2)(n-1)}-{n\choose n-3}{\phi^{(4k-2)(n-3)}}+{n\choose n-5}{\phi^{(4k-2)(n-5)}}-{n\choose n-7}{\phi^{(4k-2)(n-7)}}+ \cdots}{{n\choose n}{\phi^{(4k-2)n}}-{n\choose n-2}{\phi^{(4k-2)(n-2)}}+{n\choose n-4}{\phi^{(4k-2)(n-4)}}-{n\choose n-6}{\phi^{(4k-2)(n-6)}}+\cdots}\right)=narctan\left(\frac{F_{2k-1}\sqrt{5}}{L_{2k-1}}\right)\cdots(11)$$

$$\frac{\pi}{4}-arctan\left(\frac{1}{\phi}\right)=arctan\left(\frac{1}{3}\right)-arctan\left(\frac{1}{8}\right)+arctan\left(\frac{1}{21}\right)- arctan\left(\frac{1}{55}\right)+\cdots(12)$$

$$\frac{\pi}{4}-arctan\left(\frac{1}{\phi}\right)=arctan\left(\frac{1}{7}\right)+arctan\left(\frac{1}{18}\right)+arctan\left(\frac{1}{47}\right)+ arctan\left(\frac{1}{123}\right)+\cdots(13)$$

$$arctan\left(\frac{1}{F_{6n+5}}\right)+arctan\left(\frac{1}{F_{6n+8}}\right)= arctan\left(\frac{1}{4L_{6n}+L_{6(n-2)}+L_{6(n-4)}+\cdots+L_{2}}\right)\cdots(14)$$ Where n is even

$$arctan\left(\frac{1}{F_{6n+5}}\right)+arctan\left(\frac{1}{F_{6n+8}}\right)= arctan\left(\frac{1}{4L_{6(n-1)}+L_{6(n-3)}+L_{6(n-5)}+\cdots+L_{6}}\right)\cdots(15)$$

Where n is 0dd

$$arctan\left(\frac{1}{\phi^n}\right)=arctan\left(\frac{1}{L_{n+1}}\right)+arctan\left(\frac{1}{\phi^{n+2}}\right)\cdots(16)$$

$$arctan\left(\frac{1}{\phi^n}\right)+arctan\left(\frac{1}{\phi^{n-2}}\right)=arctan\left(\frac{\sqrt{5}}{L_{n}}\right)\cdots(16a)$$

$$arctan\left(\frac{1}{\phi^{2n}}\right)+arctan\left(\frac{1}{\phi^{6n}}\right)=arctan\left(\frac{1}{F_{2n}\sqrt{5}}\right)\cdots(17)$$

$$arctan\left(\frac{1}{F_{2n}}\right)=arctan\left(\frac{1}{\phi^{2n-1}}\right)+arctan\left(\frac{1}{\phi^{2n+1}}\right)\cdots(18)$$

$$arctan\left(\frac{1}{F_{2r+1}}\right)=\sum_{n=1}^{\infty}arctan\left({\frac{1}{n^2F_{2r-1}+nF_{2r+2}+F_{2r+2}}}\right)\cdots(19)$$

General

$$arctan\left[\frac{1}{kF_{2p-1}+zF_{2p-1}+F_{2p}}\right]=arctan\left[\frac{1}{(k+n)F_{2p-1}+zF_{2p-1}+F_{2p}}\right]+\sum_{r=1}^n arctan\left[{\frac{1}{(r+k+z)^2F_{2p-1}+(k+r+z)L_{2p}+\frac{F_{2p-2}F_{2p}+1}{F_{2p-1}}}}\right]\cdots(23)$$

$$arctan\left(\frac{1}{kF_{2p-1}+L_{2p-1}}\right)=arctan\left(\frac{1}{(k+n)F_{2p-1}+L_{2p-1}}\right)+ \sum_{r=1}^narctan\left({\frac{1}{(r+k)^2F_{2p-1}+\frac{1}{2}(r+k)(L_{2p+1}-F_{2p-6})+L_{2p-3}}}\right)\cdots(21)$$

$$arctan\left(\frac{1}{kF_{2p-1}-F_{2p-3}}\right)=arctan\left(\frac{1}{(k+n)F_{2p-1}-F_{2p-3}}\right)+ \sum_{r=1}^narctan\left({\frac{1}{(r+k)^2F_{2p-1}-\frac{1}{2}(r+k)(L_{2p+1}-F_{2p-6})+L_{2p-3}}}\right)\cdots(22)$$

Fibonacci Identities

$$L_{n+2}F_{n+2}^2+L_{n+3}F_{n+3}^2=L_{3n+6}+L_{n+1}F_{n+1}^2\cdots(58)$$(F1=1,L1=1)

$$F_{n+3}^2+F_{n+4}^2=L_{2n+5}+F_{n}F_{n+1}+F_{n+3}F_{n+4}\cdots(57)$$

$$F_{n}F_{n+1}+F_{n+2}^2=F_{2n+2}+F_{n+1}^2+(-1)^{n+1}\cdots(56)$$

$$F_{n}F_{n+1}+2F_{n+2}^2=F_{2n+4}+(-1)^{n+1}\cdots(55)$$

$$F_{n}F_{n+1}+F_{n+2}^2=F_{2n+3}+F_{n}^2-1\cdots(54)$$

$$4L_{n}L_{n+1}+L_{n-1}^2=L_{n+2}^2\cdots(53)$$

$$4F_{n}F_{n+1}+F_{n-1}^2=F_{n+2}^2\cdots(52)$$

$$F_{2n+3}+1=F_{n}^2+F_{n+2}^2+F_{n}F_{n+1}\cdots(51)$$--Where n is odd

$$F_{2n+3}+1=F_{n}^2+F_{n+2}^2+F_{n}F_{n+1}+2\cdots(50)$$--Where n is even

$$F_{n+2}F_{n+6}-F_{n}F_{n+4}=F_{2n+6}\cdots(49)$$

$$F_{n+2}F_{n+3}-F_{n}F_{n+1}=F_{2n+3}\cdots(48)$$

$$F_{2n+1}^2+F_{2n+3}^2=F_{2n+2}(F_{2n}+F_{2n+4})+2\cdots(47)$$

$$F_{2n}^2+F_{2n+2}^2+2=F_{2n+1}(F_{2n-1}+F_{2n+3})\cdots(46)$$

$$F_{n}^3+F_{n+k}^3=(F_{n}+F_{n+k})^2+(F_{n}+F_{n+k})(F_{n+k}-F_{n})^2\cdots(45)$$

$$F_{n+3}F_{n+4}+2F_{n}F_{n+2}=L_{n+3}L_{n+4}+L_{2n+7}\cdots(44)$$

$$5F_{2n+6}=F_{n+5}F_{n+6}+F_{n}F_{n+1}\cdots(43)$$

$$F_{n+5}F_{n+6}+F_{n+8}^2=L_{2n+14}-5F_{2n+6}+F_{n}F_{n+1}\cdots(42)$$

$${F_{n+1}+4F_{n+2}+9F_{n+3}+16F_{n+4}+25F_{n+5}+36F_{n+6}}= F_{n+13}+F_{n+10}+F_{n+8}+F_{n+5}+F_{n+3}+F_{n}\cdots(41)$$

$$F_{n+1}L_{n+2}+F_{n+2}^2+F_{n+3}^2=2F_{2n+4}\cdots(40)$$

$$F_{2n}^2=F_{2n-1}F_{2n}+F_{2n-2}^2-1\cdots(39)$$

$$F_{2n+1}^2=F_{2n}F_{2n+1}+F_{2n}^2+1\cdots(38)$$

$$F_{n+4}^2F_{n+3}+F_{n+1}F_{n}^2=L_{3n+7}\cdots(37)$$

$$F_{n}^5+F_{n+1}^5=F_{n+2}[(F_{n}F_{n+1}+F_{n-1}^2)^2+F_{n-1}^2F_{n}F_{n+1}]\cdots(36)$$

$$F_{n}^5+L_{n}^5=(F_{n}+L_{n})[[F_{n}L_{n}+(L_{n}-F_{n})^2]^2+(L_{n}-F_{n})^2F_{n}L_{n}]\cdots(35)$$

$$\frac{F_{n-1}^3+L_{n+1}^3}{F_{n+1}+L_{n+1}}=4F_{2n-1}-F_{n}L_{n}-F_{n-2}^2\cdots(34)$$

$$F_{n}^3+L_{n}^3=(F_{n}+L_{n})(F_{n-2}^2+F_{n}L_{n})\cdots(33)$$

$$F_{n+2}^4-F_{n+1}^4=F_{n}F_{n+3}F_{2n+3}\cdots(32)$$

$$F_{n+3}^3-F_{n+2}^3=F_{n+1}(F_{n}F_{n+1}+L_{3n})\cdots(31)$$

$$F_{n+4}^2-F_{n}^2=3F_{n+2}L_{n+3}\cdots(29)$$

$$F_{n+3}^2-F_{n}^2=4F_{n+1}F_{n+2}\cdots(28)$$

$$F_{n+2}^2-F_{n}^2=F_{n+1}L_{n+2}\cdots(27)$$

$$F_{n}^2+F_{n+1}^2+L_{n}^2=4L_{2n-1}+2(-1)^{n+1}\cdots(26)$$

$$F_{n}^2+F_{n+1}^2=2L_{n-1}F_{n}+L_{n}^2\cdots(25)$$

$$2(F_{n}^2+F_{n+1}^2)-(L_{n}^2+L_{n+1}^2)=F_{2n-4}\cdots(24)$$

$$\sqrt{F_{2n+1}F_{2n+2}}=\sqrt{F_{2n}F_{2n+3}}+\frac{1}{2\sqrt{F_{2n}F_{2n+3}}}\cdots(23)$$

$$F_{n+1}(L_{n+3})+L_{n+1}(F_{n+3})=F_{2n-2}+F_{2n+4}\cdots(22)$$

$$L_{n}^3+L_{n+1}^3=F_{n+2}(L_{n+1}^2+L_{2n-5})+(-1)^{n+1}(L_{n+1}+F_{n-7})\cdots(21)$$

$$L_{n}^4+L_{n+1}^4+L_{n+2}^4=2(L_{n}L_{n+2}+L_{n}^2)^2\cdots(20)$$

$$L_{n-2k}L_{n+2k+1}-L_{n-2k-1}L_{n+2k+2}=5(-1)^{k+1}F_{2k+2}\cdots(19)$$

$$L_{n}L_{n+1}^2=5L_{n}+L_{n-1}L_{n}^2+2L_{n}^3\cdots(18)$$

$$L_{2n+2}^2-L_{2n}^2=5F_{4n}^2\cdots(17)$$

$$L_{2n+3}^2-L_{2n+1}^2=5F_{4n+2}^2\cdots(16)$$

$$L_{n}^2+L_{n+1}^2=F_{n+2}^2+F_{n-3}^2\cdots(15)$$

$$\frac{F_{n+1}^3+F_{n+2}^3}{F_{n+3}}+\frac{F_{n+2}^3+F_{n+3}^3}{F_{n+4}}=\frac{F_{n}^3+F_{n+3}^3}{2F_{n+2}}+F_{n}F_{n+3}\cdots(14)$$

$$\frac{F_{2n+1}^2+F_{2n+3}^2}{F_{2n+2}}+\frac{F_{2n+2}^2+F_{2n+4}^2}{F_{2n+3}}-\frac{F_{2n+3}^2+F_{2n+5}^2}{F_{2n+4}}+1 =\frac{2}{F_{2n+2}}+\frac{{F_{2n+3}}-2}{F_{2n+3}}-\frac{2}{F_{2n+4}}\cdots(13)$$

$$\frac{F_{2n}^2+F_{2n+2}^2}{F_{2n+1}}+\frac{F_{2n+1}^2+F_{2n+3}^2}{F_{2n+2}}-\frac{F_{2n+2}^2+F_{2n+4}^2}{F_{2n+3}} =\frac{F_{2n+1}-2}{F_{2n+1}}+\frac{2}{F_{2n+2}}-\frac{F_{2n+3}-2}{F_{2n+3}}\cdots(12)$$

$$F_{n+2}(F_{n+1}+F_{n+5})=F_{n+3}(F_{n}+F_{n+4})\cdots(11)$$

$$F_{2n-1}F_{2n+2m-1}=F_{m}^2+F_{2n+m-1}^2\cdots(9)$$

$$L_{2n-1}^2=\frac{1+\phi^{12n-6}}{\phi^{4n-2}+\phi^{8n-4}}-1\cdots(8)$$

$$L_{2n}^2=\frac{1+\phi^{12n}}{\phi^{4n}+\phi^{8n}}+3\cdots(7)$$

$$arctan(\phi)=arccos\left(\frac{1}\sqrt\right)\cdots(6)$$

$$\pi=2arctan\left(\frac{1}{\phi}\right)+arctan\left(\frac{1}{2}\right)\cdots(5)$$

$$\frac{\pi}{2}=arctan\left(\frac{G_{k,n}}{G_{k,n+1}}\right)+arctan\left(\frac{G_{k,n+1}^2}{2G_{k,n+1}^2+[1-k(k-1)](-1)^n}\right)+arctan\left(\frac{G_{k,n+1}}{G_{k,n+2}}\right)\cdots(4)\,$$

$$ G_{k,n}= \frac{1}{\sqrt{5}}\left[\frac{1+\sqrt{5}}{2}\left(\frac{\sqrt{5}+2k+1}{3+\sqrt{5}}\right)+\frac{1-\sqrt{5}}{2}\left(\frac{\sqrt{5}-2k-1}{3+\sqrt{5}}\right)\right]\cdots(3)$$

$$ F_{n}F_{n+r}-F_{n+r-2}F_{n-2}=(-1)^nF_{r}\cdots(2)$$

$$ 3(-1)^n[F_{3n}-L_{n+1}^2F_{n}] =F_{3n-1}-L_{n}^2F_{n+1}\cdots(1)\,$$

$$L_{2n}L_{2n-1}+5F_{2n}F_{2n-1}=\frac{2\sqrt{5}({\phi^{8n-2}-1})}{{\phi^{4n}}+{\phi^{4n-2}}}\cdots(0)$$

Cube sum Of Fibonacci and Lucas series

Fibonacci series

$$S_{F}=1^3+1^3+2^3+3^3+5^3+8^3+...+F_{n+k-1}^3$$

Fibonacci cube sum formula

$$S_{F}=\frac{F_{n+k-1}F_{n+k}^2-F_{n-1}F_{n}^2+(-1)^{n+k}[F_{n+k-2}+(-1)^{k+1}F_{n-2}]}{2}\cdots(1)$$

Lucas series

$$S_{L}=2^3+1^3+3^3+4^3+7^3+11^3+...+L_{n+k-1}^3$$

Lucas cube sum formula

$$S_{L}=\frac{L_{n+k-1}L_{n+k}^2-L_{n-1}L_{n}^2+5(-1)^{n+k}[L_{n+k-2}+(-1)^{k+1}L_{n-2}]}{2}\cdots(2)$$

$$S_{L}=2^3+1^3+3^3+4^3+7^3\cdots+L_{n}$$

$$S_{L}=L_{n}^2L_{n+1}-L_2L_{3}L_{4}-L_3L_{4}L_{5}-L_4L_{5}L_{6}-\cdots-L_{n-2}L_{n-1}L_n\cdots(3)$$

$$S_{F_{m,n}}={1+1+2+3+5+8+...+F_{m}+F_{n}}$$

$$S_{F_{m,n}}=\left(\frac{\sqrt{5}+5}{10}\right)\left[\left(\frac{1+\sqrt{5}}{2}\right)^{m+1}-\left(\frac{1+\sqrt{5}}{2}\right)^{n}\right] -\left(\frac{\sqrt{5}-5}{10}\right)\left[\left(\frac{1-\sqrt{5}}{2}\right)^{m+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]\cdots(4)$$

Pythagoras Fibonacci

$${(F_{2n}F_{2n+3}+F_{2n+1}F_{2n+2}-1)}^2+[F_{2n+2}(F_{2n}F_{2n+1}-2)]^2= [F_{2n+1}(F_{2n}F_{2n+2}+2)]^2\cdots(1)$$

$${(F_{n}F_{n+k}-1)}^2+(F_{n}+F_{n+k})^2=(F_{n}F_{n+k}+1)^2\cdots(2)$$

$${(F_{2n-1}F_{2n}F_{2n+1}-2F_{2n+1})}^2+(F_{2n-1}F_{2n}+F_{2n+1}^2-1)^2=2^2+ (F_{2n-1}F_{2n}F_{2n+1}+2F_{2n})^2\cdots(3)$$

Fibonacci and Phi

$$\frac{F_{n-1}+1}{F_{n-1}-1}=\left(\frac{\phi^n+1}{\phi^n-1}\right)\left(\frac{{\phi}^{n-2}+1}{{\phi}^{n-2}-1}\right)\cdots(1)$$

Summation and approximation

$$\sum_{r=1}^{2k}L_{m+r-2}F_{n+r-1}=\sum_{r=1}^kL_{m+n+4r-4}\cdots(1)$$

$$\sum_{k=0}^m[F_{n}+(m-k)F_{n-2}]^2{m-1\choose k}(-1)^k=0\cdots(2)$$

$$\lim_{n \to \infty}\frac{\phi}{\sqrt{5}}=\frac{F_{n+1}^{n+1}+F_{n}^n}{F_{n+2}^n}\cdots(3)$$

$$\lim_{n \to \infty}{F_{2n+1}{\sqrt{}F_{2n+1}}}=L_{3n}\cdots(4)$$

$$\lim_{n \to \infty}\frac{F_{n}^2}{L_{n}}=5F_{n-4}\cdots(5)$$

$$\lim_{n \to \infty}1^4+1^4+2^4+3^4+5^4+\cdots+F_{n}^4=\frac{F_{n}^3F_{n+1}^2}{L_{n+1}}\cdots(6)$$

$$\lim_{n \to \infty}\frac{F_{2n}^2+F_{2n+2}^2}{F_{2n-2}^2+F_{2n}^2}={\phi^4}\cdots(8)$$

$$\frac{34}{747}\approx\frac{2^{2\phi}}{3^{3\phi}}\cdots(9)$$

$$\lim_{n \to \infty}\sqrt{F_{n}F_{n+1}F_{n+2}F_{n+3}}-F_{n}F_{n+3}=\frac{(-1)^n}{2}\cdots(10)$$

$$F_{n}\approx{\phi^{n-\sqrt[32]{14028000}}}\cdots(11)$$

$$\sum_{r=1}^n{F_{r}^4}\approx{11F_{2n-4}^2}\cdots(13)$$

'Other Formulae

$$\sum_{k=0}^n[a+(n-k)d]^p{n-1\choose k}(-1)^k=0\cdots(1)$$

Where n≥p+2

$$\frac{\pi}{4}=arctan\left(3+2\sqrt{2}\right)-arctan\left(\frac{1}{\sqrt{2}}\right)\cdots(2)$$

$$\frac{\pi}{6}=arctan\left(\frac{2\sqrt{3}-1}\right)-arctan\left(\frac{1}{2}\right)\cdots(3)$$

$$\frac{4e^{\frac{2}{\phi}}}{(e^{\frac{2}{\phi}}+1)^2} =1-\frac{1}{\phi^2}+\frac{2}{3\phi^4}-\frac{17}{45\phi^6}+\frac{62}{315\phi^8}-\cdots(4)$$

Phi relating to Fibonacci Lucas numbers

$$\frac=\frac{L_{n+2}}{L_{n+1}}\cdots(1)$$

Where n is odd

$$\frac=\frac{F_{n+1}}{F_{n}}\cdots(2)$$

Where is even

$$\frac{1+{\phi^{8n}}}{{\phi}^{2n}+{\phi}^{6n}}=\frac{L_{2n}^2-2}{L_{2n}}\cdots(3)$$

$$\frac{1+{\phi^{10n}}}{{\phi}^{2n}+{\phi}^{8n}}=\frac{L_{2n}^2-L_{2n}-1}{L_{2n}-1}\cdots(4)$$

$$\frac{1+{\phi^{12n}}}{{\phi}^{2n}+{\phi}^{10n}}=\frac{L_{2n}(L_{4n}-1)}{L_{4n}}\cdots(5)$$

$$\frac{1+{\phi^{8n}}}{{\phi}^{2n}+{\phi}^{4n}+{\phi^{6n}}}=\frac{L_{2n}^2-2}{L_{2n}+1}\cdots(6)$$

$$\frac{1+{\phi^{10n}}}{{\phi}^{2n}+{\phi}^{4n}+{\phi^{6n}}+{\phi^{8n}}}=\frac{L_{2n+1}^2-L_{2n+1}-1}{L_{2n+1}}\cdots(7)$$

Pi and Arctan series involve Fibonacci and Lucas numbers

$$\frac{\pi}{4}=arctan(\frac{1}{2})+arctan(\frac{1}{18})+arctan(\frac{1}{34})+arctan(\frac{1}{322})+arctan(\frac{1}{610})+\cdots$$

$$+arctan(\frac{4}{18})+arctan(\frac{4}{322})+arctan(\frac{4}{5778})+arctan(\frac{4}{103682})+\cdots(1)$$

$$\frac{\pi}{6}=arctan(\frac{1}{5})+arctan(\frac{1}{8})+arctan(\frac{1}{13})+arctan(\frac{1}{21})\cdots(2)$$

$$\frac{\pi}{4}=arctan(\frac{15L_{6}}{L_{6}^2-44})+arctan(\frac{5L_{12}}{L_{12}^2-4})+ arctan(\frac{15L_{18}}{L_{18}^2-44})+arctan(\frac{5L_{24}}{L_{24}^2-4})+\cdots(3)$$

$$\frac{\pi}{4}=arctan(\frac{3\sqrt{5}}{7})+arctan(\frac{3\sqrt{5}}{322}) arctan(\frac{3\sqrt{5}}{15177})+\cdots+arctan(\frac{3\sqrt{5}}{L_{n+8}})\cdots(4)$$

$$\frac{\pi}{4}=arctan(\frac{3}{\sqrt{5}})+arctan(\frac{3}{8\sqrt{5}}) arctan(\frac{3}{55\sqrt{5}})+arctan(\frac{3}{377\sqrt{5}})\cdots(5)$$

$$\frac{\pi}{4}=arctan(\frac{2\sqrt{5}}{4})-arctan(\frac{2\sqrt{5}}{76})+arctan(\frac{2\sqrt{5}}{1364})-\cdots- arctan(\frac{2\sqrt{5}}{L_{6n-3}})\cdots(6)$$

$$\frac{\pi}{4}=arctan(\frac{1}{2})+arctan(\frac{1}{4})+arctan(\frac{1}{24})+arctan(\frac{1}{29}) arctan(\frac{1}{1515})+arctan(\frac{1}{6884163})\cdots(1)$$

$$\frac{\pi}{4}=arctan(\frac{1}{2})+2arctan(\frac{1}{6})-arctan(\frac{1}{117})\cdots(2)$$

$$\frac{\pi}{4}=arctan(\frac{1}{2})+arctan(\frac{1}{4})+arctan(\frac{1}{16}) +arctan(\frac{1}{70})+arctan(\frac{1}{14633})\cdots(3)$$

$$\frac{\pi}{4}=arctan(\frac{1}{2})+arctan(\frac{1}{5})+arctan(\frac{1}{10}) +arctan(\frac{1}{41})+arctan(\frac{1}{3323})\cdots(4)$$

$$\frac{\pi}{4}=arctan(\frac{1}{2})+arctan(\frac{1}{5})+arctan(\frac{1}{12}) +arctan(\frac{1}{25}) +arctan(\frac{1}{810})+arctan(\frac{1}{17493})\cdots(5)$$

$$\frac{\pi}{4}=arctan(\frac{1}{2})+arctan(\frac{1}{4})+arctan(\frac{1}{14}) +arctan(\frac{1}{25})+arctan(\frac{1}{41}) +arctan(\frac{1}{735})+arctan(\frac{1}{1079717})\cdots(6)$$

$$\frac{\pi}{4}=arctan(\frac{1}{2})+arctan(\frac{1}{4})+arctan(\frac{1}{22}) +arctan(\frac{1}{32}) +arctan(\frac{1}{9193})\cdots(7)$$ Bold text