FANDOM


$ \forall k\in\mathbb{Z},\pi=16(2k+1)^4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)((2n+1)^4+4(2k+1)^4)} $

$ \forall j\in\mathbb{N},\pi=\left(\prod_{k=0}^{j}4(2k+1)^4\right)\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)\prod_{k=0}^{j}\left((2n+1)^4+4(2k+1)^4\right)} $

$ \pi=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)\cosh\left(\frac{(2n+1)\pi}{2}\right)} $

$ \frac{7\pi}{180}=\sum_{n=1}^\infty\frac{1}{n^3\tanh (n\pi)} $

$ \forall k\in\mathbb{Z},\pi=-8(2k+1)^2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)((2n+1)^2-4(2k+1)^2)} $

$ \forall k\in\mathbb{Z},\pi=\frac{16}{9}(6k+1)^2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)((2n+1)^2-\frac{4(6k+1)^2}{9})} $

$ \forall k\in\mathbb{Z},\pi=\frac{16}{9}(6k-1)^2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)((2n+1)^2-\frac{4(6k-1)^2}{9})} $

$ \forall j\in\mathbb{N},\pi=-2\left(\prod_{k=0}^{j}4(2k+1)^2\right)\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)\prod_{k=0}^{j}\left((2n+1)^2-4(2k+1)^2\right)} $

$ \pi=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)\cos\left(\frac{(2n+1)\pi}{4}\right)} $

$ \begin{align} &\forall k\in\mathbb{N},\pi=32(2k+1)^2(4k+3) \\ &\cdot\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)((2n+1)^2-4(2k+2)^2)((2n+1)^2-4(2k+1)^2)} \end{align} $

$ \forall k\in\mathbb{Z},\pi=\frac{3(-1)^k}{(2k+1)}-2(2k+1)^2\sum_{n=-\infty,\neq k,-k-1}^\infty\frac{(-1)^n}{(2n+1)((2n+1)^2-4(2k+1)^2)} $

$ \forall k\in\mathbb{Z},\pi=12(2k+1)^3\sum_{n=-\infty}^\infty\frac{(-1)^n}{(2n+1)((2n+1)^3+8(2k+1)^3)} $

$ \forall j\in\mathbb{N},\pi=12\left(\prod_{k=0}^{j}2(2k+1)^3\right)\sum_{n=-\infty}^\infty\frac{(-1)^n}{(2n+1)\prod_{k=0}^{j}\left((2n+1)^3+8(2k+1)^3\right)} $

$ \forall k\in\mathbb{Z},\pi=-384(2k+1)^6\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)((2n+1)^6-128(2k+1)^6)} $

$ \forall j\in\mathbb{N},\pi=-3\left(\prod_{k=0}^{j}2(2k+1)^6\right)\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)\prod_{k=0}^{j}\left((2n+1)^6-128(2k+1)^6\right)} $

$ \frac{8\pi}{5}-\frac{2\pi}{5\cosh\left(\frac{\pi}{2}\right)}=4\sum_{n=-\infty}^\infty\frac{(-1)^{n+1}}{(2n-1)(4n^4+1)} $

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

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

$ \forall k\in\mathbb{N},\pi=(2k+1)\sum_{n=-\infty}^\infty\frac{(-1)^n((2(2k+1)^2-4)n^2+(2k+1)^2+2)}{(4n^2+1)(4n^2-(2k+1)^2)} $

$ \frac{128\pi}{65}-\frac{36\pi}{65\coth\left(\frac{\pi}{2}\right)}=-8\sum_{n=-\infty}^\infty\frac{1}{(4n-1)(4n^4+1)} $

$ \frac{\pi}{\coth\left(\frac{\pi}{2}\right)}=2+4\sum_{n=1}^\infty\frac{1}{4n^4+1} $

$ \pi=-\frac{1}{4}\sum_{n=1}^\infty\frac{9n+14}{(4n^4+1)(4n-1)} $

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

$ \frac{\pi^2}{6}=2+2\sum_{n=1}^\infty\frac{(-1)^n(2n^2-1)}{n^2(4n^4+1)} $

$ \forall k\in\mathbb{Z},\frac{\pi^2}{3(2k+1)^2}-4\frac{\pi}{(2k+1)^3}+\frac{8}{(2k+1)^4}=4\sum_{n=1}^\infty\frac{(-1)^n}{n^2(4n^2-(2k+1)^2)} $

$ \forall k\in\mathbb{R},2\frac{\sin^3(\pi m)}{1+\cos^2(\pi m)}\sum_{n=-\infty}^\infty\frac{(-1)^n}{(n-m)^3} $

$ \forall k\in\mathbb{R},-\frac{\sin^3(\pi m)}{\cos(\pi m)}\sum_{n=-\infty}^\infty\frac{1}{(n-m)^3} $

$ \forall k\in\mathbb{Z},\pi=12(6k+3)^4\sum_{n=-\infty}^\infty\frac{(-1)^n}{(6n+1)((6n+1)^4+4(6k+3)^4)} $

$ \forall j\in\mathbb{N},\pi=3\left(\prod_{k=0}^{j}4(6k+3)^4\right)\sum_{n=-\infty}^\infty\frac{(-1)^n}{(6n+1)\prod_{k=0}^{j}\left((6n+1)^4+4(6k+3)^4\right)} $

$ \pi=6\sum_{n=0}^\infty\frac{(-1)^n}{(6n+1)\left(\cosh\left(\frac{(6n+1)\pi}{6}\right)+\cos\left(\frac{(6n+1)\pi}{6}\right)\right)} $

$ \forall k\in\mathbb{Z},\pi=-8(2k+1)^2\sum_{n=-\infty}^\infty\frac{1}{(4n+1)((4n+1)^2-4(2k+1)^2)} $

$ \forall j\in\mathbb{N},\pi=-2\left(\prod_{k=0}^{j}4(4k+1)^2\right)\sum_{n=-\infty}^\infty\frac{1}{(4n+1)\prod_{k=0}^{j}\left((4n+1)^4-4(4k+1)^2\right)} $

$ \pi=\frac{4}{3}\sum_{n=0}^\infty\frac{1}{(4n+1)\cosh\left(\frac{(4n+1)\pi}{3}\right)} $