FANDOM


$ \sum_{c:odd,c\neq 1}\frac{(-1)^\frac{c+1}{2}}{c^3-c}=\frac{\pi-3}{4} $

$ \sum_{c:odd,c\neq 1}\frac{1}{c^4-c^2}=\frac{10-\pi^2}{8} $

$ \sum_{c:even}\frac{1}{c^4-c^2}=\frac{1}{2}-\frac{\pi^2}{24} $

$ \sum_{c\equiv2(mod4)}\frac{1}{c^4-c^2}=\frac{\pi}{8}-\frac{\pi^2}{32} $

$ \sum_{c\equiv0(mod4)}\frac{1}{c^4-c^2}=\frac{1}{2}-\frac{\pi}{8}-\frac{\pi^2}{96} $

$ \sum_{k=2}^{\infty}\frac{1}{k^4-k^2}=\frac{7}{4}-\frac{\pi^2}{6} $

$ \sum_{k=2}^{\infty}\frac{(-1)^k}{k^4-k^2}=\frac{\pi^2}{12}-\frac{3}{4} $

$ \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k^2-1)^2}=\frac{1}{2}-\frac{\pi}{8} $

$ \sum_{k=1}^{\infty}\frac{4k^2}{(4k^2-1)^3}=\frac{\pi^2}{64} $

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

$ \sum_{k=1}^{\infty}\frac{1}{(2k-1)^2(2k)^2(2k+1)^2}=\frac{5\pi^2}{48}-1 $

$ \sum_{k=2}^{\infty}\frac{1}{k^3-k^2}=2-\frac{\pi^2}{6} $

$ \sum_{k=2}^{\infty}\frac{1}{k^3+k^2}=\frac{\pi^2}{6}-\frac{3}{2} $

$ \sum_{k=0}^{\infty}\frac{1}{(4k+1)^2(4k+3)(4k+5)^2}=\frac{4-\pi}{64} $

$ \sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+3)^2(4k+5)(4k+7)^2(4k+9)}=\frac{16-5\pi}{5760} $

$ \sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)}=\frac{15\pi-44}{2880} $

$ \sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)}=\frac{21\pi-64}{20160} $

$ \sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)(4k+13)}=\frac{315\pi-976}{1814400} $

$ \begin{align} &\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)} \\ &=\frac{45045\pi-141088}{14529715200} \end{align} $

$ \sum_{k=0}^{\infty}\frac{1}{k(k+1)(2k+1)^2}=10-\pi^2 $

$ \sum_{k=0}^{\infty}\frac{(n+1)2^n}{{}_{2n+4}C_{n+1}}=\frac{3}{8}\pi $

$ \sum_{k=0}^{\infty}\frac{n2^n}{{}_{2n}C_{n}}=\frac{\pi}{2} $

$ \sum_{k=0}^{\infty}\frac{2^n}{{}_{2n+1}C_{n}}=\frac{\pi}{2} $

$ \sum_{n=1}^\infty\frac{\prod_{k=1}^{n-1}p(k)^s}{\prod_{k=1}^n(p(k)^s-1)}=\zeta(s)-1 $

$ \sum_{n=1}^\infty\frac{\prod_{k=1}^{n-1}(p(k)^s-1)}{\prod_{k=1}^np(k)^s}=1-\frac{1}{\zeta(s)} $

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

$ \sum_{n=1}^\infty\frac{\prod_{k=1}^{n-1}((2n)^2-1)}{\prod_{k=1}^n(2n)^s}=\frac{2}{\pi} $