FANDOM


$ k_0=0,h_0=2,k_{n+1}=\sqrt{2+k_n},h_{n+1}=\sqrt{2-k_n} $

としたとき、

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

$ x_1=2,x_{n+1}=2^{2n+1}-2^{n+1}\sqrt{4^n-x_n} $

としたとき、

$ \lim_{n\to\infty}\sqrt{x_n+\frac{4}{3}(x_{n+1}-x_n)}=\frac{\pi}{2} $

$ k_1=60,k_{n+1}=\frac{n+1}{4n+6}k_n,l_1=\frac{1}{4},l_{n+1}=l_n+\frac{1}{(n+1)^2}\sum_{m=1}^n\frac{1}{m^2} $

としたとき、

$ \sum_{n=2}^\infty\frac{k_nl_n}{n+1}=\frac{\pi^6}{729} $

$ \sum_{n=1}^\infty \frac{1}{n^3}\left(-\frac{24}{n+1}+\frac{768}{n+2}+\frac{45}{2n+1}-\frac{1215}{2n+3}\right)=2\pi^2 $

$ \sum_{n=1}^\infty \frac{1}{n^3}\left(-\frac{7616}{n+1}+\frac{4516}{2n+1}-\frac{667}{4n+1}-\frac{10206}{4n+3}+\frac{35625}{4n+5}\right)=32\pi $

$ \sum_{n=0}^\infty \frac{\binom{2n}{n}}{2^{2n}(2n+1)}=\frac{\pi}{2} $

$ \sum_{n=1}^\infty \frac{2^n}{\binom{2n}{n}n}=\frac{\pi}{2} $

$ \sum_{n=2}^\infty \frac{2^n}{\binom{2n}{n}}=\frac{\pi}{2} $

$ \sum_{n=2}^\infty \frac{n}{\binom{2n}{n}(2n+1)}=\frac{1}{2}-\frac{2\sqrt{3}\pi}{27} $

$ \sum_{n=1}^\infty \frac{2^n}{\binom{2n}{n}n^2}=\frac{\pi^2}{8} $

$ \sum_{n=1}^\infty \frac{2^{2n}}{\binom{2n}{n}n^2}=\frac{\pi^2}{2} $

$ \sum_{n=1}^\infty \frac{1}{\binom{2n}{n}(2n+1)(2n+2)}=\frac{\pi^2}{18}-\frac{1}{2} $

$ \sum_{n=1}^\infty \frac{1}{\binom{2n}{n}(2n+1)(2n+2)(2n+3)}=\frac{\pi^2}{18}+\frac{2\sqrt{3}\pi}{3}-\frac{25}{6} $

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

$ \sum_{n=1}^\infty \left[\frac{\binom{2n}{n}}{2^{2n}(2n+1)}\cdot\sum_{m=1}^n\frac{1}{(2m-1)^2}\right]=\frac{\pi^3}{48} $

$ \sum_{n=1}^\infty \left[\frac{1}{\binom{2n}{n}(2n+1)}\cdot\sum_{m=1}^n\frac{1}{m^2}\right]=\frac{\pi^3}{3^4\cdot\sqrt{3}} $

$ \sum_{n=1}^\infty \left[\frac{2^{2n}}{\binom{2n}{n}(2n+1)(2n+2)}\cdot\sum_{m=1}^n\frac{1}{m^2}\right]=\frac{\pi^4}{96} $

$ \sum_{n=1}^\infty \left[\frac{1}{\binom{2n}{n}(2n+1)(2n+2)}\cdot\sum_{m=1}^n\frac{1}{m^2}\right]=\frac{\pi^4}{1944} $

$ \sum_{n=1}^\infty \frac{1}{n^3}\left(-\frac{24}{n+1}+\frac{768}{n+2}+\frac{45}{2n+1}-\frac{1215}{2n+3}\right)=2\pi^2 $

$ \sum_{n=1}^\infty \frac{1}{n^3}\left(-\frac{7616}{n+1}+\frac{4516}{2n+1}-\frac{667}{4n+1}-\frac{10206}{4n+3}+\frac{35625}{4n+5}\right)=32\pi $

$ \sum_{n=0}^\infty \frac{1}{\binom{2n}{n}(2n+1)^2}=\frac{8}{3}\text{G}-\frac{\pi}{3}\log(2+\sqrt{3}) $

$ \sum_{n=1}^\infty \frac{1}{(n+1)^2(n+2)^2(n+3)^2}=\frac{\pi^2}{4}-\frac{355}{144} $

$ \sum_{n=0}^\infty (-1)^n\left(\frac{1}{(4n+1)^3}+\frac{1}{(4n+3)^3}\right)=\frac{3\sqrt{2}\pi^3}{128} $

$ \sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)}=\frac{\sqrt{3}\pi}{9} $