k 0 = 0 , h 0 = 2 , k n + 1 = 2 + k n , h n + 1 = 2 − k n {\displaystyle 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、
∑ n = 0 ∞ h n ( 1 − k n 2 ) 2 n = π {\displaystyle \sum_{n=0}^\infty h_n\left(1-\frac{k_n}{2}\right)2^n=\pi}
x 1 = 2 , x n + 1 = 2 2 n + 1 − 2 n + 1 4 n − x n {\displaystyle x_1=2,x_{n+1}=2^{2n+1}-2^{n+1}\sqrt{4^n-x_n}}
lim n → ∞ x n + 4 3 ( x n + 1 − x n ) = π 2 {\displaystyle \lim_{n\to\infty}\sqrt{x_n+\frac{4}{3}(x_{n+1}-x_n)}=\frac{\pi}{2}}
k 1 = 60 , k n + 1 = n + 1 4 n + 6 k n , l 1 = 1 4 , l n + 1 = l n + 1 ( n + 1 ) 2 ∑ m = 1 n 1 m 2 {\displaystyle 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}}}}
∑ n = 2 ∞ k n l n n + 1 = π 6 729 {\displaystyle \sum _{n=2}^{\infty }{\frac {k_{n}l_{n}}{n+1}}={\frac {\pi ^{6}}{729}}}
∑ n = 1 ∞ 1 n 3 ( − 24 n + 1 + 768 n + 2 + 45 2 n + 1 − 1215 2 n + 3 ) = 2 π 2 {\displaystyle \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}}
∑ n = 1 ∞ 1 n 3 ( − 7616 n + 1 + 4516 2 n + 1 − 667 4 n + 1 − 10206 4 n + 3 + 35625 4 n + 5 ) = 32 π {\displaystyle \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 }
∑ n = 0 ∞ ( 2 n n ) 2 2 n ( 2 n + 1 ) = π 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {\binom {2n}{n}}{2^{2n}(2n+1)}}={\frac {\pi }{2}}}
∑ n = 1 ∞ 2 n ( 2 n n ) n = π 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {2^{n}}{{\binom {2n}{n}}n}}={\frac {\pi }{2}}}
∑ n = 2 ∞ 2 n ( 2 n n ) = π 2 {\displaystyle \sum _{n=2}^{\infty }{\frac {2^{n}}{\binom {2n}{n}}}={\frac {\pi }{2}}}
∑ n = 2 ∞ n ( 2 n n ) ( 2 n + 1 ) = 1 2 − 2 3 π 27 {\displaystyle \sum _{n=2}^{\infty }{\frac {n}{{\binom {2n}{n}}(2n+1)}}={\frac {1}{2}}-{\frac {2{\sqrt {3}}\pi }{27}}}
∑ n = 1 ∞ 2 n ( 2 n n ) n 2 = π 2 8 {\displaystyle \sum _{n=1}^{\infty }{\frac {2^{n}}{{\binom {2n}{n}}n^{2}}}={\frac {\pi ^{2}}{8}}}
∑ n = 1 ∞ 2 2 n ( 2 n n ) n 2 = π 2 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {2^{2n}}{{\binom {2n}{n}}n^{2}}}={\frac {\pi ^{2}}{2}}}
∑ n = 1 ∞ 1 ( 2 n n ) ( 2 n + 1 ) ( 2 n + 2 ) = π 2 18 − 1 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{{\binom {2n}{n}}(2n+1)(2n+2)}}={\frac {\pi ^{2}}{18}}-{\frac {1}{2}}}
∑ n = 1 ∞ 1 ( 2 n n ) ( 2 n + 1 ) ( 2 n + 2 ) ( 2 n + 3 ) = π 2 18 + 2 3 π 3 − 25 6 {\displaystyle \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}}}
∑ n = 2 ∞ 2 n ( 2 n n ) ( n − 1 ) 2 n = π 2 8 + π − 4 {\displaystyle \sum _{n=2}^{\infty }{\frac {2^{n}}{{\binom {2n}{n}}(n-1)^{2}n}}={\frac {\pi ^{2}}{8}}+\pi -4}
∑ n = 1 ∞ [ ( 2 n n ) 2 2 n ( 2 n + 1 ) ⋅ ∑ m = 1 n 1 ( 2 m − 1 ) 2 ] = π 3 48 {\displaystyle \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}}}
∑ n = 1 ∞ [ 1 ( 2 n n ) ( 2 n + 1 ) ⋅ ∑ m = 1 n 1 m 2 ] = π 3 3 4 ⋅ 3 {\displaystyle \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}}}}}
∑ n = 1 ∞ [ 2 2 n ( 2 n n ) ( 2 n + 1 ) ( 2 n + 2 ) ⋅ ∑ m = 1 n 1 m 2 ] = π 4 96 {\displaystyle \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}}}
∑ n = 1 ∞ [ 1 ( 2 n n ) ( 2 n + 1 ) ( 2 n + 2 ) ⋅ ∑ m = 1 n 1 m 2 ] = π 4 1944 {\displaystyle \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}}}
∑ n = 0 ∞ 1 ( 2 n n ) ( 2 n + 1 ) 2 = 8 3 G − π 3 log ( 2 + 3 ) {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{{\binom {2n}{n}}(2n+1)^{2}}}={\frac {8}{3}}{\text{G}}-{\frac {\pi }{3}}\log(2+{\sqrt {3}})}
∑ n = 1 ∞ 1 ( n + 1 ) 2 ( n + 2 ) 2 ( n + 3 ) 2 = π 2 4 − 355 144 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(n+1)^{2}(n+2)^{2}(n+3)^{2}}}={\frac {\pi ^{2}}{4}}-{\frac {355}{144}}}
∑ n = 0 ∞ ( − 1 ) n ( 1 ( 4 n + 1 ) 3 + 1 ( 4 n + 3 ) 3 ) = 3 2 π 3 128 {\displaystyle \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}}}
∑ n = 0 ∞ 1 ( 3 n + 1 ) ( 3 n + 2 ) = 3 π 9 {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(3n+1)(3n+2)}}={\frac {{\sqrt {3}}\pi }{9}}}
![{\displaystyle \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}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/008d29b3175e3ddeeeb6feaf86c9640eb0132796)
![{\displaystyle \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}}}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/0a96984227cd89a91d081b6e2d0976851e237cda)
![{\displaystyle \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}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/f271de10d906ee3f1aacdd5c1725a547b66b7078)
![{\displaystyle \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}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/bccbcfff3a117acca14a35f110ba2af68df90a9b)
