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階乗を用いた公式

113\pi+355 = \sum_{n=1}^\infty \frac{2^n(n!)^2n^4}{(2n)!}

\frac{32534}{3}\pi\sqrt{3}+29496 = \sum_{n=1}^\infty \frac{3^n(n!)^2n^4}{(2n)!n^4}

\frac{13130\pi^4}{729}+196 = \sum_{n=1}^\infty \frac{(n!)^2n^6}{(2n)!n^4}

\frac{\pi}{4}+\frac{2}{3} = \sum_{n=0}^\infty 4^n\frac{((2n)!)^2}{(4n)!}n

\frac{15\pi}{2}+26 = \sum_{n=0}^\infty 8^n\frac{((3n)!)^2}{(6n)!}(63n^2-12n+4)

\frac{105\pi}{8}+38 = \sum_{n=0}^\infty 16^n\frac{((4n)!)^2}{(8n)!}(240n^3-116n^2+53n-2)

\frac{2}{25}-\frac{6}{125}\log 2+\frac{11}{250}\pi = \sum_{n=0}^\infty\left(\frac{1}{2}\right)^n\frac{n!(2n)!}{(3n)!}

\frac{81}{625}-\frac{18}{3125}\log 2+\frac{79}{3125}\pi = \sum_{n=0}^\infty\left(\frac{1}{2}\right)^n\frac{n!(2n)!}{(3n)!}n

\frac{561}{3125}+\frac{42}{15625}\log 2+\frac{673}{31250}\pi = \sum_{n=0}^\infty\left(\frac{1}{2}\right)^n\frac{n!(2n)!}{(3n)!}n^2

\frac{\pi}{2} = \sum_{n=0}^\infty\left(\frac{1}{2}\right)^n\frac{n!(2n)!}{(3n)!}(25n-3)

16\pi\sqrt{3}+81 = \sum_{n=0}^\infty\left(\frac{8}{3}\right)^n\frac{n!(2n)!}{(3n)!}(49n+1)

162-6\pi\sqrt{3}-18\text{log}3 = \sum_{n=0}^\infty\left(\frac{8}{3}\right)^n\frac{n!(2n)!}{(3n)!}(-245n+338)

\frac{582}{5}+\frac{9}{2}\pi = \sum_{n=1}^\infty 8^n\frac{n!(3n)!}{(4n)!}(43n^2-624n+48)

-\frac{\pi}{2} = \sum_{n=0}^\infty\left(\frac{1}{2}\right)^n\frac{n!(2n)!}{(3n)!}(75n^2-115n+18)

24516-360\pi\sqrt{3} = \sum_{n=0}^\infty 9^n\frac{n!(3n)!}{(4n)!}(2743n^2-130971n-12724)

\frac{1872}{5}+8\pi\sqrt{3} = \sum_{n=0}^\infty 3^n\frac{n!(3n)!}{(4n)!}(1435n^2-3403n+96)

324+288\pi\sqrt{3}-576\text{log}2 = \sum_{n=0}^\infty\left(\frac{9}{8}\right)^n\frac{n!(3n)!}{(4n)!}(-5415n^2+6335n+5692)

7582+1008\pi\sqrt{3}-576\text{log}2 = \sum_{n=0}^\infty\left(\frac{9}{8}\right)^n\frac{n!(3n)!}{(4n)!}(18050n^2+1145n+7517)

-264+\frac{15}{2}\pi +120\text{log}2 = \sum_{n=1}^\infty\left(-\frac{1}{2}\right)^n\frac{(2n)!(3n)!}{(5n)!}(44506n^3-38681n^2+241n+1514)

20\pi\sqrt{3}+89 = \sum_{n=1}^\infty \left(-\frac{1}{3}\right)^n\frac{(2n!)^2(3n)!}{n!(6n)!}(-22100n+4123)

\frac{40}{27}\pi\sqrt{3}+9 = \sum_{n=1}^\infty (-27)^n\frac{(2n!)^2(3n)!}{n!(6n)!}(-40n+3)

15\pi\sqrt{2}+27 = \sum_{n=0}^\infty 8^n\frac{(2n!)^2(3n)!}{n!(6n)!}(350n-17)

15\pi +42 = \sum_{n=1}^\infty (-4)^n\frac{(2n!)^2(3n)!}{n!(6n)!}(-952n+201)

\frac{740025\pi - 20379280}{3} = \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n\frac{(5n)!(2n)!}{(7n)!}P(n)

ただし

\begin{align}P(n)=-885673181n^5+3125347237n^4-2942969225n^3\\
\quad+1031962795n^2-196882274n+10996648\end{align}

\frac{\pi^2}{6}-2\text{log}^22 = \sum_{n=1}^\infty \left(\frac{1}{4}\right)^n\frac{(2n)!}{(n!)^2}\frac{1}{n^2}

_4F_3\left(\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{1}{2}\\ \frac{3}{2},\frac{3}{2}\end{matrix};\frac{1}{4}\right)=\sum_{n=0}^\infty\frac{(2n)!}{16^n(n!)^2(2n+1)^3}=\frac{7}{216}\pi^3

\frac{\pi^2}{20} = \sum_{n=1}^\infty \frac{(-1)^{n-1}(2n)!}{16^n(n!)^2(2n+1)^2}

\frac{\pi^2\sqrt{2}}{12}-\frac{\sqrt{2}\log^22}{4} = \sum_{n=1}^\infty \frac{(-1)^{n-1}(2n)!}{32^n(n!)^2(2n+1)^2}

\frac{\pi\log2}{2} = \sum_{n=1}^\infty \frac{(2n)!}{4^n(n!)^2(2n+1)^2}

\pi^2\log2-\frac{7}{2}\zeta(3) = \sum_{n=1}^\infty \frac{4^n(n!)^2}{(2n)!n^3}

\frac{\pi^2\log2}{8}-\pi\text{G}-\frac{35}{16}\zeta(3) = \sum_{n=1}^\infty \frac{2^n(n!)^2}{(2n)!n^3}

\frac{\sqrt{2\pi}}{8}\Gamma^2\left(\frac{1}{4}\right) = \sum_{n=1}^\infty \frac{(-1)^{n-1}(2n)!}{4^n(n!)^2(2n+1)^2}

特殊なもの

\arcsin x = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!} \frac{x^{2n+1}}{2n+1}を利用

\begin{align}
\frac{\pi}{3} = &2 \arcsin \frac{1}{2} = 2\sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!} \frac{1}{2^{2n+1}(2n+1)}\cdots(1)\\
=&\sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!} \frac{1}{2^{2n}(2n+1)} = \sum_{n=0}^\infty \frac{((2n-1)!!)^2}{(2n)!} \frac{1}{2^{2n}(2n+1)}\\
=&\sum_{n=0}^\infty \frac{((2n-1)!!)^2}{(2n+1)!} \frac{1}{2^{2n}} = \sum_{n=0}^\infty \frac{((2n-1)!!)^2}{(4n+2)!!}\cdots(2)\\
=&\sum_{n=0}^\infty \left(\frac{1}{16}\right)^n\frac{(2n)!}{(n!)^2}\frac{1}{2n+1}
\end{align}

(1)はニュートンの公式、(2)は松永良弼の公式。

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

\sum_{n=0}^\infty \left(\frac{1}{8}\right)^n\frac{(2n)!}{(n!)^2}\frac{1}{2n+1}=\frac{\pi\sqrt{2}}{4}

\sum_{n=0}^\infty \left(\frac{3}{16}\right)^n\frac{(2n)!}{(n!)^2}\frac{1}{2n+1}=\frac{2\pi\sqrt{3}}{9}

\sum_{n=0}^\infty \left(\frac{1}{20}\right)^n\frac{(2n)!}{(n!)^2}\frac{1}{2n+1}\left(\sqrt{2}+\frac{1}{2^n}\right)=\frac{\pi\sqrt{10}}{4}

\arcsin^2 x = 2\sum_{n=0}^\infty \frac{2^{2n} (n!)^2}{(2n+2)!}x^{2n+2} = \frac{1}{2}\sum_{n=1}^\infty \frac{2^{2n-1}}{n^2{}_{2n}C_n}x^{2n}を利用

\frac{\pi^2}{9} = 4\arcsin^2 \frac{1}{2} = \sum_{n=0}^\infty \frac{(n!)^2}{{}_{2n+2}P_2}(建部賢弘の公式)

\frac{\pi^2}{18} = \sum_{n=1}^\infty \frac{(n!)^2}{(2n)!n^2}

\frac{2\pi^2}{9} = \sum_{n=1}^\infty \frac{3^n(n!)^2}{(2n)!n^2}

\frac{\pi^2}{8} = \sum_{n=1}^\infty \frac{2^n(n!)^2}{(2n)!n^2}

\frac{\pi^2}{2} = \sum_{n=1}^\infty \frac{4^n(n!)^2}{(2n)!n^2}

\frac{\pi^2}{32} = \sum_{n=1}^\infty \frac{(2-\sqrt{2})^n(n!)^2}{(2n)!n^2}

\frac{\pi^2}{72} = \sum_{n=1}^\infty \frac{(2-\sqrt{3})^n(n!)^2}{(2n)!n^2}

\frac{\pi^2}{25} = \sum_{n=1}^\infty \frac{(5-\sqrt{5})^n(n!)^2}{2^{n+1}(2n)!n^2}

\frac{\arcsin x}{\sqrt{1-x^2}} =\frac{1}{2}\left(\frac{d}{dx}\arcsin^2x\right) = \sum_{n=0}^\infty \frac{2^{n} n!}{(2n-1)!!} \frac{x^{2n+1}}{2n+1}= \sum_{n=0}^\infty \frac{2^{2n} (n!)^2}{(2n+1)!} x^{2n+1}を利用

\frac{\pi}{2} = \sqrt{2}\frac{\arcsin \frac{1}{\sqrt{2}}}{\sqrt{1-(\frac{1}{\sqrt{2}})^2}} = \sqrt{2}\sum_{n=0}^\infty \frac{2^{n} n!}{(2n-1)!!} \frac{(\frac{1}{\sqrt{2}})^{2n+1}}{2n+1} = \sum_{n=0}^\infty \frac{n!}{(2n+1)!!}

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

\frac{\pi}{5\sqrt{\phi +2}} = \frac{\arcsin \sin{\frac{\pi}{10}}}{\sqrt{1-\sin^2{\frac{\pi}{10}}}} = 2\sum_{n=0}^\infty \frac{(n!)^2}{\phi^{2n+1}(2n+1)!}

\frac{x\arcsin x}{\sqrt{1-x^2}^3}+\frac{1}{1-x^2} =\frac{1}{2}\left(\frac{d^2}{dx^2}\arcsin^2x\right) =\sum_{n=0}^\infty \frac{2^{2n} (n!)^2}{(2n)!} x^{2n}を利用

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

\frac{4\pi\sqrt{3}}{9} +3 = \sum_{n=1}^\infty \frac{3^n(n!)^2}{(2n)!}

\frac{\pi}{2} +1 = \sum_{n=1}^\infty \frac{2^n(n!)^2}{(2n)!}

\frac{6x}{(1-x^2)^2}+\frac{(6x^2+2)\arcsin x}{\sqrt{1-x^2}^5}=\frac{1}{4}\left(\frac{d^2}{dx^2}\arcsin^2x\right) =\sum_{n=0}^\infty \frac{2^{2n} (n!)^2}{(2n)!}n x^{2n-1}を利用

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

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

\pi + 3 = \sum_{n=0}^\infty 2^n\frac{(n!)^2}{(2n)!}n

\frac{\arcsin \sqrt{x}}{\sqrt{x-x^2}} =\left(\frac{d}{dx}\arcsin^2\sqrt{x}\right) = \sum_{n=1}^\infty \frac{2^{2n-1}}{n{}_{2n}C_n}x^{n-1}を利用

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

\frac{2\pi\sqrt{3}}{3} = \sum_{n=1}^\infty \frac{2^n(n!)^2}{(2n)!n}

\frac{\pi}{2} = \sum_{n=1}^\infty \frac{3^n(n!)^2}{(2n)!n}

\arctan x = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} x^{2n+1}を利用

\frac{\pi}{4} = \arctan 1= \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}(ライプニッツの公式、マーダヴァとグレゴリーが先に発見)

\pi = 6\arctan\frac{1}{\sqrt{3}} = \sqrt{12}\sum_{k=1}^\infty \frac{2(-1)^k}{(2k+1)3^k}(マーダヴァの公式、シャープの公式)

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

\frac{\pi}{6} = 2\arctan\frac{1}{3\sqrt{3}} +\arctan\frac{1}{4\sqrt{3}} = \frac{\sqrt{3}}{36}(4\sum_{n=0}^\infty \frac{(-1)^n}{27^n(2n+1)} +3\sum_{n=0}^\infty \frac{(-1)^n}{48^n(2n+1)}) \frac{5\pi}{4} = 12\arctan\frac{1}{3} + 4\arctan\frac{1}{57} - \arctan\frac{1}{239}

一般式

\frac{\pi}{4} = P\arctan\frac{1}{2} - M\arctan\frac{1}{3} + L\arctan\frac{1}{5} + K\arctan\frac{1}{7} +

(N+K+L-2M+3P-5)\arctan\frac{1}{8} + (2N+M-P+2-L)\arctan\frac{1}{17}

- (2P-3-M+L+K-N)\arctan\frac{1}{58} - N\arctan\frac{1}{239}

\frac{\pi}{4} = (N+2)\arctan\frac{1}{2} - N\arctan\frac{1}{3}-(N+1)\arctan\frac{1}{N}

(P,N,M,L,Kは正の実数)

無限和

\frac{\pi}{4} = \sum_{n=1}^\infty \arctan\frac{1}{F_{2n+1}} = \arctan\frac{1}{2} + \arctan\frac{1}{5} + \cdots =\sum_{n=1}^\infty\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)F_{2n+1}^{2k+1}}

\begin{align}
\frac{\pi}{4} &= \sum_{n=1}^\infty \arctan\frac{1}{n^2+n+1} = \arctan\frac{1}{3} + \arctan\frac{1}{7} + \cdots & \\
 &= \sum_{n=1}^\infty\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)(n^2+n+1)^{2k+1}}& \\
\end{align}

\frac{\pi}{4} = \sum_{n=1}^\infty \arctan\frac{1}{2n^2} = \arctan\frac{1}{2} + \arctan\frac{1}{8} + \cdots =\sum_{n=1}^\infty\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)(2n^2)^{2k+1}}

おまけ

(分母が連続する式)

\frac{\pi}{4} = \arctan1 + \arctan\frac{1}{2} + \arctan\frac{1}{3} - \arctan\frac{1}{4} + 4\arctan\frac{1}{5} + \arctan\frac{1}{6} +

 2\arctan\frac{1}{7} + 4\arctan\frac{1}{8} - 2\arctan\frac{1}{9} + \arctan\frac{1}{10} + \arctan\frac{1}{11} + \arctan\frac{1}{12} -

 \arctan\frac{1}{13} + \arctan\frac{1}{14} - 2\arctan\frac{1}{15} + \arctan\frac{1}{16} + \arctan\frac{1}{17} + \arctan\frac{1}{50} +

 \arctan\frac{1}{68} + \arctan\frac{1}{91} - \arctan\frac{1}{211} + \arctan\frac{1}{241}

(途中までの和を求める式)

\pi=(-1)^{n+1}\sum_{k=0}^\infty \frac{k!}{2^k(-n-\frac{1}{2})_{k+1}}-4\sum_{j=0}^n\frac{(-1)^{j}}{2j+1}

\pi=8\sum_{n=1}^\infty\frac{(-1)^{n-1}}{p^{2n-1}(2n-1)}(\sum_{k=1}^mb_kU_n^k)

ただし、U_1^k=\frac{p^2-1}{2p}a_k,U_2^k=2(U_1^k)^2-1,U_n^k=\frac{p^2-1}{p}a_kU_{n-1}^k-U_{n-2}^k

さらに、\frac{\pi}{4}=\sum_{k=1}^{m}b_k\arctan (a_k)

任意の正の実数U_0,U_1、自然数nに対して、

\pi = 4\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}((\frac{U_{n+1}}{U_{n+2}})^{2k+1}+(\frac{U_n}{U_{n+3}})^{2k+1})

ただし、U_{n+2}=U_{n+1}+U_n

証明

U_0>0,U_1>0,U_{n+2}=U_{n+1}+U_n

より、帰納的に

0<U_1<U_2<\cdots<U_n<U_{n+1}<U_{n+2}<U_{n+3}<\cdots,0<\frac{U_{n+1}}{U_{n+2}}<1,0<\frac{U_n}{U_{n+3}}<1

これより、

\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}((\frac{U_{n+1}}{U_{n+2}})^{2k+1}+(\frac{U_n}{U_{n+3}})^{2k+1})=\arctan\frac{U_{n+1}}{U_{n+2}}+\arctan\frac{U_n}{U_{n+3}}

かつ

0<\arctan\frac{U_{n+1}}{U_{n+2}}<\frac{\pi}{4},0<\arctan\frac{U_{n+1}}{U_{n+2}}<\frac{\pi}{4}

ここで

\begin{align}
&\tan(\arctan\frac{U_{n+1}}{U_{n+2}}+\arctan\frac{U_n}{U_{n+3}})=\frac{\frac{U_{n+1}}{U_{n+2}}+\frac{U_n}{U_{n+3}}}{1-\frac{U_{n+1}}{U_{n+2}}\frac{U_n}{U_{n+3}}}=\frac{U_{n+1}U_{n+3}+U_nU_{n+2}}{U_{n+2}U_{n+3}-U_nU_{n+1}} \\
&=\frac{U_{n+1}(2U_{n+1}+U_n)+U_n(U_{n+1}+U_n)}{(2U_{n+1}+U_n)(U_{n+1}+U_n)-U_nU_{n+1}}=\frac{2U_{n+1}^2+2U_nU_{n+1}+U_n^2}{2U_{n+1}^2+2U_nU_{n+1}+U_n^2}=1\\
\end{align}

0<\arctan\frac{U_{n+1}}{U_{n+2}}+\arctan\frac{U_n}{U_{n+3}}<\frac{\pi}{2}

より

\frac{\pi}{4}=\arctan \frac{U_{n+1}}{U_{n+2}}+\arctan \frac{U_n}{U_{n+3}}

\pi = 4\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}((\frac{U_{n+1}}{U_{n+2}})^{2k+1}+(\frac{U_n}{U_{n+3}})^{2k+1})

\arctan x = \frac{1}{x}\sum_{n=0}^\infty \frac{(2n)!!}{(2n+1)!!} (\frac{x^2}{1+x^2})^{n+1}を利用

\frac{\pi}{4} = 5 \arctan\frac{1}{7} + 2 \arctan\frac{3}{79} =

35\sum_{n=0}^\infty \frac{(2n)!!}{(2n+1)!!}(\frac{2}{100})^{n+1} + \frac{158}{3}\sum_{n=0}^\infty \frac{(2n)!!}{(2n+1)!!}(\frac{144}{100000})^{n+1}

(オイラーはこの公式を用いて、円周率を1時間で20桁まで求めた。)

\frac{\pi}{4} = \arctan\frac{1}{7} + 2 \arctan\frac{1}{3} =

7\sum_{n=0}^\infty \frac{(2n)!!}{(2n+1)!!}(\frac{2}{100})^{n+1} + 6\sum_{n=0}^\infty \frac{(2n)!!}{(2n+1)!!}(\frac{1}{10})^{n+1}

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