1.
1π=229801∑k=0∞(4k)!(1103+26390k)(k!)43964kπ4=∑k=0∞12k+1a0=1,b0=12,t0=14,p0=1an+1=an+bn2,bn+1=anbn,tn+1=tn−pn(an−an+1)2,pn+1=2pnπ≈(an+1+bn+1)24tn+1π=42688010005∑k=0∞(6k)!(545140134k+13591409)(3k)!(k!)3(−640320)3kπ2=21⋅23⋅43⋅45⋅65⋅67⋯arctanx=∑k=0∞(−1)kx2k+12k+1π=16arctan15−4arctan1239π=24arctan18+8arctan157+4arctan1239π=48arctan118+32arctan157−20arctan1239π=32arctan110−4arctan1239−16arctan1515π=ln−1−1π=3+17+115+11+1292+11+⋯\frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum\limits_{k=0}^{\infty}\frac{(4k)!(1103+26390k)}{(k!)^4396^{4k}}\\
\frac{\pi}{4}=\sum\limits_{k=0}^{\infty}\frac{1}{2k+1}\\ a_0=1,b_0=\frac{1}{\sqrt{2}},t_0=\frac{1}{4},p_0=1\\ a_{n+1}=\frac{a_n+b_n}{2},b_{n+1}=\sqrt{a_nb_n},t_{n+1}=t_n-p_n(a_n-a_{n+1})^2,p_{n+1}=2p_n\\ \pi\approx\frac{(a_{n+1}+b_{n+1})^2}{4t_{n+1}}\\
\Large\pi=\frac{426880\sqrt{10005}}{\sum\limits_{k=0}^{\infty}\frac{(6k)!(545140134k+13591409)}{(3k)!(k!)^3(-640320)^{3k}}}\\ \normalsize \frac{\pi}{2}=\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdots\\
\arctan{x}=\sum\limits_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\ \pi=16\arctan{\frac{1}{5}}-4\arctan{\frac{1}{239}}\\ \pi=24\arctan{\frac{1}{8}}+8\arctan{\frac{1}{57}}+4\arctan{\frac{1}{239}}\\ \pi=48\arctan{\frac{1}{18}}+32\arctan{\frac{1}{57}}-20\arctan{\frac{1}{239}}\\
\pi=32\arctan{\frac{1}{10}}-4\arctan{\frac{1}{239}}-16\arctan{\frac{1}{515}}\\ \pi=\frac{\ln{-1}}{\sqrt{-1}}\\ \huge\pi=3+\frac{1}{7+\frac{1}{15+\frac{1}{1+\frac{1}{292+\frac{1}{1+\cdots}}}}} π1 =980122 k=0∑∞ (k!)43964k(4k)!(1103+26390k) 4π =k=0∑∞ 2k+11 a0 =1,b0 =2 1 ,t0 =41 ,p0 =1an+1 =2an +bn
,bn+1 =an bn ,tn+1 =tn −pn (an −an+1 )2,pn+1 =2pn π≈4tn+1 (an+1 +bn+1 )2 π=k=0∑∞ (3k)!(k!)3(−640320)3k(6k)!(545140134k+13591409) 42688010005 2π =12 ⋅32 ⋅34 ⋅54 ⋅56 ⋅76 ⋯arctanx=k=0∑∞ (−1)k2k+1x2k+1 π=16arctan51 −4arctan2391 π=24arctan81 +8arctan571 +4arctan2391 π=48arctan181 +32arctan571
−20arctan2391 π=32arctan101 −4arctan2391 −16arctan5151 π=−1 ln−1 π=3+7+15+1+292+1+⋯1 1 1 1 1