Gao Ming's corner: 与圆周率PI有关的一些级数

注：本文版权属于Wolfram Research, Inc.本人在这里只是做了一下文字整理，而发此贴的主要目的在于测试在blogger上写 $\LaTeX$ 。

早已久仰Wolfram Alpha的大名了，刚才试了一下，果然不错。偶然间发现了一些与圆周率 $\pi$ 有关的级数，感觉挺有趣的，记录一下。其实这些级数的证明是很容易的，其中不乏一些非常有趣的结论，例如下面的第一个级数就是著名的BBP类公式，它的奇妙之处是利用它可以直接计算出 $\pi$ 的第n位二进制数字而不需要计算前n-1位数字，当然，其他公式也很重要，有些和黎曼Zeta函数有关，有些与贝努力数有关，有些与组合数有关，还有些与Fourier变换有关。这些公式如下：

$\displaystyle\pi =\sum _{k=0}^{\infty } \frac{1}{16^k}\left(\frac{8 r+4}{8k+1}-\frac{8 r}{8 k+2}-\frac{4 r}{8 k+3}-\frac{8 r+2}{8 k+4}-\frac{2r+1}{8 k+5}-\frac{2 r+1}{8 k+6}+\frac{r}{8 k+7}\right)\text{/;}r\in\text{Integers}\land r>0$

$\displaystyle\pi =2 \text{Log}[2]+4\sum _{k=0}^{\infty } \frac{(-1)^{\text{Floor}[(k+1)/2]}}{k+1}$

$\displaystyle\pi =4 \sum _{k=0}^{\infty } \frac{(-1)^k}{2 k+1}$

$\displaystyle\pi =\frac{4 }{\sqrt{2}}\sum _{k=0}^{\infty } \frac{(-1)^{\text{Floor}\left[\frac{k}{2}\right]}}{2 k+1}$

$\displaystyle\pi =16 \sum _{k=0}^{\infty } \frac{(-1)^k}{(2 k+1) 5^{2 k+1}}-4 \sum _{k=0}^{\infty } \frac{(-1)^k}{(2 k+1) 239^{2 k+1}}$

$\displaystyle\pi =-\text{Log}[2]\sqrt{3}+3\sqrt{3}\sum _{k=0}^{\infty } \frac{(-1)^k}{3k+1}$

$\displaystyle\pi =-2 \text{Log}\left[1+\sqrt{2} \right]+4\sqrt{2}\sum _{k=0}^{\infty } \frac{(-1)^k}{4k+1}$

$\displaystyle\pi = \sum _{k=0}^{\infty } \frac{(-1)^k}{4^k}\left(\frac{2}{4 k+1}+\frac{2}{4 k+2}+\frac{1}{4 k+3}\right)$

$\displaystyle\pi = \sum _{k=1}^{\infty } \frac{1}{k^3}\left(-\frac{238}{k+1}+\frac{285}{2(2k+1)}-\frac{667}{32(4 k+1)}-\frac{5103}{16 (4 k+3)}+\frac{35625}{32 (4 k+5)}\right)$

$\displaystyle\pi =\frac{1}{\sqrt{2}} \sum _{k=0}^{\infty } \frac{(-1)^k}{8^k}\left(\frac{4}{6 k+1}+\frac{1}{6 k+3}+\frac{1}{6 k+5}\right)$

$\displaystyle\pi =\frac{8}{1+\sqrt{2}}\sum _{k=0}^{\infty } \left(\frac{1}{8k+1}-\frac{1}{8k+7}\right)$

$\displaystyle\pi =\sum _{k=0}^{\infty } \frac{1}{16^k}\left(\frac{4}{8 k+1}-\frac{2}{8 k+4}-\frac{1}{8 k+5}-\frac{1}{8 k+6}\right)$

$\displaystyle\pi =2\sum _{k=0}^{\infty } \frac{(2k-1)\text{!!}}{(2k+1)(2k)\text{!!}}$

$\displaystyle\pi =\frac{5}{4} \sqrt{5} \sum _{k=0}^{\infty } \frac{\text{Binomial}[2 k,k] \text{Fibonacci}[2 k+1]16^{-k}}{2 k+1}$

$\displaystyle\pi =\sqrt{5} \sum _{k=0}^{\infty } \frac{(-1)^k 2^{2 k+3}}{(2 k+1) \left(\sqrt{5}+3\right)^{2 k+1}}\text{Fibonacci}[2 k+1]$

$\displaystyle\pi =20 \sum _{k=0}^{\infty } \frac{(-1)^k }{(2 k+1) \left(\sqrt{10}+3\right)^{2 k+1}}\text{Fibonacci}[2 k+1]^2$

$\displaystyle\pi =12 \sqrt{5} \sum _{k=0}^{\infty } \frac{(-1)^k }{2 k+1}\left(\frac{2 \left(2-\sqrt{3}\right)}{\sqrt{16 \left(2-\sqrt{3}\right)+1}+\sqrt{5}}\right)^{2 k+1}\text{Fibonacci}[2 k+1]$

$\displaystyle\pi =4 \sum _{k=1}^{\infty } \text{ArcTan}\left[\frac{1}{\text{Fibonacci}[2 k+1]}\right]$

$\displaystyle\pi =\sum _{k=1}^{\infty } \frac{\left(3^k-1\right) \text{Zeta}[k+1]}{4^k}$

$\displaystyle\pi =-3 \sqrt{3}+\frac{9}{2} \sqrt{3} \sum _{k=1}^{\infty } \frac{k}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{18 \sqrt{3}}{5}+\frac{27}{10} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^2}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{135 \sqrt{3}}{37}+\frac{81}{74} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^3}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{432 \sqrt{3}}{119}+\frac{81}{238} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^4}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{243 \sqrt{3}}{67}+\frac{81}{938} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^5}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{23814 \sqrt{3}}{6565}+\frac{243}{13130} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^6}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{42795 \sqrt{3}}{11797}+\frac{81}{23594} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^7}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{2355156 \sqrt{3}}{649231}+\frac{729 }{1298462}\sqrt{3} \sum _{k=1}^{\infty } \frac{k^8}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{48314475 \sqrt{3}}{13318583}+\frac{2187}{26637166} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^9}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{365306274 \sqrt{3}}{100701965}+\frac{2187}{201403930} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^{10}}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{99760005 \sqrt{3}}{27500287}+\frac{6561}{5005052234} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^{11}}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =-\frac{245273327208 \sqrt{3}}{67613135957}+\frac{19683}{135226271914} \sqrt{3} \sum _{k=1}^{\infty } \frac{k^{12}}{\text{Binomial}[2 k,k]}$

$\displaystyle\pi =4-2 \sum _{k=0}^{\infty } \frac{k!}{\prod _{j=0}^k (2 j+3)}$

$\displaystyle\frac{1}{\pi }=\frac{2 \sqrt{2} }{9801}\sum _{k=0}^{\infty } \frac{(4 k)! (26390 k+1103)}{k!^4 396^{4 k}}$

$\displaystyle\frac{1}{\pi }=\sum _{k=0}^{\infty } \frac{42 k+5}{2^{12 k+4}}\text{Binomial}[2 k,k]^3$

$\displaystyle\frac{1}{\pi }=12 \sum _{k=0}^{\infty } \frac{(-1)^k (6 k)! (545140134 k+13591409)}{k!^3 (3 k)! \left(640320^3\right)^{k+\frac{1}{2}}}$

$\displaystyle\pi ^2=6 \sum _{k=1}^{\infty } \frac{1}{k^2}$

$\displaystyle\pi ^2=8 \sum _{k=0}^{\infty } \frac{1}{(2 k+1)^2}$

$\displaystyle\pi ^2=12 \sum _{k=1}^{\infty } \frac{(-1)^{k-1}}{k^2}$

$\displaystyle\pi ^2=36 \sum _{k=0}^{\infty } \frac{(2 k)\text{!!}}{(2 k+1)\text{!!} 2^{2 k+2} (k+1)}$

$\displaystyle\pi ^2=\sum _{k=1}^{\infty } \frac{1}{k^3}\left(\frac{45}{2 (2 k+1)}-\frac{1215}{2 (2 k+3)}+\frac{384}{k+2}-\frac{12}{k+1}\right)$

$\displaystyle\pi ^2=18 \sum _{k=0}^{\infty } \left(\frac{1}{(6 k+1)^2}-\frac{3}{2 (6 k+2)^2}-\frac{1}{2 (6 k+3)^2}-\frac{3}{8 (6 k+4)^2}+\frac{1}{16 (6k+5)^2}\right) 64^{-k}$

$\displaystyle\pi ^3=32 \sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{(2 k-1)^3}$

$\displaystyle\pi ^3=\frac{1}{16}\sum _{k=0}^{\infty } \frac{(-1)^k}{(1024)^k} \left(\frac{32}{(1+4 k)^3}+\frac{8}{(2+4 k)^3}+\frac{1}{(3+4 k)^3}\right)+$

$\displaystyle\frac{5}{2} \sum _{k=0}^{\infty }\frac{(-1)^k}{(64)^k}\left(\frac{32}{(1+12 k)^3}-\frac{192}{(2+12 k)^3}+\frac{88}{(3+12 k)^3}-\frac{8}{(5+12k)^3}+\frac{84}{(6+12 k)^3}-\frac{4}{(7+12 k^3}+\frac{11}{(9+12 k)^3}-\frac{12}{(10+12 k)^3}+\frac{1}{(11+12 k)^3}\right)$

$\displaystyle\pi ^4=90 \sum _{k=1}^{\infty } \frac{1}{k^4}$

$\displaystyle\pi ^4=96 \sum _{k=0}^{\infty } \frac{1}{(2 k+1)^4}$

$\displaystyle\pi ^4=$

$\displaystyle\frac{27}{164}$

$\displaystyle\sum _{k=0}^{\infty } \frac{1}{2^{12 k} } \left(\frac{2048}{(1+24 k)^4}-\frac{38912}{(2+24 k)^4}+\frac{81920}{(3+24 k)^4}-\frac{2048}{(4+24k)^4}-\frac{512}{(5+24 k)^4}-\frac{23552}{(6+24 k)^4}+\frac{256}{(7+24 k)^4}-\frac{27648}{(8+24 k)^4}-\frac{10240}{(9+24 k)^4}-\frac{2432}{(10+24k)^4}-\frac{64}{(11+24 k)^4}-\right.$

$\displaystyle\left.\frac{3584}{(12+24 k)^4}-\frac{32}{(13+24 k)^4}-\frac{608}{(14+24 k)^4}-\frac{1280}{(15+24 k)^4}-\frac{1728}{(16+24 k)^4}+\frac{8}{(17+24k)^4}-\frac{368}{(18+24 k)^4}-\frac{4}{(19+24 k)^4}-\frac{8}{(20+24 k)^4}+\frac{160}{(21+24 k)^4}-\frac{38}{(22+24 k)^4}+\frac{1}{(23+24 k)^4}\right)$

$\displaystyle\pi ^4=\frac{3240}{17} \sum _{k=1}^{\infty } \frac{1}{k^4\text{Binomial}[2 k,k]}$

$\displaystyle\pi ^6=945 \sum _{k=1}^{\infty } \frac{1}{k^6}$

$\displaystyle\pi ^6=960 \sum _{k=0}^{\infty } \frac{1}{(2 k+1)^6}$

$\displaystyle\pi ^{2 n}=\frac{(-1)^{n-1} (2 n)! }{2^{2 n-1}\text{BernoulliB}[2 n]}\sum _{k=1}^{\infty } \frac{1}{k^{2 n}}\text{/;} n\in \text{Integers}\land n>0$

$\displaystyle\pi ^{2 n}=\frac{(-1)^n 2^{1-2 n} (2 n)! }{\text{BernoulliB}\left[2 n,\frac{1}{2}\right]}\sum _{k=1}^{\infty } \frac{(-1)^{k-1}}{k^{2 n}}\text{/;}n\in \text{Integers}\land n>0$

$\displaystyle\pi ^{2 n}=\frac{(-1)^{n-1} 2 (2 n)! }{\left(4^n-1\right)\text{BernoulliB}[2 n]}\sum _{k=0}^{\infty } \frac{1}{(2 k+1)^{2 n}}\text{/;} n\in\text{Integers}\land n>0$

$\displaystyle\pi ^{2 n-1}=\frac{(-1)^{n-1} 2^{2 n} (2 n-2)!}{\text{EulerE}[2 n-2]} \sum _{k=0}^{\infty } \frac{(-1)^k}{(2 k+1)^{2 n-1}}\text{/;} n\in \text{Integers}\land n>0$

$\displaystyle\pi =x+2 \sum _{k=1}^{\infty } \frac{\text{Sin}[k x]}{k}\text{/;}x\in \text{Reals}\land x>0$

$\displaystyle\pi =4 \sum _{k=0}^{\infty } \frac{(-1)^k \text{Cos}[(2 k+1) x]}{2 k+1}\text{/;}x\in \text{Reals}$

$\displaystyle\pi =4-2\sum _{k=0}^{\infty } \frac{k!}{ \prod _{j=0}^k (2j+3)}$

$\displaystyle\pi =\text{Limit}\left[2^{n+2}f[n],n\to \infty \right]\text{/;}f[0]=1\land f[n]=\frac{f[n-1]}{1+\sqrt{1+f[n-1]^2}}\land n\in \text{Integers}\land n>0$

$\displaystyle\pi ^{2 n}=(2n+1)!\sum _{k_1=1}^{\infty } \ldots \sum _{k_n=1}^{\infty } \prod _{j=1}^n \frac{1}{\left(\sum _{l=1}^n k_l\right){}^2}\text{/;} n\in \text{Integers}\land n>0$