这类级数怎么证明？

\zeta(3) 也被称为Apéry 常数 ，因为Roger Apéry第一个证明了 \zeta(3) 是无理数.所需证的第二个恒等式是由A.A.Markov于1890年首先发现，1953年被Hjortnaes重新发现，1979年被Apéry再次发现并广为宣传。不过最初的原始论文比较模糊，后续有人重新做过整理

Step 1 :

\sum_{k=1}^{K} \frac{a_{1} a_{2} \ldots a_{k-1}}{\left(x+a_{1}\right)\left(x+a_{2}\right) \ldots\left(x+a_{k}\right)}=\frac{1}{x}-\frac{a_{1} a_{2} \ldots a_{K}}{x\left(x+a_{1}\right) \ldots\left(x+a_{K}\right)} \\

其中 a_{k}^{\prime} s 是整数

A_{0}=\frac{1}{x} \text { and } A_{K}=\frac{a_{1} a_{2} \ldots a_{K}}{x\left(x+a_{1}\right) \ldots\left(x+a_{K}\right)} \\

所以

\begin{aligned} &\quad\frac{a_{1} a_{2} \ldots a_{k-1}}{\left(x+a_{1}\right)\left(x+a_{2}\right) \ldots\left(x+a_{k}\right)} \\ &=\frac{a_{1} a_{2} \ldots a_{k-1}}{x\left(x+a_{1}\right)\left(x+a_{2}\right) \ldots\left(x+a_{k-1}\right)}-\frac{a_{1} a_{2} \ldots a_{k}}{\left(x+a_{1}\right)\left(x+a_{2}\right) \ldots\left(x+a_{k}\right)} \\ &=A_{k-1}-A_{k} \end{aligned} \\

Step 2:

令 x=n^{2} ; a_{k}=-k^{2},1 \leq K \leq n-1 我们有

\begin{aligned} \sum_{k=1}^{K} \frac{-1^{2} .-2^{2} \ldots-(k-1)^{2}}{\left(n^{2}-1\right) \ldots\left(n^{2}-k^{2}\right)} &=\sum_{k=1}^{K} \frac{(-1)^{k-1} 1^{2} 2^{2} \ldots .(k-1)^{2}}{\left(n^{2}-1^{2}\right) \ldots\left(n^{2}-k^{2}\right)} \\ \Rightarrow \sum_{k=1}^{n-1} \frac{(-1)^{k-1}(k-1) !^{2}}{\left(n^{2}-1^{2}\right) \ldots\left(n^{2}-k^{2}\right)} &=\frac{1}{n^{2}}-\frac{(-1)^{n-1}(n-1) !^{2} n 2 n}{n^{2}(2 n) !^{2}} \\ &=\frac{1}{n^{2}}-\frac{(-1)^{n-1} 2 n ! n !}{n^{2}(2 n) !} \\ &=\frac{1}{n^{2}}-\frac{(-1)^{n-1} 2}{n^{2}\left(\begin{array}{c} 2 n \\ n \end{array}\right)} \end{aligned} \\

接着我们记

\epsilon_{n, k}=\frac{1}{2} \frac{(k !)^{2}(n-k) !}{k^{3}(n+k) !} \\

又因为

(-1)^{k} n\left(\epsilon_{n, k}-\epsilon_{n-1, k}\right)=\frac{(-1)^{k-1}(k-1) !^{2}}{\left(n^{2}-1^{2}\right) \ldots\left(n^{2}-k^{2}\right)} \\

展开得到

\begin{aligned} &\sum_{n=1}^{N} \sum_{k=1}^{n-1}(-1)^{k}\left(\epsilon_{n, k}-\epsilon_{n-1, k}\right)=\sum_{k=1}^{N}(-1)^{k}\left(\epsilon_{N, k}-\epsilon_{k, k}\right)\\ =&\sum_{k=1}^{N} \frac{(-1)^{k}}{2 k^{3}\left(\begin{array}{c} N+k \\ k \end{array}\right)\left(\begin{array}{l} N \\ k \end{array}\right)}+\frac{1}{2} \sum_{k=1}^{N} \frac{(-1)^{k-1}}{k^{3}\left(\begin{array}{c} 2 k \\ k \end{array}\right)}\\ =&\sum_{k=1}^{N} \frac{1}{n^{3}}-2 \sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^{3}\left(\begin{array}{c} 2 n \\ n \end{array}\right)} \end{aligned} \\

于是

\sum_{k=1}^{N} \frac{1}{n^{3}}-2 \sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^{3}\left(\begin{array}{c} 2 n \\ n \end{array}\right)}=\sum_{k=1}^{N} \frac{(-1)^{k}}{2 k^{3}\left(\begin{array}{c} N+k \\ k \end{array}\right)\left(\begin{array}{l} N \\ k \end{array}\right)}+\frac{1}{2} \sum_{k=1}^{N} \frac{(-1)^{k-1}}{k^{3}\left(\begin{array}{c} 2 k \\ k \end{array}\right)} . \\

当N \to \infty

\sum_{k=1}^{\infty} \frac{1}{n^{3}}=\frac{5}{2} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{3}\left(\begin{array}{c} 2 k \\ k \end{array}\right)}\\

\Large\square

另外一些级数形式

\begin{aligned} &\zeta(3)=\frac{\pi^{2}}{7}\left(1-4 \sum_{k=1}^{\infty} \frac{\zeta(2 k)}{2^{2 k}(2 k+1)(2 k+2)}\right)\\ &\zeta(3)=\frac{8}{7} \sum_{k=0}^{\infty} \frac{1}{(2 k+1)^{3}} \\ &\zeta(3)=\frac{4}{3} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(k+1)^{3}} \end{aligned} \\\begin{aligned} & \zeta(3)=\frac{1}{4} \sum_{k=1}^{\infty}(-1)^{k-1} \frac{(k-1) !^{3}\left(56 k^{2}-32 k+5\right)}{(2 k-1)^{2}(3 k) !} \\ & \zeta(3)=\frac{1}{64} \sum_{k=0}^{\infty}(-1)^{k} \frac{k !^{10}\left(205 k^{2}+250 k+77\right)}{(2 k+1) !^{5}} \\ & \zeta(3)=\frac{1}{24} \sum_{k=0}^{\infty}(-1)^{k} \frac{(2 k+1) !^{3}(2 k) !^{3} k !^{3}\left(126392 k^{5}+412708 k^{4}+531578 k^{3}+336367 k^{2}+104000 k+12463\right)}{(3 k+2) !(4 k+3) !^{3}} \\ & \zeta(3)=\frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^{k}(2 k) !^{3}(k+1) !^{6}\left(40885 k^{5}+124346 k^{4}+150160 k^{3}+89888 k^{2}+26629 k+3116\right)}{(k+1)^{2}(3 k+3) !^{4}} \\ \end{aligned} \\

Ramanujan 给出了如下恒等式

\zeta(3)=\frac{7}{180} \pi^{3}-2 \sum_{k=1}^{\infty} \frac{1}{k^{3}\left(e^{2 \pi k}-1\right)} \\

Simon Plouffe 在 1998 发现了如下恒等式

\zeta(3)=14 \sum_{k=1}^{\infty} \frac{1}{k^{3} \sinh (\pi k)}-\frac{11}{2} \sum_{k=1}^{\infty} \frac{1}{k^{3}\left(e^{2 \pi k}-1\right)}-\frac{7}{2} \sum_{k=1}^{\infty} \frac{1}{k^{3}\left(e^{2 \pi k}+1\right)} \\

积分表示(简单)

\begin{aligned} & \zeta(3)=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{1-x y z} \mathrm{~d } x \mathrm{~d } y \mathrm{~d } z \\ & \zeta(3)=\frac{1}{2} \int_{0}^{\infty} \frac{x^{2}}{e^{x}-1} \mathrm{~d } x \\ & \zeta(3)=\frac{2}{3} \int_{0}^{\infty} \frac{x^{2}}{e^{x}+1} \mathrm{~d } x \\ & \zeta(3)=\frac{4}{7} \int_{0}^{\frac{\pi}{2}} x \log (\sec x+\tan x) \mathrm{~d } x \\ \end{aligned} \\

积分表示(复杂)

\begin{aligned} \zeta(3)& =\pi \int_{0}^{\infty} \frac{\cos (2 \arctan x)}{\left(x^{2}+1\right)\left(\cosh \frac{1}{2} \pi x\right)^{2}} \mathrm{~d } x \\ \zeta(3)& =-\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{\log (x y)}{1-x y}\mathrm{~d } x \mathrm{~d } y\\ &=-\int_{0}^{1} \int_{0}^{1} \frac{\log (1-x y)}{x y} \mathrm{~d } x \mathrm{~d } y \\ \zeta(3)& =\int_{0}^{1} \frac{\log (x) \log (1-x)}{x} \mathrm{~d } x \\ \zeta(3) &=\int_{0}^{1} \frac{\log (x) \log (1-x)}{1-x} \mathrm{~d } x \\ \zeta(3) &=\frac{1}{2} \int_{0}^{1} \frac{\log (x) \log (1-x)}{x(1-x)} \mathrm{~d } x\\ \zeta(3) &=\frac{8 \pi^{2}}{7} \int_{0}^{1} \frac{x\left(x^{4}-4 x^{2}+1\right) \log \log \frac{1}{x}}{\left(1+x^{2}\right)^{4}} \mathrm{~d } x \\ &=\frac{8 \pi^{2}}{7} \int_{1}^{\infty} \frac{x\left(x^{4}-4 x^{2}+1\right) \log \log x}{\left(1+x^{2}\right)^{4}} \mathrm{~d }x \end{aligned} \\

试试第一个，首先，有一个已知结论，即

2\arcsin^2t=\sum_{n=1}^{\infty}\frac{4^nt^{2n}}{n^2{2n\choose n}}.

见

利用上述结论，两边同时除以 t 并积分可得

\begin{align} 2\int_0^x\frac{\arcsin^2t}{t}\mathrm{~d}t&=\int_0^x\sum_{n=1}^{\infty}\frac{4^nt^{2n-1}}{n^2{2n\choose n}}\mathrm{~d}t\\ &=\sum_{n=1}^{\infty}\frac{4^nx^{2n}}{2n^3{2n\choose n}}. \end{align}

重复一次上述操作，有

\begin{align} &\quad2\int_0^y\frac{1}{x}\int_0^x\frac{\arcsin^2t}{t}\mathrm{~d}t\mathrm{d}x\\&=\int_0^y\sum_{n=1}^{\infty}\frac{4^nx^{2n-1}}{2n^3{2n\choose n}}\mathrm{~d}x\\ &=\sum_{n=1}^{\infty}\frac{\left(4y^2\right)^{n}}{4n^4{2n\choose n}}. \end{align}

于是取 y=\frac{1}{2} ，可得

\begin{align} \sum_{n=1}^{\infty}\frac{1}{n^4{2n\choose n}}&=8\int_0^{\frac{1}{2}}\frac{1}{x}\int_0^x\frac{\arcsin^2t}{t}\mathrm{~d}t\mathrm{d}x\\ &=2\int_0^1\frac{1}{t}\arcsin^2\left(\frac{\sqrt{t}}{2}\right)\mathrm{d}t\int_t^1\frac{1}{x}\mathrm{~d}x\\ &=-2\int_0^1\frac{\ln t}{t}\arcsin^2\left(\frac{\sqrt{t}}{2}\right)\mathrm{d}t\\ &=-4\int_0^{\frac{\pi}{6}}\frac{t^2\ln\left(4\sin^2 t\right)}{\tan t}\mathrm{~d}t\\ &=-2\int_0^{\frac{\pi}{3}}t\ln^2\left[2\sin\left(\frac{t}{2}\right)\right]\mathrm{d}t\\ &=-2\mathrm{Ls_4^{(1)}}\left(\frac{\pi}{3}\right)\\ &=\frac{17\pi^4}{3240}. \end{align}

具体细节参阅冰河大佬 @神琦冰河 的文章。