巴塞尔问题和 zeta(2k)

zeta(2)（巴塞尔问题）

熟知 \(\sin\left(x\right)\) 的无穷级数展开和无穷乘积展开：

\[ \begin{aligned} \sin\left(x\right) &= \sum_{k=0}^{\infty}\dfrac{\left(-1\right)^{k}}{\left(2k+1\right)!}\cdot x^{2k+1} \\ &= x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}-\cdots \\ \sin\left(x\right) &= x\cdot\prod_{k=1}^{\infty}\left(1-\dfrac{x^2}{k^2\pi^2}\right) \\ &= x\cdot\left(1-\dfrac{x^{2}}{\pi^2}\right)\cdot\left(1-\dfrac{x^{2}}{\left(2\pi\right)^{2}}\right)\cdot\left(1-\dfrac{x^{2}}{\left(3\pi\right)^{2}}\right)\cdots \end{aligned} \]

对比系数，有

\[ \left[x^{3}\right]\sin\left(x\right)=-\sum_{k=1}^{\infty}\dfrac{1}{\left(k\pi\right)^{2}}=-\dfrac{1}{3!} \]

因此

\[ \zeta\left(2\right)=\sum_{k=1}^{\infty}\dfrac{1}{k^{2}}=\dfrac{\pi^2}{6} \]

zeta(4)

考虑 \(\left[x^{5}\right]\sin\left(x\right)\)，有

\[ \begin{aligned} \sum_{x\lt y}\dfrac{1}{\left(x\pi\right)^2\left(y\pi\right)^2} &= \dfrac{1}{5!} \\ \sum_{x\lt y}\dfrac{1}{x^2y^2} &= \dfrac{\pi^4}{120} \end{aligned} \]

又已知 \(\zeta\left(2\right)\)，因此知道

\[ \zeta\left(2\right)^{2}=\sum_{x,y}\dfrac{1}{x^2y^2}=\dfrac{\pi^4}{36} \]

因此

\[ \begin{aligned} \zeta\left(4\right) &= \sum_{k=1}^{\infty}\dfrac{1}{k^4} \\ &= \left(\sum_{x,y}\dfrac{1}{x^2y^2}\right)-2\cdot\left(\sum_{x\lt y}\dfrac{1}{x^2y^2}\right) \\ &= \dfrac{\pi^4}{36}-2\cdot\dfrac{\pi^4}{120} \\ &= \dfrac{\pi^4}{90} \end{aligned} \]

zeta(2k)

直接将牛顿恒等式作用到无穷求和上，有

\[ \boxed{ \dfrac{\zeta\left(2k\right)}{\pi^{2k}}=\dfrac{\left(-1\right)^{k+1}\cdot k}{\left(2k+1\right)!}+\sum_{i=1}^{k-1}\dfrac{\left(-1\right)^{i+1}}{\left(2i+1\right)!}\cdot\dfrac{\zeta\left(2\left(k-i\right)\right)}{\pi^{2\left(k-i\right)}} } \]

这个递推可以使用伯努利数写出封闭形式

\[ \boxed{ \zeta(2k)=-\dfrac{B_{2k}\cdot(2\pi i)^{2k}}{2\cdot(2k)!} } \]

我对伯努利数不太熟悉，这里不证了。