一个欧拉级数的证明
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
证明：
将 f(x) = |x|，在 x\in [-\pi, \pi] 上展开成傅里叶级数。
\begin{aligned}
f(x) =& \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos nx \\
a_0 =& \frac{2}{\pi}\int_0^\pi x dx \\
=& \frac{1}{\pi}x^2 \bigg|_0^\pi\\
=& \pi\\
a_n =& \frac{2}{\pi}\int_0^\pi x\cos nx dx \\
=& \frac{2}{\pi n^2}\int_0^\pi nx\cos nx dnx \\
=& \frac{2}{\pi n^2}(nx\sin nx + \cos nx)\bigg|_0^\pi \\
=& \frac{2}{\pi n^2}[(-1)^n - 1] \\
f(0) =& \frac{\pi}{2} + \frac{2}{\pi}\sum_{n=1}^\infty \frac{1}{n^2}[(-1)^n - 1] \\
=& \frac{\pi}{2} + \frac{-4}{\pi} \sum_{n=1}^\infty \frac{1}{(2n - 1)^2} = 0 \\
\frac{\pi^2}{8} =& \sum_{n=1}^\infty \frac{1}{(2n - 1)^2}\\
& \sum_{n=1}^\infty \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2} \\
=& \sum_{n=1}^\infty \frac{1}{n^2} - \frac{\pi^2}{8} \\
& \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
\end{aligned}
证毕。