将函数$f\left( x \right) = \left| x \right|$（$-\pi \le x \le \pi$）展开成Fourier级数
$$f(x)=\frac{\pi}{2}-\frac{4}{\pi }\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2} \cos (2n-1)x$$
取$x=0$得：
$$\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{\pi ^2}{8}$$
$$\sum\limits_{n=1}^\infty \frac{1}{n^2}=\sum\limits_{n=1}^\infty \frac{1}{(2n)^2}+\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2}=\frac{1}{4}\sum\limits_{n=1}^\infty \frac{1}{n^2}+\frac{\pi^2}{8}$$
$$\sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$