$\displaystyle\lim_{n\to +\infty}\sum_{i=0}^n\dfrac{\cos i\pi}{i^2}=$_______．
答案 $\dfrac{\pi^2}{12}$．
解析 根据题意, 有\[\begin{split}\lim_{n\to +\infty}\sum_{i=1}^n\dfrac{\cos i\pi}{i^2}&=\dfrac1{1^2}-\dfrac1{2^2}+\dfrac1{3^2}-\dfrac1{4^2}+\cdots\\ &=\sum_{k=1}^{+\infty}\dfrac{1}{k^2}-2\sum_{k=1}^{+\infty}\dfrac1{(2k)^2}\\ &=\dfrac 12\sum_{k=1}^{+\infty}\dfrac{1}{k^2}\\ &=\dfrac{\pi^2}{12},\end{split}\]
备注 考虑\[\dfrac{\sin x}x=1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-\dfrac{x^6}{7!}+\cdots +(-1)^n\dfrac{x^{2n}}{(2n+1)!}+\cdots ,\]由于 $y=\dfrac{\sin x}x$ 的零点为\[x=\pm {\mathrm \pi},\pm 2{\mathrm \pi},\cdots ,\pm n{\mathrm \pi},\cdots ,\]因此\[\begin{split} \dfrac{\sin x}x&=1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-\dfrac{x^6}{7!}+\cdots +(-1)^n\dfrac{x^{2n}}{(2n+1)!}+\cdots \\ &=\left(1-\dfrac{x^2}{{\mathrm \pi}^2}\right)\left(1-\dfrac{x^2}{4{\mathrm \pi}^2}\right)\cdots \left(1-\dfrac{x^2}{n^2{\mathrm \pi}^2}\right)\cdots,\end{split} \]对比上式中 $x^2$ 项的系数可得\[1+\dfrac{1}{4}+\dfrac{1}{9}+\cdots +\dfrac{1}{n^2}+\cdots = \dfrac{{\mathrm \pi}^2}6.\]