无穷多项的数列问题常常可以利用定积分的定义转化为定积分

一、题目

已知：

$$\begin{array}{rl}I& =\\ & \underset{n\to \mathrm{\infty }}{lim}(\frac{{\sin }^2\frac{\pi }{n}}{n}+\frac{{\sin }^2\frac{2\pi }{n}}{n}+\cdots +\frac{{\sin }^2\frac{n\pi }{n}}{n})\end{array}$$

则：

$$I=?$$

难度评级：

二、解析

$$\begin{array}{r}I\\ \\ & =\underset{n\to \mathrm{\infty }}{lim}(\frac{{\sin }^2\frac{\pi }{n}}{n}+\frac{{\sin }^2\frac{2\pi }{n}}{n}+\cdots +\frac{{\sin }^2\frac{n\pi }{n}}{n})\\ \\ & =\frac{1}{n}\underset{n\to \mathrm{\infty }}{lim}({\sin }^2\frac{\pi }{n}+{\sin }^2\frac{2\pi }{n}+\cdots +{\sin }^2\frac{n\pi }{n})\\ \\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{i=1}^n{\sin }^2(\pi \cdot \frac{i}{n})\\ \\ & ={\int }_0^1{\sin }^2(\pi x)\mathrm{d}x\\ \\ & =\frac{1}{\pi }{\int }_0^1{\sin }^2\pi x\mathrm{d}(\pi x)\\ \\ & \stackrel{\pi x=t}{=}\frac{1}{\pi }{\int }_0^{\pi }{\sin }^{2}t\mathrm{d}t\\ \\ & =\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}{\sin }^{2}t\mathrm{d}t\\ \\ & =\frac{2}{\pi }\cdot \frac{1}{2}\cdot \frac{\pi }{2}\\ \\ & =\frac{1}{2}\end{array}$$