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：

$$\begin{array}{rl}I& =\\ & \underset{n\to \mathrm{\infty }}{lim}(\frac{1+\sqrt{n^2-1^2}}{n^2}+\frac{2+\sqrt{n^2-2^2}}{n^2}+\cdots +\frac{n+\sqrt{n^2-n^2}}{n^2})\end{array}$$

则：

$$I=?$$

难度评级：

二、解析

首先：

$$\begin{array}{r}I\\ & =\underset{n\to \mathrm{\infty }}{lim}(\frac{1+\sqrt{n^2-1^2}}{n^2}+\frac{2+\sqrt{n^2-2^2}}{n^2}+\cdots +\frac{n+\sqrt{n^2-n^2}}{n^2})\\ \\ & =\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^n\frac{i+\sqrt{n^2-i^2}}{n^2}\\ \\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{i=1}^n\frac{i+\sqrt{n^2-i^2}}{n}\\ \\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{i=1}^n[\frac{i}{n}+\sqrt{1-{\left(\frac{i}{n}\right)}^2}]\\ \\ & \stackrel{x=\frac{i}{n}}{=}{\int }_0^1(x+\sqrt{1-x^2})\mathrm{d}x\\ \\ & =\frac{1}{2}+{\int }_0^1\sqrt{1-x^2}\mathrm{d}x\\ \\ & \stackrel{x=\sin t}{=}\frac{1}{2}+{\int }_0^{\frac{\pi }{2}}{\cos }^{2}t\mathrm{d}t\\ \\ & =\frac{1}{2}+\frac{1}{2}\cdot \frac{\pi }{2}\\ \\ & =\frac{1}{2}+\frac{\pi }{4}\end{array}$$